CONTOUR INTEGRATION AND

CAUCHY’S THEOREM

CHRISTOPHER M. COSGROVE

The University of Sydney

These Lecture Notes cover Goursat’s proof of Cauchy’s theorem, together with some intro-ductory material on analytic functions and contour integration and proofs of several theoremsin the complex integral calculus that follow on naturally from Cauchy’s theorem.

These notes are primarily intended as introductory or background material for the third-year unit of study MATH3964 Complex Analysis, and will overlap the early lectures wherethe Cauchy-Goursat theorem is proved. The treatment is in finer detail than can be done inlectures. These notes would be especially useful for students who are attempting MATH3964without the recommended prerequisite MATH2962.

For further reading, the student is referred to

• Lars V. Ahlfors: Complex Analysis , McGraw-Hill NY (3rd edition 1979).

• Einar Hille: Analytic Function Theory, Volume I , Chelsea NY (reprint 1982 of 1st edi-tion 1959).

• E. C. Titchmarsh: The Theory of Functions , Oxford University Press (reprint 1979 of2nd edition 1939).

For real-variable support, the student is referred to

• Tom M. Apostol: Mathematical Analysis, Addison-Wesley, Reading MA (1957).

c© C. M. Cosgrove 2006, 2009, 2012. These Lecture Notes are intended for use by stu-dents enrolled in MATH2917 Working Seminar B (Special Studies Program) andMATH3964 Complex Analysis, Pure/Applied Mathematics 2 and 3, the University ofSydney, and no part may be reproduced without permission.

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Contents:

1 Analytic functions and power series 3

2 Contour integration 15

3 Cauchy’s theorem and extensions 31

4 Cauchy’s integral formula 46

5 The Cauchy-Taylor theorem and analytic continuation 63

6 Laurent’s theorem and the residue theorem 76

7 Maximum principles and harmonic functions 85

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1 Analytic functions and power series

The subject of complex analysis and analytic function theory was founded by Augustin Cauchy(1789–1857) and Bernhard Riemann (1826–1866). Karl Weierstrass (1815–1897) placed both realand complex analysis on a rigorous foundation, and proved many of their classic theorems.

Suppose that f(z) is a function of a single complex variable z whose domain D is a nonemptypath-connected subset of the complex plane C. Such a function can be endowed with additionalproperties such as being continuous or differentiable at a particular point or on a set. The simplefunctions z (complex conjugate of z) and |z|2 (square of the modulus, same as zz), for example,are everywhere continuous and the latter example has a derivative at the origin.

In complex analysis, the most important objects of study are analytic functions. Anothername for these functions is holomorphic. Briefly, a function is analytic if it is differentiable on anopen set. The natural domain (or partial domain) of an analytic function is a particular type ofopen set called a region:

Definition 1.1. A region (or open region) in C is a subset of C that is open, connected andnonempty.

Definition 1.2. A function f(z) is differentiable, or possesses a derivative, at a particularpoint z0 in the interior of its domain if the limit of the difference quotient,

f ′(z0) = limζ→z0

f(ζ) − f(z0)

ζ − z0,

exists independently of the path of approach to z0.

Definition 1.3.

• A function f(z) is analytic (or holomorphic) at a particular point z0 ∈ C if it is differen-tiable at z0 as well as on some open neighbourhood of z0.

• A function f(z) is analytic (or holomorphic) on a region D ⊆ C if it is differentiableeverywhere on D. Thus the derivative f ′(z) is also a function defined on D.

• A function f(z) is entire if it is analytic everywhere in C.

• A function f(z) may be called locally analytic to indicate that it is analytic on some regionbut is not necessarily entire.

• A point at which a locally analytic function f(z) fails to be analytic is a singularity of f(z).

There is an obvious bijection between the complex z-plane and the real xy-plane:

C → R2, z = x+ iy 7→ (x, y).

A region D ⊂ C has an image in R2 under this bijection, and conversely. Without any serious riskof confusion, we can use the same symbol D interchangeably for both regions.

A function f : D → C can be split into its real and imaginary parts:

f(z) = u(x, y) + iv(x, y), z = x+ iy.

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A necessary condition for f(z) to be differentiable at z0 ∈ D is that the limit of the differencequotient be the same on all radial paths to z0. In particular, the two limits parallel to the axes shouldbe equal, which requires u and v to have first partial derivatives at (x0, y0), where z0 = x0 + iy0.These limits are

f ′(z0) = limx→x0

f(x+ iy0) − f(x0 + iy0)

x− x0= ux + ivx

∣

∣

∣

(x0,y0),

f ′(z0) = limy→y0

f(x0 + iy) − f(x0 + iy0)

i(y − y0)= vy − iuy

∣

∣

∣

(x0,y0).

We see that a necessary condition for f(z) to be differentiable at z0 is that u and v satisfy theCauchy-Riemann equations,

vy = ux, vx = −uy,

at (x0, y0).

Observe that the very simple function f(z) = z fails this test of differentiability at every point.The function |z|2 is differentiable only at the origin. It is therefore not analytic there because it isnot differentiable on an open set covering the origin.

If we make the additional assumption that f(z) be twice differentiable at z0, then the equalityof mixed derivatives implies that u and v satisfy the two-dimensional Laplace equation,

uxx + uyy = 0, vxx + vyy = 0,

at (x0, y0). Functions of two (or more) real variables that satisfy Laplace’s equation on a regionare called harmonic. If two harmonic functions u and v of two real variables are related by theCauchy-Riemann equations, then v is the conjugate harmonic function of u and −u is the conjugateharmonic function of v. Some properties of harmonic functions are explored in Chapter 7.

Given that the Cauchy-Riemann equations hold at (x0, y0), we will see that a sufficient conditionfor f(z) to be differentiable at z0 is that u(x, y) and v(x, y) be differentiable at (x0, y0). This meansthat their graphs are surfaces in R3 having tangent planes at the points of contact where x = x0

and y = y0. It would be natural to assume that these conditions together are also necessary. But,in the case of analyticity at a point or on a region, a function f(z) can be proved analytic undermuch weaker hypotheses on u(x, y) and v(x, y) (see, for example, Theorem 3.11 below).

A function u(x, y) is differentiable at (x0, y0) if it defined on a neighbourhood of (x0, y0), haspartial derivatives ux and uy at (x0, y0), and

u(x0 + h, y0 + k) = u(x0, y0) + ux(x0, y0)h+ uy(x0, y0)k + ǫ(h, k)√

h2 + k2 ,

where ǫ(h, k) → 0 as (h, k) → (0, 0) on all paths. There are alternative, but equivalent, ways towrite the remainder term in this definition. A sufficient (but not necessary) condition for a functionof two real variables to be differentiable at a point is that both partial derivatives exist at thatpoint and one is continuous there (which requires the latter to exist on a neighbourhood of thepoint as well).

Theorem 1.4. A function f(z) = f(x + iy) = u(x, y) + iv(x, y) defined on a region D is differ-entiable at an interior point z0 = x0 + iy0 in D whenever u and v are differentiable at (x0, y0) andsatisfy the Cauchy-Riemann equations at (x0, y0).

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Proof. Let (x0 + h, y0 + k) be a point in a neighbourhood of (x0, y0) that is contained in D.Then differentiability of u and v at (x0, y0) implies respective functions ǫ1(h, k) and ǫ2(h, k) as inthe above definition that tend to zero as (h, k) → (0, 0) on all paths. Let z = z0 + (h+ ik) andconsider the difference quotient:

f(z) − f(z0)

z − z0=

u(x0 + h, y0 + k) + iv(x0 + h, y0 + k) − u(x0, y0) − iv(x0, y0)

h+ ik

=ux(x0, y0)h+ uy(x0, y0)k + ǫ1(h, k)

√h2 + k2

h+ ik

+ ivx(x0, y0)h+ vy(x0, y0)k + ǫ2(h, k)

√h2 + k2

h+ ik

= ux(x0, y0) + ivx(x0, y0) +(

ǫ1(h, k) + iǫ2(h, k))

√h2 + k2

h+ ik,

where we have used the Cauchy-Riemann equations to eliminate y-derivatives. Now the complexnumber

√h2 + k2/(h+ ik) has unit modulus. Hence,

∣

∣

∣

∣

f(z) − f(z0)

z − z0−

(

ux(x0, y0) + ivx(x0, y0))

∣

∣

∣

∣

=√

ǫ1(h, k)2 + ǫ2(h, k)2 .

This proves that the difference quotient tends to the unique limit ux(x0, y0) + ivx(x0, y0) as z → z0on every path. In other words, f(z) is differentiable at z0 and the value of the derivative is

f ′(z0) = ux(x0, y0) + ivx(x0, y0) = vy(x0, y0) − iuy(x0, y0).

Corollary 1.5. A function f(z) = u(x, y) + iv(x, y) is analytic on a region D ⊂ C wheneveru and v are differentiable on the corresponding region D ⊂ R2 and satisfy the Cauchy-Riemannequations on D.

Remarks.

• A locally analytic function is continuous on its domain of analyticity and is bounded on allcompact (that is, closed and bounded) subsets.

• If a function is analytic on some non-open set such as a closed disc or an interval of the realaxis, it is automatically analytic on some larger open set that covers the given set.

• So far, the definition of analyticity does not impose any niceness properties on f ′(z) apartfrom its pointwise existence on an open set. Thus, continuity of f ′(z) is not to be assumeduntil we have proven it. This is in contrast to other branches of analysis, where the conceptof analyticity is much stricter. For example, a function of one or more real variables isreal-analytic if it is differentiable to all orders on an open interval or connected open setand is locally the sum of its own convergent Taylor series. A remarkable fact, which willbecome a theorem in Chapter 4, is that complex analytic functions automatically possess allof that additional structure without any additional hypotheses. In fact, we could substantiallyweaken our definition of analyticity of f(z) without affecting the class of functions that arecovered.

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• At this early stage, the reader should be aware that analytic continuation of a locallyanalytic function outside a given domain is well-defined and unique. A precise definition anduniqueness theorem will be provided in Chapter 5. Analytic functions, when they first arisein a context, are often not presented on their maximal domains of analyticity, especially whenthe definition involves an infinite series or improper integral, which have their own domainsof validity. If the larger domain is simply connected except for possible holes already in theoriginal domain, the analytic continuation will be unique. However, if a function extendsto a domain that is multiply connected, it may end up becoming a multi-valued function,which is therefore allowed in complex analysis. The preferred way to view such a domain isas a multi-sheeted covering of the complex plane called a Riemann surface, onto which theanalytic function continues uniquely.

• Singularities of analytic functions can take many shapes. For example, the origin z = 0 is asingularity of 1/z3 and cot z because these functions are unbounded near that point. This typeof singularity is called a pole. A more complicated type of singularity, called an essential

singularity, occurs in the examples e1/z and tan(1/z) at z = 0. Another type of singularityis a branch point, near which analytic continuation of the function forces it to becomemulti-valued, or else requires its domain to be artificially cut to force it to be single-valued.More will be said about the classification of singularities later.

Analytic functions are abundant and easily constructed. Nearly all of the well-known elementaryand special functions of a real variable have well-defined natural extensions into the complex domain.Exceptions are such things as step functions and artificial periodic functions, where analytic piecesare joined discontinuously or non-smoothly. Many functions defined by Fourier series are onlydefined in the domain of real variables, a famous example being the Weierstrass nondifferentiablefunction (1880),

W (x) =∞∑

n=0

cos(3nπx)

2n.

The easiest way to construct analytic functions is as sums of convergent power series. Anypower series in the complex domain,

f(z) :=

∞∑

k=0

ak(z − z0)k,

z0 ∈ C, ak ∈ C, k = 0, 1, 2, . . . , having a positive or infinite radius of convergence R, convergesto an analytic function in the interior of its disc of convergence, |z − z0| < R. A selection ofelementary theorems on power series is included at the end of this chapter. The main result is thatevery convergent power series is differentiable term by term in the interior of its disc. Then onederivative implies infinitely many, and all derived power series have the same radius of convergence.The nth derivative of f(z) at z = z0 is

f (n)(z0) = n! an,

which shows that every convergent power series is identical to its own Taylor series about the centreof its disc.

So far, the statement that the sums of convergent power series are analytic is in one directiononly. The converse is the Cauchy-Taylor theorem, which will be stated precisely and proved in

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Chapter 5. Briefly, it asserts that every analytic function is the sum of its own convergent Taylorseries about every point of analyticity, and the radius of convergence is best possible.

Any function of a real variable x which can be expressed as the sum of a convergent Tayloror power series on an open interval has an automatic complex extension to an open disc in C

having that interval as its diameter. One simply changes x to z in the power series. Absoluteconvergence is guaranteed by an elementary comparison theorem for infinite series. A function of areal variable x is real-analytic on an open interval if that interval is a union of open subintervalson each of which the function is the sum of its own convergent Taylor series. Any function that isreal-analytic on an open interval has an automatic extension to a complex analytic function on aregion in C that covers that interval of the real axis.

In this way, functions such as sin z, cos z, ez, z−νJν(z) and 1F1(α; γ; z), for example, are definedas entire functions by their standard power series expansions, each of which has an infinite radius ofconvergence. All the trigonometric identities, such as addition theorems, extend to sin z and cos zbecause they can be proved directly from their power series definitions and also because analyticcontinuation preserves identities that are valid on the real axis. The identity, eiz = cos z + i sin z,is an immediate consequence of the power series expansions of the three functions appearing.

Other functions such as log(1+z), sin−1 z, tan−1(z) and (1+z)α, for example, can be defined ina finite disc (here the open unit disc |z| < 1 plus possibly parts of the boundary) by their standardpower series about z = 0. Such constructions will preserve identities such as sin(sin−1 z) = z andelog(1+z) = 1 + z. Thereafter, in general, they can be analytically extended beyond their discs ofconvergence. But when the radius of convergence of a power series is finite, there must be atleast one singularity of the sum function on the circle of convergence. In these four examples,the singularities are branch points (except when α is an integer), and so the maximal domains ofanalyticity of these functions are Riemann surfaces.

Analytic functions can be constructed in many other ways. For example, the definite integral,

Γ(z) :=

∫ ∞

0tz−1e−t dt,

defines the gamma function as an analytic function of z in the half-plane Re z > 0. Analyticityis guaranteed by the uniform convergence of the integral on all compact subsets of the open righthalf-plane. Afterwards, on account of either of the two functional relations,

Γ(z + 1) = zΓ(z), Γ(z)Γ(1 − z) =π

sin(πz),

the gamma function analytically extends to the left half-plane except for simple poles at zero andthe negative integers.

Another example is the Riemann zeta function defined in the half-plane Re z > 1 by theinfinite series,

ζ(z) = 1 +1

2z+

1

3z+

1

4z+ . . . .

The analyticity of ζ(z) is guaranteed by the uniform convergence of the series in half-planesRe z ≥ 1 + δ, δ > 0. This function also analytically continues to the left half-plane. The exten-sion to the larger half-plane Re z > 0, except for a pole at z = 1, can be achieved by either of the

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formulae,

ζ(z) =1

1 − 21−z

1 − 1

2z+

1

3z− 1

4z+ . . .

,

ζ(z) =1

z − 1+

1

2− z

∫ ∞

1

t− [t] − 12

tz+1dt,

where [t] denotes the greatest integer ≤ t. Thereafter, the analytic continuation to the rest of theplane is given by Riemann’s functional relation,

ζ(1 − z) = 21−z π−z cos 12πz Γ(z) ζ(z),

or, equivalently,ξ(1 − z) = ξ(z), ξ(z) := z(z − 1)π−z/2 Γ(z/2) ζ(z).

There are also contour integral formulae and other expressions that are able to define the gammaand zeta functions globally. The functional relations just given are best proved using contourintegration methods.

Certain standard theorems for sequences and series of functions of a real variable have automaticextensions to functions of a complex variable, simply because the latter can be split into their realand imaginary parts. For example, suppose each member of a sequence of functions

fn(z)

iscontinuous on a (non-empty) connected set E ⊂ C. If the sequence converges uniformly on E to alimit function f(z), then f(z) is also continuous on E. A similar conclusion applies to uniformlyconvergent series. The proof is very similar to the real variable case, and is left as an exercise.

Consider a sequence of analytic (that is, differentiable) functions

fn(z)

on an open region D.Based on the real-variable theorem giving sufficient conditions for the interchange of derivative andlimit (Theorem 4.13 in Chapter 4), we would expect the limit function to exist and be analytic on Dif the derived sequence

f ′n(z)

converges uniformly to a limit function on all compact subregionsof D and if the original sequence converges at (at least) one point in D. That is certainly true,but it turns out that sequences and series of analytic functions obey a stronger theorem, namely,the Weierstrass limit theorem, stated and proved in Chapter 4 below. When complex analyticfunctions are involved, it is not actually necessary to examine the uniform convergence of thederived sequence or series. The uniform convergence of the original sequence or series is enough.This is a significant difference between the real and complex theories.

Other examples of real-variable results that lift up in a natural way to the complex domainare conditions for the validity of interchanging limits with integrals (Arzela’s bounded convergencetheorem), the properties of functions defined by integrals with a parameter, differentiation underthe integral sign, and the handling of improper integrals and double integrals. In general, we will notsupply proofs of such results unless there are new features to be observed in the complex domain.We assume that the student knows these real-variable results, or knows where to look them up. Arecommended text is Tom M. Apostol: Mathematical Analysis, Addison-Wesley (1957).

These notes will often mention Riemann surfaces without attempting a precise definition. Con-sequently, any theorems in these notes that mention Riemann surfaces are intended to be viewedas descriptive rather than rigorous. Except in a few simple cases, we will not be looking for the fullRiemann surface that forms the global domain of analyticity of any particular function. Rather, wewill be interested in local analytic extensions of analytic functions where the extension runs onto

8

a domain that covers all or part of the complex plane more than once. Some examples of globalRiemann surfaces appear at the end of Chapter 4.

Let us give two examples of global Riemann surfaces now because of their importance. Thelogarithm function takes the form,

log z = log(reiθ) = log r + iθ = log |z| + i arg z, z 6= 0,

where we have put z in polar form,

z = x+ iy = r(cos θ + i sin θ) = reiθ.

The real part of log z is just a real-variable logarithm and is defined and single-valued in the complexplane minus the origin. On the other hand, the multi-valuedness of arg z implies that the imaginarypart of log z takes infinitely many values for each nonzero z ∈ C. So log z itself takes infinitely manyvalues for each nonzero z, each differing by multiples of 2πi. The preferred viewpoint is that log z issingle-valued and analytic on the infinite-sheeted covering of the complex plane (minus the origin)parametrised by r > 0 and −∞ < θ <∞. This is the full Riemann surface for log z. It ishomeomorphic to R2, where homeomorphic means related by a continuous bijection. The origin isa branch point for log z because log z is necessarily multi-valued in a deleted neighbourhood ofthe origin.

The principal value of log z is assigned by restricting the argument θ to its principal range(−π, π]. The domain of the principal value of log z is just the ordinary complex plane C minus theorigin. Its range is a half-open horizontal strip. The function so restricted has a jump discontinuityacross the negative real axis, which (together with the origin) constitutes a choice of branch cutfor the logarithm.

Complex powers are defined by zα = eα log z. The principal value of zα is the value where log ztakes its principal value. When α is irrational (which includes non-real), zα has infinitely manyvalues and lives on the same Riemann surface as log z (except that we can include the origin as apoint of continuity, but not analyticity, when Reα > 0). When α is rational, write α = p/q, wherep and q are integers having no common factor and q > 0. In that case zp/q has exactly q valueswhen z 6= 0 and lives on a Riemann surface that covers the complex plane q times, because zp/q doesnot change when log z is replaced by log z + 2qπi. In particular, square roots of nonzero complexnumbers have two values and z1/2 is single-valued on a two-sheeted Riemann surface, cube rootshave three values and are single-valued on a three-sheeted Riemann surface, and so on.

Elementary power series theorems. Most of the power series theorems have elementary proofs,which will be omitted. (Some of the Tauberian theorems are difficult.) We are primarily interestedin the proof that a power series is differentiable term by term to all orders, in view of its importanceto our claim that the sum of a convergent power series is an analytic function in the interior of itsdisc of convergence. Unless otherwise stated, the power series under consideration is

∞∑

n=0

an(z − z0)n, a0, a1, a2, . . . ∈ C, z0 ∈ C.

The point z0 can be called the centre of the power series, being the centre of its disc of convergence.

Notation. Let D(z0, r) denote the open disc |z − z0| < r and [D](z0, r) its closure |z − z0| ≤ r.Let C(z0, r) denote the bounding circle |z − z0| = r.

9

The bar notation D(z0, r) is frequently used for the closure [D](z0, r). We are not using the barbecause it sometimes prints faintly or does not display clearly on computer screens, especially ontop of capital letters. (The bar will be used occasionally for complex conjugates and changes ofvariable.)

Lemma 1.6. If a power series converges absolutely at any point z1 6= z0, then it converges absolutelyand uniformly in the closed disc [D](z0, r), where r = |z1 − z0|.

Lemma 1.7. If a power series converges conditionally or diverges at any point z1 6= z0, thenit diverges everywhere outside the aforementioned closed disc. It may converge conditionally ordiverge elsewhere on the boundary.

Lemma 1.8. Every power series has a radius of convergence R, 0 ≤ R ≤ ∞, with the followingproperties:

• If R is finite and nonzero, the power series converges absolutely in the open disc of conver-

gence D(z0, R), and diverges outside its closure. It may or may not converge on the circle

of convergence |z − z0| = R and convergence may be conditional there. Convergence is ab-solute and uniform on the closed disc [D](z0, R) whenever convergence is absolute at any onepoint of the circle of convergence. Otherwise convergence is uniform at least on all compactsets interior to the disc of convergence.

• If R = ∞, the series converges absolutely for all z ∈ C and uniformly on all compact sets.

• If R = 0, the series converges trivially at z = z0 but diverges everywhere else.

Theorem 1.9. Abel’s theorem. If a power series converges in any manner at a point z1 on itscircle of convergence, then convergence is uniform and the sum of the series is continuous on everyinward facing closed sector with vertex at z1. (It is understood that the sector stops at a positivedistance before reaching the other side of the disc of convergence.)

Remark. Abel’s theorem is most valuable when the convergence at z1 is conditional, for otherwisethe stronger statement Lemma 1.6 would supersede Abel’s theorem.

Proof. Let the power series be∑∞

n=0 an(z − z0)n and let it converge in any manner at the point

z = z1 on the circle of convergence |z − z0| = R, where R = |z1 − z0| > 0. Let Sn denote the nthpartial sum,

Sn =

n∑

j=0

aj(z1 − z0)j .

So Sn tends to a finite limit S as n→ ∞. Let ǫ > 0 be given. Then there exists N ∈ Z+ such that|Sn − S| < ǫ whenever n ≥ N . Define

ζ =z − z0z1 − z0

.

We are allowing z to be interior to the circle of convergence or else take the value z1 on the circleitself. Thus |ζ| ≤ 1 with equality only at ζ = 1. The definition of Sn gives

an(z − z0)n = (Sn − Sn−1)ζ

n.

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Then, for M > N + 1,

M∑

n=N+1

an(z − z0)n =

M∑

n=N+1

(

(Sn − S) − (Sn−1 − S))

ζn

=

M∑

n=N+1

(Sn − S)ζn −M−1∑

n=N

(Sn − S)ζn+1

= (SM − S)ζM+1 − (SN − S)ζN+1 +

M∑

n=N+1

(Sn − S)ζn(1 − ζ).

The triangle inequality gives, for |ζ| < 1,

∣

∣

∣

∣

M∑

n=N+1

an(z − z0)n

∣

∣

∣

∣

≤ ǫ

(

2 + |1 − ζ|M∑

n=N+1

|ζ|n)

≤ ǫ

(

2 +|1 − ζ|1 − |ζ|

)

.

At ζ = 1, the corresponding bound is ǫ by construction.

We are looking for a domain of uniform convergence of the power series inside and on the circle ofconvergence, where the point z = z1 is the only point on the circle itself. Any domain containingpoints near z = z1 where |1 − ζ|/(1 − |ζ|) is bounded by a positive constant, plus the point z = z1itself, is a domain of the required type. Let ζ = 1 − ρeiθ, ρ > 0. Then

|1 − ζ|1 − |ζ| =

1 + |ζ|2 cos θ − ρ

≤ 2

2 cos θ − ρ.

This has a constant upper bound in the truncated sector,

− 12π + δ ≤ θ ≤ 1

2π − δ, 0 ≤ ρ ≤ 2 cos θ − sin δ,

where 0 < δ < π/2. We have attached the vertex ρ = 0 since we know separately that the boundis applicable there. This is an inward facing closed sector of angle π − 2δ, vertex at z1 included,that stops before reaching the other side of the disc of convergence. We have just shown that it isa domain of uniform convergence of the power series. The sum of the series is continuous on thisdomain. In particular, the sum of a power series is continuous up to a point of convergence on itscircle of convergence from within any inward facing sector.

Example. The power series, z − z2/2 + z3/3 − z4/4 + . . . , converges conditionally on its circleof convergence |z| = 1 except at z = −1. Hence, it converges to log(1 + z) uniformly on the closeddisc |z| ≤ 1 except for an indent of positive radius at z = −1.

Theorem 1.10. Tauber’s theorem (improved by Littlewood 1911). If nRnan is bounded and thesum of the power series approaches a limit L as z approaches a point z1 on the circle of convergenceof radius R from within an inward facing sector, then the power series converges at z1 to L.

The proof of Littlewood’s version of Tauber’s theorem can be found in Titchmarsh, Theory of

Functions, Chapter 7.

Lemma 1.11. Formula for R. The radius of convergence is given by either of the equivalentexact formulae:

R = lim infn→∞

|an|−1/n,1

R= lim sup

n→∞|an|1/n.

11

Remark. When discussing supremums, infimums, limit supremums and limit infimums, it isunderstood that the sets of numbers involved are members of the extended real number systemR∗ = [−∞,∞] = R ∪ −∞,∞. Then every subset of R∗ has a supremum and infimum, andevery sequence in R∗ has a limit supremum and limit infimum. In particular, both formulae for theradius of convergence of a power series always give a unique result in the interval 0 ≤ R ≤ +∞.

Proof. Let R be defined by the first formula and suppose, first, that R is finite and nonzero. Letǫ > 0 be given. From the definition of a limit infimum, there exists N depending on ǫ such thatfor all n > N , |an|−1/n > R− ǫ. Then for such n, |an| < (R− ǫ)−n. Then, by comparison with ageometric series, the power series converges absolutely in the disc D(z0, R− ǫ). Since ǫ is arbitrary,the power series converges absolutely in D(z0, R). In the case R = ∞, the role of R− ǫ is playedby an arbitrarily large real number.

Again, by the definition of a limit infimum, there are infinitely many values of n such that|an|−1/n ≤ R+ ǫ. For such n, |an| ≥ (R+ ǫ)−n. Then for all z outside D(z0, R + ǫ), a subse-quence of terms of the power series does not tend to zero. Hence, the series diverges. Since ǫ isarbitrary, the series diverges outside [D](z0, R). The case R = 0 is included with the understandingthat [D](z0, 0) denotes the single point z0.

Theorem 1.12. Every power series with a positive or infinite radius of convergence is differentiableterm by term to all orders in the interior of its disc of convergence.

Proof. Without loss of generality, put z0 = 0. Let

f(z) =∞

∑

n=0

anzn,

for z ∈ D(0, R) and possibly also for some or all points on the boundary. Let g(z) denote theformal term-by-term derivative:

g(z) :=

∞∑

n=1

nanzn−1.

The radius of convergence of the derived series is

lim infn→∞

|nan|−1/n = limn→∞

n−1/n lim infn→∞

|an|−1/n = R limn→∞

e−(log n)/n = R,

which is the same as for the original series. Our job is to prove that g(z) is the actual derivativeof f(z), not just the formal derivative, in the interior of the common disc of convergence. (Notethe theorem does not make any claim about convergence of the derived series on the circle ofconvergence. The example of the logarithm series shows that convergence of a power series on itscircle of convergence may be lost after differentiation.)

Choose ρ < R, with ρ arbitrarily close to R. (If R is infinite, then ρ is arbitrarily large.) Thenanρ

n is bounded, say, |anρn| ≤ K. Choose z ∈ D(0, ρ) and thence h such that |z| + |h| < ρ. Then

f(z + h) − f(z)

h− g(z) =

∞∑

n=0

an

(z + h)n − zn

h− nzn−1

.

12

The modulus of the expression multiplying an is

∣

∣

∣

∣

(z + h)n − zn

h− nzn−1

∣

∣

∣

∣

=

∣

∣

∣

∣

(

n

2

)

zn−2h+

(

n

3

)

zn−3h2 + . . . + hn−1

∣

∣

∣

∣

≤(

n

2

)

|z|n−2|h| +(

n

3

)

|z|n−3|h|2 + . . . + |h|n−1

=(|z| + |h|)n − |z|n

|h| − n|z|n−1,

where we used the triangle inequality for complex numbers. Then,

∣

∣

∣

∣

f(z + h) − f(z)

h− g(z)

∣

∣

∣

∣

≤∞∑

n=0

|an|

(|z| + |h|)n − |z|n|h| − n|z|n−1

≤∞∑

n=0

K

ρn

(|z| + |h|)n − |z|n|h| − n|z|n−1

.

On the right-hand side, we have three convergent series with elementary sums:

∞∑

n=0

(|z| + |h|)nρn

=ρ

ρ− |z| − |h| ,

∞∑

n=0

|z|nρn

=ρ

ρ− |z| ,

∞∑

n=0

n|z|n−1

ρn=

ρ

(ρ− |z|)2 .

The first two are geometric series. In the third case, multiply both sides by |z|/ρ and subtract fromthe original. The result is a geometric series proportional to the second case. Now,

∣

∣

∣

∣

f(z + h) − f(z)

h− g(z)

∣

∣

∣

∣

≤ K

|h|

ρ

ρ− |z| − |h| −ρ

ρ− |z|

− Kρ

(ρ− |z|)2

=Kρ|h|

(ρ− |z|)2(ρ− |z| − |h|) .

The right-hand side tends to zero as h→ 0 independently of the path. Hence f ′(z) exists in D(0, ρ)and

f ′(z) = g(z).

In other words, the actual derivative equals the formal derivative in D(0, ρ).

Since ρ can be arbitrarily close to R from below, it follows that f(z) is analytic in D(0, R) andf ′(z) = g(z). Since the derived power series is another power series with the same radius of con-vergence, the process of differentiation can be repeated indefinitely. Also, we can move the centreback to z0. Hence the kth derivative of the sum of any convergent power series equals the sum ofthe series formed by differentiating the original series term by term k times in the interior of itsdisc of convergence.

13

In complex analysis, it is useful to consider power series that contain negative powers as wellas positive powers. A series of the form,

∞∑

k=−∞

ak(z − z0)k,

is known as a Laurent series, and an important theorem concerning such series is given in Chap-ter 6. These have an inner and an outer radius of convergence, say, R1 and R2, respectively. Theouter radius R2 is the radius of convergence of the nonnegative powers, which form an ordinarypower series. The inner radius R1 is the reciprocal of the radius of convergence of the power series∑∞

k=1 a−kζk in the complex ζ-plane. If there are only finitely many negative powers, the inner

radius is R1 = 0. Then the point z0 is a pole of the sum function if there is at least one negativepower. (Of course, it is an ordinary point of analyticity if there are no negative powers.)

For the whole Laurent series to converge anywhere, we need R1 ≤ R2 with R1 finite. The case0 ≤ R1 < R2 ≤ ∞ is of greatest interest in the theory of analytic functions. The above theoremson power series carry over to this case. In particular, Laurent series are differentiable term by termin the annulus R1 < |z − z0| < R2 and their sum functions are therefore analytic in that annulus.Suppose R1 ≤ R ≤ R2 with R finite and nonzero. By putting z = z0 +Reiθ, the Laurent seriesbecomes a Fourier series in the real variable θ. The case R1 = R2, R1 finite and nonzero, whichgives a circle along which the series may or may not converge, is the case of greatest interest in thetheory of Fourier series since the sum functions of Fourier series are not expected to analyticallycontinue off the real axis, in general.

While it is possible to build a respectable theory of analytic functions based on power series,the real power of analytic functions will become apparent when we turn to the complex integralcalculus, with particular reference to Cauchy’s theorem and its numerous consequences.

14

2 Contour integration

A contour in the complex plane is just a curve, finite or infinite, that is not too irregular, andwhich has an arrow or orientation. Contours will play the role of paths of integration. We wishto assign a meaning to the contour integral,

∫

Cf(z) dz,

where C is a contour and f(z) is a function which is defined and piecewise continuous along C.In the following definitions, we restrict attention to bounded contours. Integrals can be definedon unbounded contours by a process similar to the construction of improper Riemann integrals oninfinite or semi-infinite intervals. Certain types of unbounded discontinuities in f(z) can be also beadmitted by similar constructions.

A contour can be parametrised by a real parameter, say, t. Its equation will take the form,

z = z(t) = x(t) + iy(t), a ≤ t ≤ b,

where z(t) is a continuous complex-valued function of the real variable t on [a, b], and its orientationis the direction of increasing t. (We can always set a = 0 and b = 1, but this would cause unnecessaryinconvenience when we join arcs end to end.) A change of parametrisation can be effected with acontinuous strictly increasing function t = T (t) (where the bar here just denotes a new variable,not complex conjugation).

A contour C (or curve if an orientation is not required) can be endowed with one or more ofthe following characteristics. The background space can be either C or R2, where z denotes x+ iyor (x, y), respectively.

• C is a closed curve if its endpoints coincide, that is, z(b) = z(a). The common endpointcan be anywhere on the curve, depending on the parametrisation.

• C is a simple arc or Jordan arc if it never self-intersects, that is, z(t1) = z(t2) impliest1 = t2. In other words, the map z : [a, b] → C or R2 is injective. Equivalently, z(t) is aone-to-one function from [a, b] onto its image in C or R2.

• C is a simple closed curve or Jordan curve if it is a closed curve that never self-intersectsexcept for the obvious requirement that the endpoints coincide. In this case, z(t) is one-to-oneon [a, b) and z(b) = z(a). The Jordan curve theorem (not proved in these notes) statesthat a simple closed curve C in the plane separates the plane into two disjoint open sets calledthe inside or interior of C and the outside or exterior of C, the inside being bounded andsimply connected, the outside being unbounded and connected, and both sides having C astheir complete boundary.

• A Jordan curve contour is positively oriented if its inside is on the left when followed inthe direction of its arrow.

• A region is a Jordan region if it is the inside (or interior) of a Jordan curve. Jordan regionsare bounded and simply connected. A closed Jordan region is the union of a Jordan regionand its boundary.

15

• C is rectifiable or, equivalently, has a finite arc length, if z(t) is a function of bounded

variation on [a, b], by which we mean that x(t) and y(t) are real-valued functions of boundedvariation on [a, b]. (A definition of arc length appears below.)

• C is smooth if it has a continuously turning tangent, in which case a parametrisation z(t) canbe given such that z′(t) is continuous and nonzero on [a, b], with one-sided derivatives beingunderstood at the endpoints. A smooth closed curve has a smooth join at its commonendpoint.

• C is piecewise smooth if it consists of a finite number of smooth arcs joined end to end.The joins may be (but are not limited to) smooth joins, corner points, cusps, or reversals.

• C is a polygonal line if it consists of a finite number of straight line segments joined end toend, in which case z(t) is piecewise linear. It can self-intersect or be simple. It can be closedor not closed. A closed polygonal line is also called a polygon.

• C is a Jordan curve on a Riemann surface if C is contained in a bounded region E on theRiemann surface such that there is a continuous bijection from E to the open unit disc, andthe image of C under the bijection is a Jordan curve inside the unit disc. (This definitionpermits branch points inside or on C, but many common applications would forbid them.)

In these notes, we aim to prove the main theorems that are of a complex analytic character,but will inevitably introduce some unproved statements. Some will be assumed knowledge fromthe theory of functions of a real variable and the Riemann integral. Others will be of a geometricor topological character, such as statements about point sets, polygonal lines, connectivity, and soon. Some of these statements may appear to be self-evident, but that does not necessarily meanthat their proofs are always easy. The Jordan curve theorem is notoriously difficult to prove.(Jordan recognised that the apparently obvious statement required proof, but his own proof wasincomplete, the first undisputed proof being due to Veblen (1905).)

We assume that the student is familiar with elementary point set concepts in n-dimensionalEuclidean space. For example, a set can be open (all points of the set are interior points), closed

(all limit points are included in the set), neither open nor closed, bounded (the set fits inside afinite n-cube or n-ball), convex (all pairs of points in the set are joined by line segments belongingto the set), compact (closed and bounded, Heine-Borel property), connected, path-connected,simply connected (no holes in the case of sets in the plane), and so on. Given a point set inRn or Cn, we can form its complement, interior, closure, or boundary. Some of these concepts aredefined or described in the next few paragraphs.

In the case of open sets, connected and path-connected are equivalent. Otherwise path-connected is stronger than connected. A set in Rn or Cn is connected if it cannot be coveredby two disjoint open sets. In the xy-plane, for example, the graph of y = sin(1/x), x ∈ (0, 1],together with the vertical line segment [−1, 1] on the y-axis, forms a closed connected set that isnot path-connected. In these notes, we intend connected to always mean path-connected and willusually use the latter term in the case of sets that are not necessarily open.

A path-connected set is simply connected if every closed curve in the set can be continuouslydeformed to a point without leaving the set. In the plane (that is, R2 or C), but not in mosthigher-dimensional spaces, a bounded path-connected set is simply connected if and only if it has aconnected complement. This is a way of saying that the bounded set has no holes. It also applies to

16

subsets of a 2-sphere. It does not necessarily apply to unbounded sets in the plane, as straight linesand strips are simply connected while their complements are not connected. By contrast, in R3,the 2-sphere is bounded and simply connected, but its complement is not connected.

The word compact is used in these notes to refer to a set in Rn or Cn that is closed andbounded. In general topology, the word compact refers to a particular type of subset of a topologicalspace (possibly the whole space) having the Heine-Borel property that every covering of the givenset by open sets has a finite subcover. The Heine-Borel Theorem states that subsets of Rn arecompact in this sense if and only if they are closed and bounded. This theorem is of fundamentalimportance in real and complex analysis.

The word convex can be used in several ways:

Definition 2.1.

• A set E ⊂ Rn or Cn is convex if the line segment joining any two points of E is also containedin E. In other words, λx+ (1− λ)y is a member of E for every x,y ∈ E and every λ ∈ [0, 1].

• A Jordan curve in the plane (R2 or C) is convex if the line segment joining any two pointsof the curve is inside or on the curve. (This definition extends to simple closed hypersurfacesin Rn.)

• A function f : E → R, where E is a path-connected set in Rn, is a convex function if

f(λx + (1 − λ)y) ≤ λf(x) + (1 − λ)f(y),

for every line segment λx + (1 − λ)y in E, where λ runs from 0 to 1. Equivalent names areconvex down and concave up. If the set E itself is convex, then the function f : E → R

is convex if every line segment joining two points on the graph of f is on or above the graph.

Convex functions are continuous, but may not be differentiable. (However, a remarkable the-orem of Aleksandrov states that they are twice differentiable almost everywhere.) If a convexfunction of one variable f(x) is twice differentiable at a point, then f ′′(x) ≥ 0 at that point. Cir-cles, ellipses and convex polygons are special cases of convex Jordan curves. Convex sets are simplyconnected. The interior and closure of a convex set is convex. In the plane, bounded convex setshave boundaries that are convex Jordan curves.

The following is an important generalisation of the first two usages of the word convex:

Definition 2.2.

• A set E in Rn or Cn is starlike with respect to an interior point P if the half-open linesegment [PQ) is interior to E for every Q ∈ E distinct from P .

• A Jordan curve in the plane is starlike with respect to a point P on its inside if the unionof the curve and the Jordan region inside is a starlike set with respect to P according to thepreceding definition.

Starlike sets are simply connected. The interior of any starlike set is also starlike about thesame point. However, the closure of a starlike set is not necessarily starlike about the same point

17

because the boundary may contain a radial line segment. Convex sets are starlike about everyinterior point.

In the plane (R2 or C), compact starlike sets have starlike Jordan curve boundaries. Theirinteriors are Jordan regions. A Jordan curve is starlike with respect to the origin if and only ifit has a polar coordinate equation r = f(θ), where f(θ) is positive and continuous on [0, 2π] andf(2π) = f(0).

A neighbourhood of a point z0 ∈ C is generally understood to be an open disc with z0 at itscentre, but a more general definition is sometimes intended. Introduce the notation,

N(z0, δ) :=

z : |z − z0| < δ

, N ′(z0, δ) :=

z : 0 < |z − z0| < δ

.

In the first case, N(z0, δ) is a neighbourhood of z0 consisting an open disc with centre z0 andradius δ. This notation can be used interchangeably with the earlier notation D(z0, δ) for an opendisc, and context will determine the preferred usage. In the latter case, the centre z0 is punchedout, and we say that the region is a deleted neighbourhood or a punctured neighbourhood of z0.

The more general definition of a neighbourhood of a point z0 is any path-connected set in C

that contains a disc N(z0, δ) for some δ > 0. So a neighbourhood does not need to be either openor simply connected, although the latter restrictions often occur naturally in practice. Similarly, adeleted neighbourhood of z0 is any neighbourhood of z0 with the point z0 itself punched out.

A contour integral is just a type of line integral in the complex plane. The natural paths ofintegration for line integrals are rectifiable curves, which are more general than we usually need incomplex analysis. Because Cauchy’s theorem, when we get to it, will allow us to deform contours toa certain extent without changing the value of a contour integral, it follows that irregular contourscan nearly always be deformed into smooth contours. Since many useful contours have cornerpoints, the natural paths of integration for complex contour integrals are piecewise smooth. Allsuch integrals can be written as Riemann integrals with respect to the parameter t (or improperintegrals if we allow unbounded integrands or paths).

So nearly all of the contours actually used in applications of contour integration, such as eval-uation of real definite integrals, are piecewise smooth. Typical examples are closed semicircles,closed semicircles with one or more indents, rectangles with or without indents, circular sectorswith the vertex possibly truncated, full circles with or without a detour around a branch cut, andso on. Among the results that can be derived by contour integration are the functional relationsfor the gamma, Riemann zeta and Dedekind eta functions, and the formula for the surface area ofa triaxial ellipsoid.

Despite the preference for piecewise smooth contours, we nevertheless intend to give the basicdefinition of a contour integral for rectifiable curves. Piecewise smooth curves are just a special case.Some proofs that start from the basic definitions are cleaner and more transparent when rectifiablecurves are assumed. An example will be the proof of Cauchy’s theorem. The underlying integralinvolved is the Riemann-Stieltjes integral, but the student will not need any previous contactwith this type of integral to understand the definitions below or their applications. A familiaritywith the ordinary Riemann integral as a limit of a sum will certainly be helpful. These notes aimto be mostly self-contained when referring to Riemann-Stieltjes integrals and their theory.

Let the curve C have the equation z = z(t), a ≤ t ≤ b, as above, where z(t) is continuous and

18

of bounded variation on [a, b], and let P be any partition of the interval [a, b] defined by

P =

t0, t1, t2, . . . , tn

, t0 = a, tn = b,

∆tj := tj − tj−1 > 0, ‖P‖ := max1≤j≤n

∆tj ,

∆zj := z(tj) − z(tj−1).

The curve C can be approximated by the polygonal line joining the points z(tj) in turn along C.The length of the polygonal line is

L(P,C) =

n∑

k=1

|∆zk|.

As the partition is refined by the addition of extra points, the length L(P,C) does not decrease.Hence, it is natural to define the arc length of a curve by the following supremum:

Definition 2.3. The arc length of the curve C is

L(C) = supP

L(P,C),

where the supremum is taken over all possible partitions P of [a, b].

Every curve has a well-defined arc length if +∞ is admitted as a possible value. Zero arc lengthcan only occur if the curve degenerates to a single point. The curve C is rectifiable when L(C) isfinite. A necessary and sufficient condition for C to be rectifiable is that z(t) be a complex-valuedfunction of bounded variation on [a, b]. When C is piecewise smooth, an elementary theorem inreal-variable calculus provides the well-known integral formula,

L(C) =

∫ b

a

∣

∣

∣

∣

dz

dt

∣

∣

∣

∣

dt =

∫ b

a

√

(dx

dt

)2+

(dy

dt

)2dt.

(Even if C is not piecewise smooth, but is rectifiable, this integral exists in the Lebesgue sensebecause a function of bounded variation is differentiable almost everywhere and its derivative isintegrable. However, in that case, the value of the integral may be less than L(C).)

As in the real domain, two types of line integral can be formed. The most important oneis defined as follows. Suppose that C is rectifiable and f(z) is defined and piecewise continuousalong C, which implies that f(z) is also bounded on C (since the subintervals of continuity areclosed and finite in number). With the partition P above, form the Riemann sum,

S(f, P,C) :=n

∑

k=1

f(z(t∗k))∆zk .

where t∗k is any point in the closed subinterval [tk−1, tk].

Definition 2.4. The function f(z) is integrable over the contour C if, given arbitrary ǫ > 0,there exists a partition Pǫ of [a, b] and a complex number A such that

|S(f, P,C) −A| < ǫ

19

for every partition P that is a refinement of Pǫ and for every choice of sample point t∗j in the jthsubinterval of P . Then

A = limn→∞‖P‖→0

S(f, P,C),

where n is the number of subintervals in P and the limit is in the manner just described.

Definition 2.5. The contour integral of f(z) with respect to z over C is the number A justconstructed. That is,

∫

Cf(z) dz = lim

n→∞‖P‖→0

S(f, P,C).

The contour integral is exactly the same as the complex-valued Riemann-Stieltjes integral off(z(t)) with respect to z(t) on [a, b]. The notation for the latter appears on the right-hand side of

∫

Cf(z) dz =

∫ b

af(z(t)) dz(t).

Here f(z(t)) is the integrand and z(t) is the integrator. The definition given is equally applicableto both kinds of integral.

In the real domain, the most useful sufficient condition for the existence of a Riemann-Stieltjesintegral is that the integrand be continuous and the integrator be a function of bounded variation.A brief explanation appears in the proof of the next theorem. When applied to line integralsand contour integrals, this condition implies that line and contour integrals exist whenever theintegrand is continuous and the path of integration is rectifiable. But Riemann-Stieltjes integralshave a reciprocity theorem, called integration by parts, that allows the integrand and integrator tobe swapped. So Riemann-Stieltjes integrals also exist when the integrand is of bounded variationand the integrator is continuous. This case is less useful for line and contour integrals because, ifthe path is not rectifiable, then the integrand is not, in general, going to be a function of boundedvariation because it is being evaluated on a non-rectifiable path.

Theorem 2.6. The contour integral of a continuous function over a rectifiable contour exists.

Sketch of Proof. If z = x+ iy and f(z) = u(x, y) + iv(x, y), the contour integral can be ex-pressed in terms of line integrals in the plane according to

∫

Cf(z) dz =

∫

C(u dx− v dy) + i

∫

C(v dx+ u dy),

where, on the right-hand side, C is understood to be the corresponding oriented rectifiable curvein the real xy-plane. Thus the contour integral has been reduced to a linear combination of fourreal-variable Riemann-Stieltjes integrals of the form,

∫ b

ag(t) dα(t) = lim

n→∞‖P‖→0

S(g, P, α), S(g, P, α) =n

∑

k=1

g(t∗k)∆αk ,

∆αk = α(tk) − α(tk−1), ∆tk = tk − tk−1 , ‖P‖ = max1≤k≤n

∆tk ,

where P is the partition above, g(t) is continuous, and the integrator α(t) is continuous and ofbounded variation on [a, b]. Since a real-valued function of bounded variation can be expressed as a

20

difference of two strictly increasing functions, we can write a contour integral as a linear combinationof eight real-variable Riemann-Stieltjes integrals with continuous integrands and strictly increasingintegrators. So let α(t) be strictly increasing. For each of these separate real-variable integrals,upper and lower Riemann sums can be formed, and the existence of their common limit follows thesame argument as for ordinary Riemann integrals. Because both the integrand and integrator arecontinuous, other reasonable rules for the refinement or otherwise of partitions will not make anydifference to the limit.

Theorem 2.7. If the contour C is piecewise smooth, the contour integral reduces to the Riemannintegral,

∫

Cf(z) dz =

∫ b

af(z(t))

dz

dtdt.

Remarks. This formula, which is easy to remember, will serve as the definition of the contour inte-gral when we are restricting attention to piecewise smooth contours. In all reasonable applicationsof contour integrals, this is the formula we will need. As already mentioned, the earlier constructionusing rectifiable paths will be used in the proofs of basic theorems from first principles, especiallyCauchy’s theorem. In the Riemann integral on the right, a change of parametrisation induces anintegration by substitution. If C is unbounded or if f(z) has an unbounded discontinuity on C,the integral can be interpreted as an improper Riemann integral in the usual ways.

Proof of Theorem 2.7. There is no loss of generality in assuming that C is smooth, becausea piecewise smooth contour can be partitioned into a finite number of smooth arcs. As in thepreceding proof, express the contour integral as a linear combination of real-variable Riemann-Stieltjes integrals with strictly increasing integrators. The smoothness of C implies that eachintegrator α(t) has a positive continuous derivative on [a, b]. The mean-value theorem yields apoint sk ∈ (tk−1, tk) such that

∆αk = α′(sk)∆tk.

In the Riemann sum, we can set t∗k = sk. Then the sum reduces to an ordinary Riemann sum whoselimit is the Riemann integral on the right-hand side of

∫ b

ag(t) dα(t) =

∫ b

ag(t)

dα

dtdt.

When applied to all the real-variable integrals that comprise the contour integral, we get the resultof the theorem.

Contour integrals satisfy elementary identities that they inherit from their definitions as lineintegrals or Riemann integrals:

∫

C

f(z) + g(z)

dz =

∫

Cf(z) dz +

∫

Cg(z) dz,

∫

Cλf(z) dz = λ

∫

Cf(z) dz.

If −C denotes the contour C with its orientation reversed, then

∫

−Cf(z) dz = −

∫

Cf(z) dz.

21

Suppose n rectifiable arcs γ1, γ2, . . . , γn are joined end to end with consistent orientation. Let Cdenote the combined arc γ1 + γ2 + . . . + γn, where overlapping pieces are counted separately. Then

∫

Cf(z) dz =

∫

γ1

f(z) dz + . . . +

∫

γn

f(z) dz.

Exactly the same notation can be used regardless of whether or not the γj are joined endto end. The γj can be disjoint, intersecting, or copied on top of themselves with the same orreverse orientation. A closed set K formed as a linear combination,

K = k1γ1 + k2γ2 + . . . knγn , kj ∈ Z,

of oriented arcs γj is called a chain. A linear combination of chains is also a chain. A linearcombination of oriented closed curves is a cycle and a linear combination of cycles is also a cycle.A chain or cycle is rectifiable or piecewise smooth if the γj forming it all have that property. It iseasy to define the line or contour integral of f(z) over a chain or cycle K according to

∫

Kf(z) dz =

n∑

j=1

kj

∫

γj

f(z) dz.

In many problems, we would like to break a closed contour or cycle into smaller closed contours γj ,j = 1, 2, . . . , n, by adding suitable auxiliary arcs that are followed in both directions. The latterdo not contribute to the value of the contour integral.

The second main type of line integral is the integral with respect to arc length s. Let s(t)denote the arc length along C from the initial point t = a to the moving point t = t. Observe thatds(t) = |dz(t)|. Modify the above Riemann sum to

T (f, P,C) :=

n∑

k=1

f(z(t∗k))|∆zk|,

and take the limit as n → ∞ and ‖P‖ → 0 in the manner described above. The result is the lineintegral,

∫

Cf(z) |dz| = lim

n→∞‖P‖→0

T (f, P,C) =

∫ b

af(z(t)) ds(t).

Here, reversal of orientation does not produce a sign change. The arc length of C is just

L(C) =

∫

C1 |dz|.

On a piecewise smooth contour, the second type of line integral can also be written as a Riemannintegral according to

∫

Cf(z) |dz| =

∫ b

af(z(t))

∣

∣

∣

∣

dz

dt

∣

∣

∣

∣

dt.

The main application of the line integral with respect to arc length in complex analysis is to placebounds on the preceding type of line or contour integral, as in the next theorem.

Lemma 2.8. Triangle inequality for contour integrals:∣

∣

∣

∣

∫

Cf(z) dz

∣

∣

∣

∣

≤∫

C|f(z)| |dz|.

22

Proof. According to the triangle inequality for complex numbers,

|S(f, P,C)| ≤n

∑

k=1

|f(z(t∗k))| |∆zk|.

The result follows by taking the limit as n→ ∞ and ‖P‖ → 0 on both sides in the manner describedabove. Initially, the partition Pǫ will, in general, be different on each side. Replace the separatepartitions by their common refinement, and let the latter play the role of Pǫ.

It is very important to be able to place bounds on contour integrals. Often, we want to be able toshow that a certain contour integral around a large semicircle or rectangle tends to zero as the radiusor edge length tends to infinity. Similarly, we want to be able to show that indents around sometypes of singularities do not contribute to the value of a contour integral. The triangle inequality

is a powerful tool for placing bounds on contour integrals, including absolutely convergent improperintegrals. In many applications, however, we can get a sufficiently accurate bound with a simplercorollary of the triangle inequality. The two most frequently used are the ML formula andJordan’s lemma.

Lemma 2.9. The ML formula. If a contour C has arc length L and if |f(z)| ≤M along C, then

∣

∣

∣

∣

∫

Cf(z) dz

∣

∣

∣

∣

≤ ML.

Proof. According to the triangle inequality for contour integrals,

∣

∣

∣

∣

∫

Cf(z) dz

∣

∣

∣

∣

≤∫

C|f(z)| |dz| ≤

∫

CM |dz| = ML.

Alternatively, if C is piecewise smooth, we can use the triangle inequality for Riemann integrals,

∣

∣

∣

∣

∫ b

af(z(t))

dz

dtdt

∣

∣

∣

∣

≤∫ b

a|f(z(t))|

∣

∣

∣

∣

dz

dt

∣

∣

∣

∣

dt ≤ M

∫ b

a

∣

∣

∣

∣

dz

dt

∣

∣

∣

∣

dt = ML.

Lemma 2.10. Jordan’s lemma. Let CR be all or part of the semicircular contour in the upperhalf-plane z = Reiθ, where the parameter θ runs from 0 to π (or part of the way). Suppose f(z)is continuous, not identically zero and |f(z)| ≤M(R) on CR, and suppose also that λ is a positivereal number. Define

I(R) =

∫

CR

f(z) eiλz dz.

Then,

|I(R)| < π

λM(R).

Remarks.

• Jordan’s lemma requires the integrand to have an exponential factor that is decaying on all ofthe circular-arc path as R→ ∞ except at and near one or both endpoints. For paths in thelower half-plane, the exponential factor should be e−iλz. For paths in the right half-plane, itshould be e−λz, and so on.

23

• In typical applications, the numerical factor π/λ is of no consequence. Besides, it can belowered. Thus the essential feature of Jordan’s lemma is the O-bound,

I(R) = O(M(R)) as R→ ∞.

In particular, I(R) → 0 as R→ ∞ whenever f(z) → 0 uniformly as |z| → ∞ in the upperhalf-plane (including the real axis).

• The factor eiλz is of order 1 near the ends of the semicircle on the real axis, but is exponentiallydecaying away from the ends. So only short arcs at the ends of the semicircle make a significantcontribution to the integral. Jordan’s lemma takes this into account. So Jordan’s lemma givesa sharper bound than the ML formula, which would give I(R) = O(RM(R)) as R→ ∞,because the ML formula would use the arc length of the whole path. (Of course, if the pathis bounded away from the real axis at both ends, the exponential factor would then decay onthe whole path, and the ML formula would be better.)

• If an integrand contains a factor similar to eiλzn

, one should change variable, z = ζ1/n,dz = (1/n)ζ1/n−1 dζ, and map the path into the complex ζ-plane before deciding whether touse Jordan’s lemma, the ML formula, or the triangle inequality.

Proof of Jordan’s lemma. On the semicircle CR,

z = Reiθ, dz = Rieiθ dθ, |dz| = Rdθ,

eiλz = eiλR(cos θ+i sin θ), |eiλz| = e−λR sin θ,

0 ≤ θ ≤ π. The triangle inequality gives

|I(R)| ≤∫

CR

|f(z)| |eiλz | |dz|

=

∫ π

0

∣

∣f(Reiθ)∣

∣e−λR sin θRdθ

≤ RM(R)

∫ π

0e−λR sin θ dθ

= 2RM(R)

∫ π/2

0e−λR sin θ dθ.

On the interval [0, π/2], sin θ ≥ 2θ/π, with equality only at the endpoints. Hence, if f(z) is notidentically zero, we have the strict inequality,

|I(R)| < 2RM(R)

∫ π/2

0e−2λRθ/π dθ

= 2RM(R)

[

− π

2λRe−2λRθ/π

]π/2

0

=πM(R)

λ

(

1 − e−λR)

<π

λM(R).

24

If CR is not the complete semicircle, then a smaller bound can be given. For example, if CR is aquadrant or less, the given bound can be halved.

It is handy to have the following very simple contour integrals evaluated directly from thedefinition before we introduce more general methods.

Lemma 2.11. If C[A,B] is any rectifiable contour joining z = A to z = B, then∫

C[A,B]1 dz = B −A,

∫

C[A,B]z dz = 1

2 (B2 −A2).

If C is any rectifiable closed contour, then∫

C1 dz = 0,

∫

Cz dz = 0.

Proof. Use the parametrisation z = z(t) as above with A = z(a) and B = z(b). In the casef(z) = 1, the Riemann sum for the partition P is a telescoping sum,

S(1, P,C[A,B]) =

n∑

k=1

z(tk) − z(tk−1)

= z(b) − z(a) = B −A.

The result is independent of the partition and of the choice of sample points t∗k. Thus,∫

C[A,B]1 dz = B −A,

regardless of whether or not the curve C[A,B] is rectifiable. So the corresponding Riemann-Stieltjesintegral exists even when the integrator is not of bounded variation. But, in this case, the integratoris continuous while the integrand is of bounded variation (trivially), so one of the standard sufficientconditions for existence of the integral is satisfied.

For the case f(z) = z, we must assume that C[A,B] is rectifiable or otherwise the integral maynot exist according to our definition. (For example, the integral does not exist on the spiral pathC[0, 1] with equation z(t) = t1/3 exp(2πi/t).) According to the integration-by-parts formula,

∫

C[A,B]z dz =

∫ b

az(t) dz(t) =

[

z(t)z(t)

]b

a

−∫ b

az(t) dz(t) = B2 −A2 −

∫

C[A,B]z dz.

This gives the desired result. However, we promised that our treatment of Riemann-Stieltjes inte-grals would be self-contained as far as possible, while integration by parts is a relatively nontrivialtheorem in that theory. So let us provide a direct proof from our definitions.

A Riemann sum for the partition P is

S(z, P,C[A,B]) =

n∑

k=1

z(t∗k)∆zk.

Since t∗k is any point in the subinterval [tk−1, tk], we are free to choose both endpoints and formtwo Riemann sums:

S1(z, P,C[A,B]) =

n∑

k=1

z(tk)∆zk,

S2(z, P,C[A,B]) =n

∑

k=1

z(tk−1)∆zk.

25

Their sum telescopes according to

S1 + S2 =

n∑

k=1

z(tk) + z(tk−1)

∆zk =

n∑

k=1

(

z(tk))2 −

(

z(tk−1))2

= B2 −A2.

Once again, the result is independent of the partition, and we get, in the appropriate limit,∫

C[A,B]z dz = 1

2(B2 −A2).

In both these cases, the value of the integral depends only on the endpoints and not on the detailsof the path between the endpoints. A closed contour C is just a case having B = A. Alternatively,C can be formed by taking one path C[A,B] from A to B and returning by another such path withits orientation reversed. Either way, the integrals of 1 and z around a closed contour vanish.

The preceding lemma shows that the following working gives the correct result:

∫ B

Az dz =

[

z2

2

]B

A

=B2

2− A2

2.

The path of integration is any rectifiable path from z = A to z = B. Observe that z2/2 is anantiderivative of z. We would expect this result to generalise to a suitable analogue of the funda-mental theorem of calculus in the theory of the Riemann integral. Thus if F (z) is an antiderivativeof f(z), then

∫ B

Af(z) dz =

[

F (z)

]B

A

= F (B) − F (A).

The next theorem gives sufficient conditions for the validity of this working.

Theorem 2.12. Fundamental theorem of calculus for contour integrals. Suppose that f(z)is continuous and is the derivative of an analytic function F (z) in a region D. Thus F (z) is anantiderivative or primitive of f(z). If C[A,B] is a rectifiable contour in D joining z = A toz = B, then

∫

C[A,B]f(z) dz = F (B) − F (A).

In particular, if C is a closed contour in D, then∫

Cf(z) dz = 0.

Proof. Parametrise the path C := C[A,B] according to

z = z(t), a ≤ t ≤ b, A = z(a), B = z(b).

The complex-valued function z(t) of the real variable t is a continuous function of bounded variationon [a, b]. Together with the continuity of f(z), these properties guarantee the existence of thecontour integral.

If the path is piecewise smooth, the proof is easy, for then

d

dtF (z(t)) = f(z(t))

dz

dt,

26

except at its isolated corner points and cusps, if any. Then∫

C[A,B]f(z) dz =

∫ b

af(z(t))

dz

dtdt =

[

F (z(t))]b

a= F (B) − F (A).

If the path is not piecewise smooth, we look for a suitable Riemann sum for the contour integral.The differentiability of F (z) implies that, given ǫ > 0, there exists δ > 0, depending on ǫ and z,such that

∣

∣

∣

∣

F (ζ) − F (z)

ζ − z− F ′(z)

∣

∣

∣

∣

< ǫ,

whenever |ζ − z| < δ. (We assume that δ is small enough that the latter disc is contained in D.)Recall that F ′(z) = f(z) and that f(z) is continuous in D. Define

ǫ1(ζ, z) :=F (ζ) − F (z)

ζ − z− f(z).

Then ǫ1(ζ, z) is continuous on D × D and vanishes on the complex line ζ = z. If z is allowed torun over a compact set, then we can choose δ uniformly with respect to z. The arc C = C[A,B] issuch a compact set. Hence, for z ∈ C[A,B], given ǫ > 0, there exists δ > 0, depending only on ǫ,such that

F (ζ) − F (z) = f(z)(ζ − z) + ǫ1(ζ, z)(ζ − z), |ǫ1(ζ, z)| <ǫ

2L,

whenever |ζ − z| < δ, where L is the finite arc length of C.

Next, the integrability of f(z) on C implies that there exists a partition Pǫ of [a, b] such that, forevery partition P finer than Pǫ, and every choice of sample point t∗k on the kth subinterval of P ,

∣

∣

∣

∣

∫

Cf(z) dz − S(f, P,C)

∣

∣

∣

∣

<ǫ

2,

where ǫ is the arbitrary positive number already introduced. The Riemann sum takes the form,

S(f, P,C) :=

n∑

k=1

f(z(t∗k))∆zk , ∆zk := zk − zk−1, zk = z(tk), zk−1 = z(tk−1),

t∗k ∈ [tk−1, tk], where the partition P is t0, t1, t2, . . . , tn, t0 = a, tn = b. By refining the partitionPǫ, if necessary, we can ensure that all the |∆zk| formed from Pǫ are less than δ, and so all the |∆zk|formed from P are less than δ. We are also free to make the restriction t∗k = tk−1, k = 1, 2, 3, . . . , n.

Then, by the earlier inequality for F (z), we have

S(f, P,C) =

n∑

k=1

f(zk−1)(zk − zk−1)

=

n∑

k=1

F (zk) − F (zk−1) − ǫ1(zk, zk−1)(zk − zk−1)

= F (zn) − F (z0) −n

∑

k=1

ǫ1(zk, zk−1)(zk − zk−1)

= F (B) − F (A) −n

∑

k=1

ǫ1(zk, zk−1)(zk − zk−1),

27

where |ǫ1(zk, zk−1)| < ǫ/(2L) for all subintervals of partitions P finer than Pǫ. The remainder termhas a modulus bounded by

∣

∣

∣

∣

n∑

k=1

ǫ1(zk, zk−1)(zk − zk−1)

∣

∣

∣

∣

<ǫ

2L

n∑

k=1

|zk − zk−1| ≤ǫ

2.

Hence, by the triangle inequality,∣

∣

∣

∣

∫

Cf(z) dz − F (B) − F (A)

∣

∣

∣

∣

≤∣

∣

∣

∣

∫

Cf(z) dz − S(f, P,C)

∣

∣

∣

∣

+

∣

∣

∣

∣

S(f, P,C) − F (B) − F (A)∣

∣

∣

∣

<ǫ

2+ǫ

2= ǫ.

Since ǫ is arbitrary, this completes the proof that∫

Cf(z) dz = F (B) − F (A),

when C is any rectifiable path joining A to B.

Remark. This result shows that, for such f(z), the value of the integral depends only on theendpoints and not on the details of the path in between. If we wished, we could define the integralover a non-rectifiable path by this formula. There does not seem to be any compelling reason toartificially generalise the contour integral in this fashion.

The fundamental theorem of calculus has an important converse:

Theorem 2.13. Suppose that D is a bounded region in C and f(z) is continuous in D. Supposealso that

∫

Γf(z) dz = 0

for every closed contour Γ belonging to any one of the following three classes:

(a) all piecewise smooth closed curves in D,(b) all closed polygonal lines in D, or(c) all closed polygonal lines in D consisting of horizontal and vertical line segments only.

Then f(z) is the derivative of an analytic function F (z) in D.

Proof. The strongest statement of the theorem is (c). Statement (c) implies (b), and (b) im-plies (a). It is therefore enough to prove (c). We will give a proof for both (b) and (c) because theyillustrate different techniques.

Begin with case (b). We do not need f(z) to be continuous in D, but just continuous on all linesegments (endpoints included) in D. Let Γ[A,B] denote an arbitrary polygonal line in D joiningz = A to z = B. Then

∫

Γ[A,B]f(z) dz

can only depend on the endpoints A and B of the path. Hence,

F (z) :=

∫

Γ[A,z]f(ζ) dζ

28

is a well-defined function of z in D. Choose a particular point z0 ∈ D and let z0 + h be a point ina neighbourhood N(z0, δ) in D. Then

F (z0) =

∫

Γ[A,z0]f(ζ) dζ, F (z0 + h) =

∫

Γ[A,z0+h]f(ζ) dζ.

We are free to restrict the latter path to the form,

Γ[A, z0 + h] = Γ[A, z0] + γ,

where γ is a polygonal line joining z0 to z0 + h in N(z0, δ). Then,

F (z0 + h) − F (z0)

h=

1

h

∫

γf(ζ) dζ.

We wish to take the limit as h→ 0 in any manner. The integral on the right is the same for allpolygonal paths γ joining z0 to z0 + h. In particular, we can let γ be the single straight line segmentparametrised by

z = z0 + ht, dz = hdt, 0 ≤ t ≤ 1.

Then, the difference quotient is

F (z0 + h) − F (z0)

h=

∫ 1

0f(z0 + ht) dt.

Continuity of f(z) along γ now guarantees that F (z) is differentiable at z0 and that

F ′(z0) = limh→0

F (z0 + h) − F (z0)

h= f(z0).

Since z0 is any point in D, we have proved that F (z) is analytic in D and that F ′(z) = f(z). Inother words, f(z) is the derivative of an analytic function. (The latter is unique up to an additiveconstant.)

Next, consider case (c), which implies (a) and (b). Define

F (z) :=

∫

Γ[A,z]f(ζ) dζ,

where Γ[A, z] is a polygonal line in D joining the point ζ = A to ζ = z and consisting only ofhorizontal and vertical line segments. The integral is independent of the choice of polygonal lineand so the function F (z) is well-defined. We can restrict attention to paths where the line segmentsare alternately horizontal and vertical. Let z = x+ iy and let z vary over a neighbourhood of aparticular point z0 in D. The last line segment of Γ[A, z] is either horizontal or vertical. If it ishorizontal, let it run from x0 + iy to x+ iy, where x0 is fixed, and let the immediately precedingline segment run from x0 + iy1 to x0 + iy. All the earlier line segments forming the path are fixed.Then the integral over the path minus the last line segment is a function G1(y) of y alone, and

F (z) = G1(y) +

∫ x

x0

f(u+ iy) du.

Taking a partial derivative with respect to x gives

∂F (z)

∂x= f(z).

29

If the last line segment is vertical, let it run from x+ iy0 to x+ iy, where y0 is fixed. The sameargument gives

F (z) = G2(x) +

∫ y

y0

f(x+ iv)i dv.

Taking a partial derivative with respect to y gives

∂F (z)

∂y= if(z).

So F (z) satisfies the partial differential equation,

∂F

∂x+ i

∂F

∂y= 0,

which is equivalent to the Cauchy-Riemann equations for the real and imaginary parts of F (z).The directional derivative of F (z) parallel to the real axis is f(z). Since f(z) is continuous in D bythe hypotheses of the theorem, F (z) has continuous partial derivatives and so is continuously dif-ferentiable in D. This, together with the Cauchy-Riemann equations, implies that F (z) is analyticin D and F ′(z) = f(z).

Remarks. In the last two theorems, the domain D does not need to be simply connected. This issignificant because not every analytic function has a single-valued primitive in a multiply connecteddomain. The wording of these theorems guarantees that F (z) is single-valued. If we let D besimply connected, the proof of the Theorem 2.13 can be modified to apply to all triangles in Dor all rectangles in D with sides parallel to the axes. This theorem will soon be superseded by astronger theorem known as Morera’s theorem, Theorem 4.9.

30

3 Cauchy’s theorem and extensions

Cauchy’s theorem is the pivotal result that separates complex analysis from other branches of anal-ysis. Without it, complex analysis would just be real analysis in the plane with various convenientshort cuts made available by strategically using the square root of minus one. Of course, complexanalysis is rich enough to have important branches, such as conformal mapping, that do not appearat first glance to rely much on Cauchy’s theorem. Conversely, real analysis has branches, such asharmonic function theory, that rely heavily on theorems that are real-variable analogues or corol-laries of Cauchy’s theorem. In fact, the proof that every nonconstant polynomial in one variableover the field of real numbers factorises into real linear and quadratic factors is best handled withthe complex integral calculus.

Cauchy’s theorem provides sufficient conditions for the vanishing of a contour integral arounda closed curve. The actual statement of the theorem depends on what hypotheses we are willingto impose on the class of functions appearing in the integrand, the class of closed curves, andthe connectivity of the domain holding the curves. It is best to formulate a basic statement ofreasonable generality and build extensions upon it later.

Theorem 3.1. Cauchy’s theorem, also called the Cauchy-Goursat theorem (Cauchy 1825,Goursat 1900, Moore 1900, Pringsheim 1901, Knopp 1913). If f(z) is analytic (that is, point-wisedifferentiable) on a simply connected open region D ⊂ C and if C is any rectifiable closed contouror cycle in D, then

∫

Cf(z) dz = 0.

Our proof of Cauchy’s theorem will proceed in several steps. We begin by establishing Cauchy’stheorem for triangles. Another possible starting point is rectangles with sides parallel to the axes.There are in fact several quite distinct ways to arrive at the generality in the statement of Cauchy’stheorem just given, and of the extensions to follow. Some of these alternative approaches will beexamined briefly.

Recall that, at this stage, analytic just means pointwise differentiable on a region. The addi-tional smoothness that complex analytic functions necessarily possess is the subject of theoremsthat follow on after Cauchy’s theorem.

The curve C is not necessarily simple. It may self-intersect in any manner, provided it is closedand rectifiable. Rectifiable includes piecewise smooth, which is the class of contours of greatestinterest in complex analysis. Rectifiable paths are more natural for real-variable line integrals, whichmakes them more natural to work with in some of the proofs in the complex domain. Nevertheless,most textbooks restrict attention to piecewise smooth paths, but their proofs are usually easy toadapt to the rectifiable case. Some also restrict attention to simple closed curves, which is a moresignificant loss of generality. Generalisations to domains that are not simply connected and tofunctions that may not be analytic will be considered after the proof of Theorem 3.1.

Cauchy’s original proof employed Green’s theorem in the plane. He needed to impose stricterhypotheses on f(z) and C than are needed in Goursat’s later proof. Suppose, for the purposes ofthis paragraph only, that f ′(z) is continuous in D and that C is piecewise smooth and simple. LetE denote the inside of C plus C itself. Use the same symbols C, D and E for the correspondingsets in the real xy-plane. Put z = x+ iy and f(z) = u(x, y) + iv(x, y). Under these hypotheses, u

31

and v satisfy the Cauchy-Riemann equations, ux = vy and uy = −vx, the partial derivatives beingcontinuous in D in the present context. Then, two applications of Green’s theorem give

∫

Cf(z) dz =

∫

C(u dx− v dy) + i

∫

C(v dx+ u dy)

= −∫∫

E

(

vx + uy

)

dx dy + i

∫∫

E

(

ux − vy

)

dx dy

= 0.

This quick proof appears in many textbooks, and is frequently done in lectures when the purposeis to get through Cauchy’s theorem quickly on the way to applications.

The quick proof is unsatisfactory for a couple of reasons. The obvious reason is that we haveimposed unnecessary restrictions on C and f(z). A more potent objection is that Green’s theoremis at least at the same mathematical depth as Cauchy’s theorem, and so one should not claim aproof of one on the basis of the other unless the other has been proved rigorously under decenthypotheses. The proof of Green’s theorem for rectifiable Jordan curves, for example, is not easyand was first done by D. H. Potts (1951). The latter proof is included in the first edition of TomM. Apostol: Mathematical Analysis (Addison-Wesley 1957). Also Green’s theorem is poorly suitedto self-intersecting contours.

Edouard Goursat (1900) proved Cauchy’s theorem for piecewise smooth Jordan curves by par-titioning the interior into small squares, with residual fragments next to the boundary. He did notrequire f ′(z) to be continuous. Ahlfors’ textbook Complex Analysis contains a proof of a generalform of Cauchy’s theorem (Theorem 3.18 below) that includes a stage where are region is parti-tioned in this fashion. Alfred Pringsheim (1901) simplified and strengthened Goursat’s proof bystarting with the special case where C is a triangle. This approach allows for upgrades in stages toself-intersecting rectifiable contours.

Our proof is modelled on the proofs by Konrad Knopp in the fifth edition of his book, Funk-

tionentheorie I (1937, originally 1913) (English language Dover editions 1945 and 1996), and byEinar Hille, Analytic Function Theory, Volume I (1959, reprinted by Chelsea NY 1982), whichfollow Pringsheim’s approach.

Some textbooks begin by proving Cauchy’s theorem for a circular or rectangular contour. Fromthere, they deduce Cauchy’s integral formula for circular or rectangular contours, which then impliesthe differentiability of f(z) to all orders, the Cauchy-Taylor theorem (circular contours preferred),and many of the incidental theorems and corollaries that we will meet in the coming pages, butwith the contours restricted. Having obtained all this extra information about analytic functions insimple domains, they can then turn their attention to more general contours and more complicateddomains of analyticity.

Another approach is to prove that an analytic function in an open disc has a primitive function.This can be deduced from a proof of Cauchy’s theorem for rectangles with sides parallel to the axes.A primitive function implies Cauchy’s theorem for arbitrary rectifiable closed contours accordingto Theorem 2.12, but the domain D is restricted to a disc. The next step is to prove Cauchy’sintegral formula for Jordan curves in a disc and use it to step up to more general formulations ofCauchy’s theorem.

Let us begin the proof of Theorem 3.1.

32

Lemma 3.2. Pringsheim’s lemma (Cauchy’s theorem for triangles). Suppose that f(z) isanalytic in a region D. Let ∆ be any triangle in D whose interior is also contained in D. Then

∫

∆f(z) dz = 0.

Proof. Suppose, on the contrary, that

I0 :=

∫

∆f(z) dz 6= 0.

Let A, B and C denote the vertices of ∆ in the order following its orientation and let L, M and Ndenote the midpoints of the sides AB, BC and CA, respectively. Form the four smaller orientedcongruent triangles, δ01 = ALN , δ02 = LBM , δ03 = NMC and δ04 = MNL. Then the internaledges are followed in both directions and so

∆ = ABC =

4∑

k=1

δ0k ,

I0 =4

∑

k=1

∫

δ0k

f(z) dz.

Let I1 denote the integral on the right-hand side with largest modulus and let ∆1 denote thecorresponding triangle (one of the δ0k). If two or more integrals share the same largest modulus,choose the one with lowest k. Then, according to the triangle inequality for complex numbers,

|I0| ≤4

∑

k=1

∣

∣

∣

∣

∫

δ0k

f(z) dz

∣

∣

∣

∣

≤ 4|I1|.

Next, the triangle ∆1 can be subdivided in the same manner into four congruent triangles δ1k,k = 1, 2, 3, 4, and one of those triangles ∆2, around which the integral of f(z) is I2, can beselected so that

|I1| ≤ 4|I2|.This process can be iterated indefinitely. The sequence of integrals, I0, I1, I2, . . . , around therespective triangles, ∆0 := ∆, ∆1, ∆2, . . . , satisfies

|In−1| ≤ 4|In|, |I0| ≤ 22n|In|.

If the perimeter of ∆ is L0, the perimeter of ∆n is Ln = 2−nL0. Let En denote the closed setconsisting of the triangle ∆n and its interior. The sequence of closed sets, E0, E1, E2, . . . , satisfies

E0 ⊃ E1 ⊃ E2 ⊃ . . . ⊃ En ⊃ . . . .

There is exactly one limit point z0 common to all of the En. Because z0 ∈ ∆, it is also containedin D.

We now make use of the differentiability of f(z) at z0. Given ǫ > 0, there exists δ > 0 such that,for all z in the open disc |z − z0| < δ,

f(z) = f(z0) + f ′(z0)(z − z0) + η(z)(z − z0)

33

with |η(z)| < ǫ. For sufficiently large n, the triangle ∆n and all later triangles in the sequence arecontained within the open disc |z − z0| < δ. Since z0 is inside or on ∆n, this triangle will alwaysbe inside the disc if Ln < 2δ. Now

In =

∫

∆n

f(z) dz

=(

f(z0) − z0f′(z0)

)

∫

∆n

1 dz + f ′(z0)

∫

∆n

z dz +

∫

∆n

η(z)(z − z0) dz

=

∫

∆n

η(z)(z − z0) dz,

where we used Lemma 2.11 to set the integrals of 1 and z to zero. In the last integrand, we havethe bounds, |η(z)| < ǫ and |z − z0| < 1

2Ln, for z ∈ ∆n. Hence, according to the ML formula,

|In| =

∣

∣

∣

∣

∫

∆n

η(z)(z − z0) dz

∣

∣

∣

∣

≤ 12ǫ(Ln)2 = 2−2n−1ǫ(L0)

2.

Then|I0| ≤ 22n|In| ≤ 1

2ǫ(L0)2.

Because ǫ is arbitrary, the right-hand side is arbitrarily small. This contradicts our initial assump-tion that I0 6= 0. Hence I0 = 0. In other words,

∫

∆f(z) dz = 0

under the stated hypotheses.

Corollary 3.3. Goursat’s lemma (Cauchy’s theorem for rectangles). Suppose that f(z) isanalytic in a region D. Let R be any rectangle in D whose interior is also contained in D and,optionally, whose sides are parallel to the axes. Then

∫

Rf(z) dz = 0.

Proof. Any oriented rectangle can be formed by gluing together two oriented right-angled trian-gles along their common hypotenuse, where the two copies of the hypotenuse will have oppositeorientations and cancel each other. Of course, the proof can be given directly from first principlesby following an argument almost identical to the argument just given for triangles.

The next two lemmas are geometric in character and will allow us to upgrade Cauchy’s theoremfrom triangles to closed polygonal lines, which may self-intersect in any manner. After that, wewill complete the proof of Cauchy’s theorem by upgrading from closed polygonal lines to rectifiableclosed curves and cycles.

Lemma 3.4. Every oriented simple closed polygon can be triangulated.

Proof. Let the polygon have n ≥ 4 sides. (There is nothing to prove when n = 3.) First, a convexpolygon is very easily triangulated. Just add the n− 3 diagonals that join all the nonadjacentvertices to one particular vertex. The orientation of each triangle so formed is induced by the

34

arrow on the side or sides in common with the original polygon. In the contour sum of all thetriangles, all the auxiliary diagonals are traversed in both directions and so cancel out.

In the case of a nonconvex but simple closed polygon, proceed by induction on n, the result for thesimplest case n = 4 being obvious. One just needs to find one diagonal of the polygon that is interiorto the polygon except for its endpoints. The diagonal then divides the polygon into two polygonswith ≤ n− 1 sides, and is traversed both ways in the sum of the two polygons. The inductionhypothesis now guarantees the required triangulation. To construct a suitable diagonal, select avertex A at which the interior angle of the polygon is reentrant (exceeds π). Begin by consideringthe ray that bisects this interior angle and then rotate the ray anticlockwise. Let B be the nearestintersection point to A of the ray with the polygon. As the ray is rotated, a suitable diagonal willoccur if B reaches a vertex or if the interior of AB contacts another reentrant vertex. (If more thatone vertex is contacted, pick the nearest one to A.) The only way that this process can fail is if ABdoes not meet a vertex before it coincides with one of the two produced edges of the polygon withendpoint A. But if that happens, it cannot happen again if the ray is rotated clockwise, becausethe angle at A is reentrant. So a suitable diagonal exists. Note that this construction yields atriangulation consisting of exactly n− 2 suitably oriented triangles.

Lemma 3.5. Every oriented closed polygonal line Γ is a finite sum of oriented simple closedpolygons and digons, where a digon is a straight line segment traversed both ways. Hence everyoriented closed polygonal line is a sum of oriented triangles and digons.

Proof. Let Γm, m ≥ 2, denote any closed polygonal line (or polygon) having m vertices and medges (not necessarily distinct). Let Pm, m ≥ 3, denote a Γm which is a simple closed polygon.Let P2 denote a digon. Let the given Γ be a Γn. Label the vertices A0, A1, A2, . . . , An, withAn = A0. The single edge [A0, A1] is a Jordan arc, whereas the full polygon [A0, A1, . . . , An] isnot. Let r, 1 ≤ r ≤ n− 1, be the largest integer such that [A0, A1, . . . , Ar] is a Jordan arc. Then[A0, A1, . . . , Ar+1] must self-intersect in one of the following three ways:

• r + 1 = n, in which case Γ is a simple closed polygon Pn or a digon P2;

• [Ar−1, Ar, Ar+1] is a digon or contains a digon;

• (Ar, Ar+1] intersects or overlaps with [A0, A1, . . . , Ar−1) at least once.

In the first case, we have nothing more to prove. In the second case, either Ar+1 = Ar−1 or Ar−1 isan interior point of [Ar, Ar+1] or Ar+1 is an interior point of [Ar−1, Ar]. Either way, Γ = Γm + P2,where m is either n− 1 or n− 2.

In the third case, let B be the intersection of (Ar, Ar+1] with [A0, . . . , Ar−1) nearest to Ar. ThenB occurs on a unique [Aq−1, Aq), 1 ≤ q ≤ r − 1, because [A0, . . . , Ar] is a Jordan arc. Now[B,Aq, . . . , Ar, B] is a simple closed polygon Pr−q+2. The remainder of Γ, with the auxiliaryvertex B if B 6= Aq−1, is either a Γn+q−r or a Γn+q−r−1. In all cases so far, Γ is either a Pn or

Γ = Γm + Pk ,

where 2 ≤ m ≤ n − 1 and 2 ≤ k ≤ n − 1. The orientations of Γm and Pk are induced in anobvious way from the arrows on the edges they have in common with Γ. No auxiliary edges havebeen added. If Γm is not a Pm, repeat the construction on Γm and get Γm = Γm2

+ Pk2. Since

n > m > m2 > . . . , the successive constructions must terminate with Γ expressed as a finite sum

35

of Pk’s only, each with a uniquely defined orientation. Then, according to the previous lemma, Γcan be triangulated in the sense that Γ is a finite sum of oriented triangles and digons.

Lemma 3.6. Cauchy’s theorem for closed polygonal lines. Suppose that f(z) is analytic ina simply connected region D. Let Γ be any closed polygonal line in D. Then

∫

Γf(z) dz = 0.

Proof. By the previous two lemmas, Γ can be expressed as a finite sum of oriented trianglesand digons. The integral of any integrable function over a digon is zero. Because D is simplyconnected, all the triangles and their interiors are contained in D. According to Pringsheim’slemma (Lemma 3.2), the integral of f(z) vanishes around each triangle of the triangulation of Γ.Hence, the integral of f(z) around Γ vanishes. Thus Cauchy’s theorem is proved for closed polygonallines.

Completion of the proof of Cauchy’s theorem. It remains to upgrade the proof from closedpolygonal lines to closed rectifiable curves.

Let C be the closed curve parametrised by

z = z(t), a ≤ t ≤ b, z(b) = z(a).

The complex-valued function z(t) is continuous on [a, b]. The curve C is rectifiable if and only ifz(t) is of bounded variation on [a, b]. Because f(z) is analytic in D, f(z(t)) is at least continuouson C. These conditions are sufficient to guarantee the existence of the Riemann-Stieltjes integralthat defines the contour integral:

∫

Cf(z) dz =

∫ b

af(z(t)) dz(t).

Thus, given arbitrary ǫ > 0, there exists a partition Pǫ of [a, b], such that, for all finer partitionsP = t0, t1, . . . , tn, t0 = a, tn = b, the Riemann sum,

S(f, P,C) =

n∑

k=1

f(z(t∗k))∆zk ,

∆zk := zk − zk−1, zk := z(tk), zk−1 := z(tk−1), t∗k ∈ [tk−1, tk],

satisfies∣

∣

∣

∣

∫

Cf(z) dz − S(f, P,C)

∣

∣

∣

∣

<ǫ

2.

Since t∗k is any point in the subinterval [tk−1, tk], we are free to choose the endpoint t∗k = tk.

Because C is a subset of a region D, every point of C is an interior point of D, and so every pointof C is contained in an open disc neighbourhood within D. These neighbourhoods form an opencover of C. Because C is closed and bounded, the Heine-Borel theorem guarantees that we canselect a finite subcover, the smallest neighbourhood of which has a positive radius. This provesthat C is bounded away from the boundary ∂D of D. Define

ρ :=1

2inf

|z1 − z2| : z1 ∈ C, z2 ∈ ∂D

.

36

Let E be the closed set,E =

z : |z − z1| ≤ ρ, z1 ∈ C

.

Then C ⊂ interior of E ⊂ E ⊂ D. Note that E may not be simply connected. If desired, we couldconstruct a closed simply connected set E1 such that E ⊂ E1 ⊂ D, but we do not need E1 in thisproof. We only need to know that D is simply connected.

Let Γ be the closed oriented polygonal line [z(t0), z(t1), . . . , z(tn)], where the tk are the subdivisionpoints of the partition P above. By refining the partition P if necessary, we can guarantee thatall the line segments of Γ are contained in E. Since f(z) is analytic in D, it is continuous andtherefore uniformly continuous in the closed subset E. This means that, with the same arbitrary ǫthat we have already chosen, we can find δ > 0 depending only on ǫ such that

∣

∣f(z1) − f(z2)∣

∣ <ǫ

2L,

whenever |z1 − z2| < δ, z1 ∈ E, z2 ∈ E, where L is the arc length of C. Again, by refining thepartition P if necessary, we can guarantee that |∆zk| < δ for each k, where ∆zk = z(tk)− z(tk−1).In other words, all the line segments forming Γ have lengths less than δ.

Next, with t∗k = tk, an application of the triangle inequality for contour integrals and the definitionof arc length as a supremum give

∣

∣

∣

∣

∫

Γf(z) dz − S(f, P,C)

∣

∣

∣

∣

=

∣

∣

∣

∣

n∑

k=1

∫ z(tk)

z(tk−1)f(z) dz −

n∑

k=1

f(z(tk))

∫ z(tk)

z(tk−1)1 dz

∣

∣

∣

∣

=

∣

∣

∣

∣

n∑

k=1

∫ z(tk)

z(tk−1)

f(z) − f(z(tk))

dz

∣

∣

∣

∣

≤n

∑

k=1

∫ z(tk)

z(tk−1)|f(z) − f(z(tk))| |dz|

<ǫ

2L

n∑

k=1

∫ z(tk)

z(tk−1)1 |dz|

≤ ǫ

2LL

=ǫ

2.

Finally, the triangle inequality for complex numbers gives

∣

∣

∣

∣

∫

Cf(z) dz −

∫

Γf(z) dz

∣

∣

∣

∣

≤∣

∣

∣

∣

∫

Cf(z) dz − S(f, P,C)

∣

∣

∣

∣

+

∣

∣

∣

∣

S(f, P,C) −∫

Γf(z) dz

∣

∣

∣

∣

37

<ǫ

2+ǫ

2

= ǫ.

But, by the previous lemma,∫

Γf(z) dz = 0.

Hence,∣

∣

∣

∣

∫

Cf(z) dz

∣

∣

∣

∣

< ǫ.

Since ǫ is arbitrary, it follows that∫

Cf(z) dz = 0,

for every rectifiable closed contour C in D. This completes the proof of Cauchy’s theorem forrectifiable closed contours. The upgrade to rectifiable cycles is a trivial consequence of the definitionof a cycle as a linear combination of closed contours.

The fundamental theorem of calculus for contour integrals Theorem 2.12 and its converseTheorem 2.13 immediately imply:

Corollary 3.7. Suppose that f(z) is analytic in a simply connected region D. Then f(z) has ananalytic antiderivative (or primitive function) F (z) in D, unique up to an additive constant. Itsatisfies F ′(z) = f(z) and

∫

C[A,B]f(z) dz = F (B) − F (A),

where C[A,B] is any rectifiable contour in D joining the point A to the point B.

Proof. Take

F (z) =

∫

C[A0,z]f(ζ) dζ + constant,

where the path of integration is any rectifiable curve in D joining the fixed point A0 to z. Cauchy’stheorem guarantees that the integral is independent of the path, and so F (z) is a well-defined func-tion of z, unique up to an additive constant which depends on the position of A0. The paths fromA0 to z include the special paths in Theorem 2.13, in which case F (z) is known to be an antideriva-tive (or primitive function) of f(z). The main result follows from Theorem 2.12. Alternatively, wecan apply Cauchy’s theorem to the closed path C[A0, A,B,A0] = C[A0, A] + C[A,B] + C[B,A0]and get

0 =

∫

C[A,B]+

∫

C[A0,A]−

∫

C[A0,B]=

∫

C[A,B]f(z) dz + F (A) − F (B).

Extensions of Cauchy’s theorem. Cauchy’s theorem can be extended or generalised in severaldifferent directions. Some of these extensions are important and are depended upon routinely incomplex analysis. First, inspection of the above proof shows that it works with only minor changesof wording in the following corollary:

Corollary 3.8. Suppose f(z) is analytic in an open region D which is not necessarily simplyconnected, and C is a closed rectifiable contour in D. Then

∫

Cf(z) dz = 0

38

if either (a) C is contained in a simply connected subregion of D, or (b) C forms the orientedboundary of a simply connected subregion of D.

An example of case (b) is where C is the oriented boundary of an open annulus with a transversalcut. The contour C includes two copies of the transversal cut followed in opposite directions.

More commonly occurring is the case where f(z) has a jump discontinuity across such a transver-sal cut, but where f(z) analytically continues across the cut from both sides. The domain of ana-lyticity of f(z) has now run onto a two-sheeted (or more) covering of the complex plane, in otherwords, a local Riemann surface. We do not need the full Riemann surface for f(z), just enough ofit to place the contour C interior to a domain of analyticity of f(z) along with a simply connectedregion D that covers C.

The situation just described occurs routinely in applications of contour integration to the eval-uation of real definite integrals, where one wants to wrap a contour around a branch cut. Sowhile f(z) is not single-valued along C, f(z(t)) is a single-valued function of the parameter t thatparametrises C.

Given that locally analytic functions can and sometimes do analytically continue onto localRiemann surfaces, it is handy to have the following formal extension of Cauchy’s theorem:

Corollary 3.9. Cauchy’s theorem holds whenever D is any bounded simply connected region(homeomorphic to an open disc in C) on a Riemann surface and f(z) is analytic in D.

As already mentioned, without a precise definition of a Riemann surface, theorems that mentionRiemann surfaces are intended to be viewed as descriptive rather than rigorous. Many later resultsextend in a similar fashion to local Riemann surfaces, and we will not remind the reader on everyoccasion.

It is understood that the branch points of f(z) that join different Riemann sheets are not pointsof analyticity of f(z) and should not be included in D. It may be possible to subdivide a regionD on a Riemann surface into overlapping simply connected regions that each fit on one copy ofthe complex plane. It may also be possible to change variable in a contour integral so that theimage of the contour and the domain D that encloses it lives on a single copy of the complex plane.Such a change of variable may also cancel out some branch points, thereby allowing certain typesof branch points to be included in the original domain.

The following so-called strong form of Cauchy’s theorem is the work of several authors, beginningwith V. V. Golubev (1916) and S. Pollard (1923). Briefly, it asserts that Cauchy’s theorem holdsif the contour or cycle C runs onto the boundary of D at places where f(z) is continuous fromwithin. By shrinking D away from boundary points where f(z) is not continuous, if any, we cangive D a boundary onto which f(z) is continuous from within at all points. The essential idea isthat we are not asking f(z) to be analytic on the contour or cycle C, as in the main statementof Cauchy’s theorem. It is therefore possible for some types of essential singularities and branchpoints (but not poles) of f(z) to reside on the boundary and satisfy the continuity requirements.(See Chapters 4 and 5 for a partial classification of singularities.)

A simple example is where the closure of D is the semicircle x2 + y2 ≤ R2, y ≥ 0, and f(z)is

√−iz or ze−2πi/z . If f(z) is analytic on the boundary of D except for a small number of

singularities, the usual method is to indent the path into D around these singularities and takea limit using the ML formula or triangle inequality. This method does not require f(z) to be

39

continuous at or even bounded near the exceptional points on the boundary.

Let the oriented boundary of D be denoted Γ. So f(z) is analytic in D and continuous on theclosed set D ∪ Γ. We can allow Γ to be a non-Jordan curve along which f(z) is multi-valued asa function of z but single-valued and continuous (and continuous from within D) as a function ofthe parameter t that parametrises Γ. We do not need to burden the statement of the theorem withthis complication as it belongs to the formal extension of the theorem to local Riemann surfaces.

Theorem 3.10. Cauchy’s theorem for closed regions (based on Pollard 1923). Let D ⊂ C

be a Jordan region whose boundary is a rectifiable Jordan curve Γ. Let f(z) be a function whichis analytic in D and continuous on [D] := D ∪ Γ. Then

∫

Γf(z) dz = 0.

More generally, if C is any rectifiable closed contour or cycle in the closed set [D], then

∫

Cf(z) dz = 0.

We will not prove this theorem in the generality presented. Reasonably simple proofs areavailable in the following separate subcases:

• [D] is a starlike set with respect to one interior point (see Definition 2.2). Equivalently Γ isa starlike Jordan curve.

• f(z) is a function of bounded variation on Γ, which means that f(z(t)) is a function ofbounded variation of the parameter t that parametrises Γ.

• Γ is piecewise smooth.

We will give the proof in the first case, based on the proof in Einar Hille: Analytic Function

Theory Volume I . Starlike sets are necessarily simply connected and their interiors are Jordanregions. Starlike Jordan curves include all convex Jordan curves, such as circles, ellipses, triangles,rectangles and convex polygons. These are, in fact, starlike with respect to every interior point.Many non-starlike closed regions can be partitioned into a finite number of starlike or convex sets.

The second case is a case where standard theorems on the continuity of a Riemann-Stieltjesintegral with respect to a parameter can be called upon. The third case is proved in K. Kodaira:Complex Analysis (Cambridge 1984 and 2007).

Proof of Theorem 3.10 for starlike closed sets. The proof relies on a simple continuityargument. It does not require f(z) to obey any additional restrictions on Γ. Because [D] = D ∪ Γis a compact set, the continuity of f(z) on [D] implies its uniform continuity on [D].

Let Γ be parametrised by

z = z0(t), 0 ≤ t ≤ 1, z0(0) = z0(1),

and let L be its finite arc length. Under the original hypotheses of the theorem, z0(t) is continuousand of bounded variation on [0, 1] and f(z0(t)) is continuous on [0, 1]. These properties are more

40

than are needed to guarantee the existence of the Riemann-Stieltjes integral on the right of

∫

Γf(z) dz =

∫ 1

0f(z0(t)) dz0(t).

Similarly, the hypotheses of the theorem guarantee the existence of

∫

Cf(z) dz,

where C is any rectifiable cycle on [D]. The method of proof of Cauchy’s theorem for the path Γworks equally well for C, and so we will just show the proof for Γ.

The plan of the proof is to exhibit a one-parameter family of positively oriented rectifiable Jordancurves Γλ, 0 < λ < 1, that are interior to D and that tend continuously and uniformly to Γ0 := Γas λ→ 0+. Subject to suitable additional hypotheses, we wish to ensure that

∫

Γλ

f(z) dz →∫

Γf(z) dz,

as λ → 0+. But, for positive λ, the integral on the left is identically zero by the earlier version ofCauchy’s theorem, because Γλ is interior to D. That would complete the proof that

∫

Γf(z) dz = 0.

Suppose that [D] is starlike with respect to an interior point z1. In that case, the function of tworeal variables,

z(t, λ) := zλ(t) := λz1 + (1 − λ)z0(t), 0 ≤ λ ≤ 1, 0 ≤ t ≤ 1,

maps the indicated closed unit square surjectively onto the set [D]. With t fixed and λ varying, thegraph of this function is a closed line segment from z1 to the point z0(t) on Γ. According to thedefinition of starlike, every point on this line segment except the endpoint z0(t), where λ = 0, isinterior to D. For fixed λ, 0 ≤ λ < 1, let Γλ be the contour parametrised by z = z(t, λ), t varyingfrom 0 to 1. This is a positively oriented rectifiable Jordan curve of arc length (1 − λ)L. Whenλ = 0, it is Γ itself. (Optionally, we can define Γ1 to be the single point z1.) The contour integralof f(z) on Γλ is

∫

Γλ

f(z) dz = (1 − λ)

∫ 1

0f(λz1 + (1 − λ)z0(t)) dz0(t).

In the Riemann-Stieltjes integral on the right, the integrand is uniformly continuous on the closureof the unit square, and the integrator z0(t) is of bounded variation on [0, 1]. These are standardsufficient conditions to guarantee the continuity of the integral with respect to λ on [0, 1]. Sincethe integral vanishes by Cauchy’s theorem when 0 < λ < 1 (and trivially when λ = 1), we haveproved that the integral must also vanish when λ = 0. This, together with a similar argument forcycles C, completes the proof of Theorem 3.10 for starlike closed sets.

Many of the theorems in these lecture notes could be modified to take account of this extensionof Cauchy’s theorem to closed regions. In some cases, theorems that refer to functions that areanalytic on open regions and continuous on their closures do not actually need this stronger form

41

of Cauchy’s theorem. Examples are the Weierstrass limit theorem Theorem 4.14 and the maximummodulus theorem Theorem 7.1.

Another type of generalisation of Cauchy’s theorem involves weakening the definition of ana-lyticity.

Theorem 3.11. (H. Looman 1923, improved by D. Menchoff 1935 and G. Tolstoff 1942). Supposethat f(z) is locally bounded on a region D. Let z = x+ iy and f(z) = u(x, y) + iv(x, y). If u(x, y)and v(x, y) have partial derivatives existing pointwise in the corresponding set in the real xy-plane(also to be called D), and if they satisfy the Cauchy-Riemann equations everywhere in D, thenf(z) is analytic in D.

A discussion of this theorem and its extensions and a proof for the case of continuous f(z) canbe found in the paper: J. D. Gray and S. A. Morris, American Mathematical Monthly 85:246–256(1978).

The proof of Theorem 3.11 uses ideas from real analysis and measure theory. A function islocally bounded on a region if it is bounded on every compact subset. All continuous functions arelocally bounded. Note that some sort of restriction on f(z) such as continuity or local boundednessis necessary. For example, the function f(z) = exp(−1/z4), with f(0) = 0, satisfies the Cauchy-Riemann equations everywhere, but it is unbounded near z = 0 and therefore not analytic there.

The most important extensions of Cauchy’s theorem for practical purposes are those that takeaccount of exceptional points and multiple connectivity of the domain of analyticity. All complexfunctions described as “analytic” are understood to be locally analytic, which means analytic onsome region. Except in the case of entire functions, locally analytic functions have one or moresingularities. Their full domains of analyticity may not be known and may not be needed tobe known. Domains or partial domains of analytic functions can be either simply or multiplyconnected, or run onto Riemann surfaces. Furthermore, we need to be able to recognise whenapparent exceptional points are actually removable singularities.

Consider the function f(z) = 1/z, which has a pole at z = 0. Let C be the positively orientedcircle z = Reiθ, 0 ≤ θ ≤ 2π. Then

∫

C

1

zdz =

∫ 2π

0

1

ReiθRieiθ dθ =

∫ 2π

0i dθ = 2πi.

This nonzero result tells us that we cannot expect Cauchy’s theorem to always apply to contoursthat enclose a singularity or hole in the domain. On the other hand, the integrals of 1/zn, n =2, 3, . . . around the same circle are all zero. The latter functions have a single-valued primitive inthe punctured domain, and so the fundamental theorem of calculus (Theorem 2.12) guarantees thattheir integrals around all closed contours will vanish (provided they do not run across the pole).But 1/z is different. It has the multi-valued primitive log z, whose global domain is a Riemannsurface. Hence, if we wish to apply the fundamental theorem of calculus, we must unwrap the circleinto a Jordan arc on the Riemann surface for log z with endpoints R and Re2πi (or, equivalently,Re−iπ and Reiπ), now treated as distinct.

An important point to notice is that the integral just described is independent of the radius ofthe circle. More generally,

∫

C

1

zdz = 2πi,

42

for every closed rectifiable contour C that winds around the origin exactly once in the positivesense. This will be a consequence of the next theorem, but the reader should verify the result bydirect integration for some particular curves (for example, rectangle, circle with different centre).

Theorem 3.12. Deformation of contours. Suppose that f(z) is analytic in a region D (notnecessarily simply connected) and that C1 and C2 are two positively oriented rectifiable Jordancurves in D, with C1 properly inside C2. Suppose also that the annular region between C1 and C2

is also contained in D. Then∫

C1

f(z) dz =

∫

C2

f(z) dz.

Proof. If f(z) is analytic throughout the inside of C1, there is nothing to prove because bothintegrals vanish by Cauchy’s theorem. So the theorem is interesting when C1 encloses one or moreholes in the domain of analyticity. If the shapes of C1 and C2 are not too irregular, we see that aclosed oriented curve can be formed if we reverse the orientation of C1 and cut the annular regionby a suitable arc that is traversed both ways. Then the combined curve is the oriented boundary ofa simply connected domain of analyticity of f(z). The original version of Cauchy’s theorem can beapplied if we make two cuts. Then the cycle C2−C1 is the sum of two positively oriented rectifiableJordan curves C3 and C4 that are each inside a simply connected domain of analyticity of f(z).Thus, with the common integrand f(z),

∫

C2

−∫

C1

=

∫

C2−C1

=

∫

C3

+

∫

C4

= 0.

To complete the proof, we just need to show that two cuts of the required type are always possible.

Let z0 be any point in the inside of C1. It does not matter if z0 is in D or in a hole of D. Draw ahorizontal line through z0. It must intersect both the Jordan curves C1 and C2 at least twice. Ifit intersects them both exactly twice, then we are done. However, it may intersect them infinitelymany times, or coincide with straight pieces. Because both the line and the curves are closed sets,there will be a last intersection point A with C1 as we follow the line from z0 to the right. Aswe continue further to the right, there will be a first subsequent intersection point B with C2. (Itdoes not matter if C2 was intersected before A.) The line segment AB, being fully contained in theannular region between C1 and C2 except for its endpoints is a cut of the required type. A secondcut can be formed by similarly following the line to the left from z0. Conclude the proof as in theprevious paragraph.

The theorem just proved required one contour to be properly inside the other. The theoremcan be extended to the case where the contours C1 and C2 intersect each other, but care is neededto ensure that one can be deformed into the other while remaining in D. The simplest way toensure this is to have available a third contour C3 such that C1 and C3 satisfy the conditions ofthe theorem, and similarly for C2 and C3.

Three straightforward corollaries follow immediately.

Corollary 3.13. Exterior Cauchy’s theorem. If f(z) is analytic outside and on a positivelyoriented Jordan curve C and if zf(z) → 0 uniformly as z → ∞, then

∫

Cf(z) dz = 0.

43

Proof. In Theorem 3.12, take C1 = C and let C2 be a large positively oriented circle |z| = R.By the ML formula, the modulus of the integral over C2 is bounded above by 2πRM(R), whereM(R) is the maximum of |f(z)| on the circle. The hypothesis that zf(z) → 0 uniformly as z → ∞implies that RM(R) → 0 as R→ ∞. Hence, the integral over C2 can be made arbitrarily small bychoosing R sufficiently large. This proves the corollary.

Corollary 3.14. Suppose that f(z) is analytic in a region D that has a finite number of puncturesor holes, each of which is a closed bounded simply connected set (possibly just an isolated point orJordan arc). Let C be a positively oriented rectifiable Jordan curve in D that winds around N ofthe holes. Then

∫

Cf(z) dz =

N∑

k=1

∫

γk

f(z) dz,

where γk is a rectifiable Jordan curve that winds around the kth hole and only the kth hole oncein the positive sense.

Definition 3.15. Let C be any rectifiable closed contour or cycle that does not run across thepoint z0. Then

n(C, z0) :=1

2πi

∫

C

1

z − z0dz

is an integer called the winding number of C about z0.

The complement of C is a nonconnected open set which is a disjoint union of a finite or countablenumber of components or regions, one of which is unbounded. The winding number is constant asz0 runs over each separate component, and is zero if z0 is in the unbounded component. Hence aclosed rectifiable contour in a multiply connected region has a well-defined winding number abouteach of the holes in the region.

Definition 3.16. A closed contour or cycle C in a (multiply connected) region D is homologous

to zero in D when the winding number of C about every point not in D is zero. In particular,every closed contour or cycle in a simply connected region is homologous to zero.

Example. The Pochhammer contour is a double figure-eight contour that winds around two holesin a region in such a way that each hole is wound around twice, once each in the positive andnegative directions. The Pochhammer contour is homologous to zero. It plays a role in the analytictheory of the hypergeometric function.

Corollary 3.17. Let f(z), D and the γk be as in Corollary 3.14. Let C be any rectifiable closedcontour or cycle in D and let nk be the winding number of C about the kth hole, k = 1, 2, . . . , N .Then

∫

Cf(z) dz =

N∑

k=1

nk

∫

γk

f(z) dz,

The following is a special case of this result:

Theorem 3.18. Cauchy’s theorem for homologous to zero paths. If f(z) is analytic in a(possibly multiply connected) region D and if C is any rectifiable closed contour or cycle in D thatis homologous to zero in D, then

∫

Cf(z) dz = 0.

44

An important extension of Cauchy’s theorem is to the case where f(z) has a finite number ofwhat appear to be mild types of singularities inside C. This will lead to an extremely useful theorem(Theorem 4.11 in the next chapter), which we call the removable singularities theorem. Themain result of the latter theorem is that exceptional points of the sort postulated in the presenttheorem cannot exist. But before we can prove that, we need to show that such exceptional points,if they do occur, do not affect Cauchy’s theorem.

Theorem 3.19. Suppose that f(z) is analytic in a simply connected region D except possibly fora finite number of exceptional points z1, z2, . . . , zn at which

limz→zk

(z − zk)f(z) = 0, k = 1, 2, . . . , n.

(These include points at which f(z) is continuous but possibly not differentiable.) If C is a rectifiableclosed contour or cycle in D that does not cross any of the exceptional points, then

∫

Cf(z) dz = 0.

Remark. The function (z − z1)−1/2 has the right growth rate near z1 but fails to qualify because

it is not single-valued near z1.

Proof. In view of Corollaries 3.14 and 3.17, there is no loss of generality in assuming thatthere is only one exceptional point z1 ∈ D and that C is a positively oriented rectifiable Jordancurve enclosing z1. Let γ be a small positively oriented circle with centre z1. Then, according toTheorem 3.12,

∫

Cf(z) dz =

∫

γf(z) dz.

Given ǫ > 0, choose δ > 0 such that

|z − z1| |f(z)| < ǫ

7

whenever 0 < |z − z1| < δ. Let the radius of the circle γ be ρ < δ. Then the triangle inequality forcontour integrals yields

∣

∣

∣

∣

∫

Cf(z) dz

∣

∣

∣

∣

=

∣

∣

∣

∣

∫

γf(z) dz

∣

∣

∣

∣

≤∫

γ|f(z)| |dz|

≤∫

γ

ǫ

7|z − z1||dz|

=ǫ

7ρ

∫

γ1 |dz|

=ǫ

7ρ2πρ

=2π

7ǫ

< ǫ.

Since ǫ is arbitrary, the contour integral vanishes.

45

4 Cauchy’s integral formula

Cauchy’s integral formula is of tremendous importance in complex analysis. It shows that ananalytic function is uniquely determined inside a Jordan curve by its values on the curve itself.This result will lead directly to the proof that every analytic function is differentiable to all orders(this chapter) and is the sum of its own convergent Taylor series (next chapter).

Theorem 4.1. Cauchy’s integral formula. Suppose that f(z) is analytic in a simply connectedregion D and that C is a positively oriented rectifiable Jordan curve in D. Then

1

2πi

∫

C

f(z)

z − z0dz =

f(z0), z0 inside C

0, z0 outside C.

Proof. First, if z0 is outside C, the integrand of the Cauchy integral is analytic inside and on C.In that case, the integral vanishes by Cauchy’s theorem. If z0 is inside C, write

1

2πi

∫

C

f(z)

z − z0dz =

1

2πi

∫

C

f(z) − f(z0)

z − z0dz +

f(z0)

2πi

∫

C

1

z − z0dz.

Now the function,

g(z) :=f(z) − f(z0)

z − z0,

has a removable discontinuity at z = z0. It becomes continuous at z0 if we define g(z0) = f ′(z0).(We do not assume that g(z) is analytic at z0, which is a consequence of a later theorem.) Thus z0qualifies as an exceptional point in the sense of Theorem 3.19 and so

1

2πi

∫

C

f(z) − f(z0)

z − z0dz = 0.

Then,

1

2πi

∫

C

f(z)

z − z0dz =

f(z0)

2πi

∫

C

1

z − z0dz =

f(z0)

2πi

∫

circle centre z0

1

z − z0dz = f(z0).

This completes the proof.

The strong form of Cauchy’s theorem for closed regions implies a corresponding strong form ofCauchy’s integral formula:

Theorem 4.2. Cauchy’s integral formula for closed Jordan regions. Let D ⊂ C be asimply connected open region which forms the inside of a rectifiable Jordan curve Γ. Let f(z) bea function which is analytic in D and continuous on [D] := D ∪ Γ. Then

1

2πi

∫

Γ

f(z)

z − z0dz =

f(z0), z0 inside Γ

0, z0 outside Γ.

Corollaries 3.14 and 3.17 imply the following generalisation of Cauchy’s integral formula:

Corollary 4.3. Suppose that f(z) is analytic in a simply connected region D and that C is arectifiable closed contour or cycle in D. Let z0 be any point in the complement of C. Then

1

2πi

∫

C

f(z)

z − z0dz = n(C, z0)f(z0),

46

where n(C, z0) is the winding number of C about z0.

Proof. As above, if C is any cycle in D and z0 6∈ C, Theorem 3.19 yields

1

2πi

∫

C

f(z)

z − z0dz =

1

2πi

∫

C

f(z) − f(z0)

z − z0dz +

f(z0)

2πi

∫

C

1

z − z0dz =

f(z0)

2πi

∫

C

1

z − z0dz.

The right-hand side equals n(C, z0)f(z0) according to the definition of winding number. Theoriginal statement of Cauchy’s integral formula is the special case n(C, z0) = 1 when z0 is inside Cand n(C, z0) = 0 when z0 is outside C.

Now the Cauchy integral,∫

C

f(ζ)

ζ − zdζ,

defines an analytic function of z on each of the separate disjoint components of the complementof C, regardless of whether or not the function f(ζ) in the integrand is analytic on C. It is sufficientthat f(ζ) be integrable on C. In fact the Cauchy integral can be differentiated to all orders underthe integral sign. This is a direct consequence of standard theorems in the theory of Riemann orRiemann-Stieltjes integrals. Because of the great importance of this result in the case when theCauchy integral is the one appearing in Cauchy’s integral formula, we will derive the first derivativefrom first principles. Higher derivatives will be similar.

Theorem 4.4. Analyticity of Cauchy integrals. Let C be a rectifiable closed contour or cycleand let D be any one of the regions whose disjoint union is the complement of C. Suppose thatf(z) is defined on C (and possibly only on C) and is integrable on C. Then the function,

g(z) :=1

2πi

∫

C

f(ζ)

ζ − zdζ,

is differentiable to all orders in D and is therefore analytic in D. Its first, second and nth derivativesare

g′(z) =1

2πi

∫

C

f(ζ)

(ζ − z)2dζ,

g′′(z) =1

πi

∫

C

f(ζ)

(ζ − z)3dζ,

g(n)(z) =n!

2πi

∫

C

f(ζ)

(ζ − z)n+1dζ, n ≥ 0.

Proof. An integrable function (as we have defined it) is bounded, so let |f(z)| ≤ M on C. LetL be the arc length of the contour or cycle C. Since D is an open set in the complex z-plane, anypoint z0 ∈ D is inside an open disc that is a subset of D, and we can bound the disc away from theboundary of D. Choose ρ > 0 such that the closed discs,

D1 :=

z : |z − z0| ≤ ρ

, D2 :=

z : |z − z0| ≤ 2ρ

,

are both contained in D. Let h be any complex number such that |h| < ρ. Then z0 + h is in theinterior of D1. Define

g1(z) :=1

2πi

∫

C

f(ζ)

(ζ − z)2dζ.

47

Now |ζ − z| > ρ for every ζ ∈ C and for every z ∈ D1 and |ζ − z0| > 2ρ. Then

g(z0) =1

2πi

∫

C

f(ζ)

ζ − z0dζ,

g(z0 + h) =1

2πi

∫

C

f(ζ)

ζ − z0 − hdζ,

g(z0 + h) − g(z0)

h=

1

2πi

∫

C

f(ζ)

(ζ − z0)(ζ − z0 − h)dζ,

g(z0 + h) − g(z0)

h− g1(z0) =

h

2πi

∫

C

f(ζ)

(ζ − z0)2(ζ − z0 − h)dζ.

According to the ML formula,

∣

∣

∣

∣

g(z0 + h) − g(z0)

h− g1(z0)

∣

∣

∣

∣

≤ |h|2π

ML

4ρ3.

As h → 0 on any path, the right-hand side tends to zero. Hence, g(z) is differentiable at z = z0and g′(z0) = g1(z0). Since z0 is any point in the region D, we have proved that g(z) is analyticin D and

g′(z) = g1(z) =1

2πi

∫

C

f(ζ)

(ζ − z)2dζ.

Higher derivatives can be calculated from first principles in the same fashion, and the general resultcan be proved by induction on n. The results agree with differentiation under the integral sign,and we find

g(n)(z) =n!

2πi

∫

C

f(ζ)

(ζ − z)n+1dζ, n ≥ 0.

In Theorem 4.4, the Cauchy integral was analytic in z merely because the kernel 1/(ζ − z) wasanalytic in z for each fixed ζ outside the z-domain. This is a special case of the following moregeneral result:

Theorem 4.5. Contour integrals with a parameter. Suppose that Φ(ζ, z) is a continuousfunction of two complex variables whose domain is C ×D, where C is a rectifiable contour in thecomplex ζ-plane and D is a region in the complex z-plane. Suppose also that Φ(ζ, z) is analyticin D for each fixed ζ ∈ C. Define

g(z) :=

∫

CΦ(ζ, z) dζ.

Then g(z) is analytic in D and

g′(z) =

∫

C

∂Φ

∂zdζ.

Proof. As in the case of Theorem 4.4, differentiation under the integral sign is permitted under thestated hypotheses by standard theorems for Riemann or Riemann-Stieltjes integrals. By replacingΦ(ζ, z) by its Cauchy integral,

Φ(ζ, z) =1

2πi

∫

γ

Φ(ζ, z)

z − zdz,

48

where γ is the positively oriented circle |z − z| = ρ, one can deduce Theorem 4.5 directly fromTheorem 4.4. We leave this as an exercise. The theorem can be extended to unbounded integrandsor infinite paths whenever the improper integral defining g(z) is uniformly convergent with respectto z on all compact subsets of D.

The following theorem gives the boundary values on C of the analytic functions defined by aCauchy integral. We need to impose some restrictions on the path C and the function f(z) definedalong C. This result is of importance in the solution of Riemann-Hilbert and Wiener-Hopf

problems, where two analytic functions in different domains are to be found such that they satisfygiven junction conditions across a common boundary. Such problems occur in the Fourier analysisof certain linear partial differential equations and the inverse scattering transform for soliton-typenonlinear partial differential equations. They also occur in the analysis of a class of special functionsknown as Painleve transcendents.

Theorem 4.6. Plemelj formulae. Suppose that C is a piecewise smooth positively orientedJordan curve in the complex plane and that f(z) is any continuous function defined on C (andpossibly only on C) that satisfies a uniform Holder condition,

∣

∣f(z1) − f(z2)∣

∣ ≤M∣

∣z1 − z2∣

∣

α, 0 < α ≤ 1,

for every two points z1 and z2 on C. Define the Cauchy integrals,

1

2πi

∫

C

f(ζ)

ζ − zdζ =

f+(z), z inside C

f−(z), z outside C

f0(z), z on C,

where, in the third case, the integral is a Cauchy principal value integral. Then f+(z) and f−(z)are analytic in their specified domains and, for any point z0 ∈ C where C is smooth, the one-sidedlimits are given by the Plemelj formulae:

f+(z0) := limz→z0

f+(z) = f0(z0) + 12f(z0), limit from inside C,

f−(z0) := limz→z0

f−(z) = f0(z0) − 12f(z0), limit from outside C.

At a corner point where the interior angle is θ, 0 ≤ θ ≤ 2π, the respective limits are

f+(z0) = f0(z0) +2π − θ

2πf(z0), f−(z0) = f0(z0) −

θ

2πf(z0).

Either way, f+(z0) − f−(z0) = f(z0).

More generally, we can let f(z) be merely integrable along C and be continuous and Holderjust at the point z0 ∈ C. Then f+(z) and f−(z) are analytic functions defined by the above Cauchyintegrals on their respective domains, and f0(z) is defined at z0. The one-sided limits in the Plemeljformulae apply to approaches to z0 from within closed sectors that are properly inside or outside C,as appropriate, except at their vertex z0.

Proof. The proof of the more general form of the theorem plus extensions to handle discontinuitieson the contour, more complicated contours, matrix function integrands, the solution of Riemann-Hilbert and Wiener-Hopf problems, and applications to the electrified disc problem and other

49

half-range boundary-value problems, are topics in the Fourth-Year course, Integral Transforms

and Asymptotic Methods.

When applied to Cauchy’s integral formula, Theorem 4.4 proves that every analytic function isdifferentiable to all orders. The consequences for Taylor series expansions of analytic functions willbe explored in the next chapter.

Theorem 4.7. Higher derivatives of an analytic function. Suppose that f(z) is analyticin a simply connected region D and that C is a positively oriented rectifiable Jordan curve in D.Then, for all z inside C, differentiation under the integral sign is permitted and we get

f(z) =1

2πi

∫

C

f(ζ)

ζ − zdζ,

f ′(z) =1

2πi

∫

C

f(ζ)

(ζ − z)2dζ,

f ′′(z) =1

πi

∫

C

f(ζ)

(ζ − z)3dζ,

f (n)(z) =n!

2πi

∫

C

f(ζ)

(ζ − z)n+1dζ, n ≥ 0.

It follows that an analytic function is differentiable to all orders throughout its global domain ofanalyticity, which may be a Riemann surface, regardless of connectivity.

Proof. The main statement is a direct corollary of Theorems 4.1 and 4.4. The final statement istrivially true because any point in the global domain of analyticity is inside a local simply connectedopen neighbourhood contained within the larger domain, and the contour C can be drawn in thelocal neighbourhood.

Now that we know that an analytic function is expressible by Cauchy’s integral formula andis differentiable to all orders, some of the lemmas and theorems in the first three chapters can betightened up.

Corollary 4.8. A function f(z) has an antiderivative in a simply connected region D if and onlyif f(z) is analytic in D.

Proof. The “if” part has already been proved. So suppose that F (z) exists such that F ′(z) = f(z).Because F (z) is analytic, we know that F ′′(z) also exists. Hence f ′(z) exists and so f(z) is analyticin D.

The next theorem is an important converse of Cauchy’s theorem. Note that minimal hypotheseson f(z) and the paths of integration make this a stronger theorem.

Theorem 4.9. Morera’s theorem (Morera 1886). Suppose thatD is a bounded simply connectedregion in C and f(z) is continuous in D. Suppose also that

∫

∆f(z) dz = 0

for every closed contour ∆ belonging to any one of the following three classes:

50

(a) all piecewise smooth closed curves in D,(b) all triangles interior to D, or(c) all rectangles interior to D with sides parallel to the axes.

Then f(z) is analytic in D.

Proof. Statements (b) and (c) are strong versions of Morera’s theorem. They both imply (a),which is the version that is often stated in textbooks.

In the triangle case, the geometric lemmas 3.4 and 3.5 in the proof of Cauchy’s theorem imply that∫

Γf(z) dz = 0,

for every closed polygonal line Γ in D. (If we begin with this hypothesis, then D does not needto be simply connected.) Since D is open and bounded, any two points in D can be joined by apolygonal line in D, where the polygonal line is understood to have a finite number of line segments.Hence,

F (z) :=

∫

Γ[z0,z]f(ζ) dζ

is the same for all polygonal lines Γ[z0, z] in D joining z0 to z, and so is a well-defined function of zin D. Under these conditions, part (b) of Theorem 2.13 guarantees that F (z) is an antiderivative(or primitive) of f(z). Thus F (z) is analytic in D and F ′(z) = f(z). Theorem 4.7 now guaranteesthat f(z) itself is analytic in D.

In the rectangle case, an oriented closed polygonal line consisting only of horizontal and vertical linesegments is a union of a finite number of oriented rectangles and digons. (This can be proved in asimilar fashion to lemmas 3.4 and 3.5.) Hence, the integral of f(z) around such a closed polygonalline is zero, and

F (z) :=

∫

Γ[z0,z]f(ζ) dζ

is a well-defined function of z, where Γ[z0, z] is any polygonal line running from z0 to z consistingof horizontal and vertical line segments only. Part (c) of Theorem 2.13 guarantees that F (z) is anantiderivative of f(z), and so F (z) and f(z) are both analytic, as before.

Corollary 4.10. Suppose that f(z) is analytic in a region D except possibly for a rectifiableJordan arc C ⊂ D, or any set of exceptional points through which such an arc can be drawn, andsuppose that f(z) is continuous across C. (The case where C separates D into a finite or countablenumber of disjoint pieces is included.) Then f(z) is analytic across C.

Proof. Let ∆ be any triangle interior to D. If ∆ does not cross or touch C, then∫

∆f(z) dz = 0

by Cauchy’s theorem. If ∆ touches C on one side in any manner, then the integral vanishesaccording to Theorem 3.10. If ∆ crosses C, then C may leave the interior of ∆ connected or mayseparate it into two or more disjoint regions. Either way, ∆ can be written as the contour sum of oneor more oriented Jordan curves built from parts of ∆ and C, with the latter parts being traversedin both directions in the sum. The integrals around these Jordan curves vanish by Theorem 3.10.Hence,

∫

∆f(z) dz = 0,

51

regardless of the manner in which ∆ intersects C. Now Morera’s theorem guarantees that f(z) isanalytic everywhere in D, including all the points of C.

Theorem 4.11. Removable singularities theorem. Suppose that f(z) is analytic in a region Dexcept possibly at the point z1 ∈ D. At z1, suppose that

limz→z1

(z − z1)f(z) = 0.

Then a value of f(z1) can be assigned so that f(z) becomes analytic at z1. In particular, a point ofcontinuity z0 of f(z) in a deleted neighbourhoodN ′(z0, δ) of analyticity is also a point of analyticity.

Proof. The point z1 qualifies as an exceptional point in the sense of Theorem 3.19. Cauchy’stheorem is not affected by such exceptional points. Neither is Cauchy’s integral formula, for theinside of the contour C in Theorem 4.1 can be subdivided, by adding a suitable arc that is traversedboth ways, into a region containing z0 and another containing all the exceptional points. Theo-rem 4.1 applies to the former region and Theorem 3.19 to the latter. Choose a simply connectedneighbourhood N(z1, δ) ⊂ D and draw a positively oriented circle C ⊂ N(z1, δ), with z1 inside C.Define

f1(z) :=1

2πi

∫

C

f(ζ)

ζ − zdζ,

for z inside C. Then f1(z) = f(z) inside C except possibly at z1. But f1(z) is analytic inside Cby Theorem 4.4. So by defining f(z1) = f1(z1), f(z) becomes analytic at z1 as well. Thus z1is a removable singularity under the stated hypotheses, and exceptional points in the sense ofTheorem 3.19 cannot exist.

Remark. In view of the previous two results, removable singularities are generally ignored incomplex analysis. The default is to consider such singularities as having been automatically removedwithout explanation and treated as ordinary points of analyticity.

For example, the origin z = 0 is an ordinary point of analyticity of the functions,

f(z) =sin z

z, g(z) = cos

√z , h(z) = (1 + z)1/z .

Their values at the origin are

f(0) = 1, g(0) = 1, h(0) = e.

Just to emphasize that the origin is not just a point of continuity, we give their first, second andtenth derivatives:

f ′(0) = 0, g′(0) = − 1

2, h′(0) = − 1

2e,

f ′′(0) = − 1

3, g′′(0) =

1

12, h′′(0) =

11

12e,

f (10)(0) = − 1

11, g(10)(0) =

1

670442572800, h(10)(0) =

145644286955

101376e.

These three examples are examined a little more closely in Chapter 5.

The first example f(z) is a special case of the following lemma:

52

Lemma 4.12. If f(z) is analytic at z = a, so also is the function,

g(z) :=f(z) − f(a)

z − a.

Then g(a) = f ′(a), g′(a) = f ′′(a)/2 and, more generally, g(n)(a) = f (n+1)(a)/(n + 1) for n =0, 1, 2, 3, . . . .

Proof. The analyticity of f(z) at z = a implies that the following limit exists:

f ′(a) = limz→a

f(z) − f(a)

z − a.

This limit states that g(z) has a removable discontinuity at z = a, and the discontinuity is removedby defining g(a) = f ′(a). Now a becomes a point of analyticity of g(z) and we are able to takederivatives. Differentiate the identity,

(z − a)g(z) = f(z) − f(a),

k times, k ≥ 1, giving(z − a)g(k)(z) + kg(k−1)(z) = f (k)(z).

Put z = a to get kg(k−1)(a) = f (k)(a). The cases k = 1, 2 and n+ 1 read

g(a) = f ′(a), g′(a) = f ′′(a)/2, g(n)(a) = f (n+1)(a)/(n + 1).

Cauchy’s integral formula provides stronger theorems for the validity of the interchange oflimit processes than some of their counterparts in the real domain. An important example is thecondition for the validity of term-by-term differentiation of a sequence or series of functions. In thedomain of one real variable, the standard theorem is:

Theorem 4.13. Suppose un(x), n = 1, 2, 3, . . . , is a sequence of functions of a real variable, eachdifferentiable on an open interval (a, b) ∈ R, such that the series

∑∞k=1 uk(x) converges at at least

one point of that interval. Suppose also that the derived series∑∞

k=1 u′k(x) converges to a function

g(x) on (a, b), uniformly on all closed subintervals. Then the original series converges on (a, b) toa function f(x), uniformly on all closed subintervals, and f ′(x) = g(x). (A corresponding theoremholds with series replaced by sequence.)

This theorem is proved in standard textbooks on advanced calculus. See, for example, Tom M.Apostol: Mathematical Analysis, Addison-Wesley (1957).

The need to examine the uniform convergence of the derived series rather than the originalseries is illustrated by the sequence fn(x) = sin(nx)/

√n , n = 1, 2, 3, . . . . Here fn(x) converges

uniformly to the zero function f(x) = 0 as n → ∞ on R. Each term of the sequence, as well asthe limit function, is differentiable. However, the derived sequence f ′n(x) =

√n cos(nx) diverges

everywhere.

In the corresponding theorem in the complex domain, there is one very noticeable difference. Itis not necessary to take derivatives before testing for uniform convergence. Uniform convergence ofthe original sequence or series is enough to guarantee term-by-term differentiability to all orders.Indeed, we can get a strong result by just assuming uniform convergence on the boundary of asimply-connected region. (Some textbooks do not give this theorem a name. You would need to look

53

for a section on uniformly convergent series of analytic functions. Some name it after Weierstrass,because Weierstrass proved an equivalent theorem on double series. The name “Weierstrass limittheorem” is occasionally used and has the advantage of clearly identifying the theorem.)

Theorem 4.14. Weierstrass limit theorem. Suppose un(z), n = 1, 2, 3, . . . , is a sequence offunctions, each analytic in a simply connected open region D which forms the inside of a rectifiableJordan curve Γ, and suppose that the un(z) are all continuous onto Γ from within. Suppose alsothat the series

∑∞j=1 uj(z) converges uniformly on Γ. Then

• the series∑∞

j=1 uj(z) converges uniformly to a function f(z) on [D] = D ∪ Γ;

• f(z) is analytic in D and continuous on [D];

• the series f(z) =∑∞

j=1 uj(z) is differentiable term by term to all orders in D;

• all the derived series f (n)(z) =∑∞

j=1 u(n)j (z) are uniformly convergent on all closed sets

interior to D.

(A corresponding theorem holds with series replaced by sequence.)

Proof. The first statement is best handled with the maximum modulus theorem, Theorem 7.1,which is stated and proved in Chapter 7. It is an elementary result that does not require the complexintegral calculus. The main part of that theorem implies that a function g(z) which is analyticin D and continuous on [D] attains its maximum modulus on the boundary Γ. Now the uniformconvergence of the series

∑∞j=1 uj(z) on Γ implies that, given arbitrary ǫ > 0, there exists N

depending only on ǫ such that∣

∣

∣

∣

M∑

j=N

uj(ζ)

∣

∣

∣

∣

< ǫ,

for every M > N and every ζ ∈ Γ. But for all z ∈ [D], the maximum modulus theorem impliesthat

∣

∣

∣

∣

M∑

j=N

uj(z)

∣

∣

∣

∣

≤∣

∣

∣

∣

M∑

j=N

uj(ζ)

∣

∣

∣

∣

max on Γ

< ǫ,

which proves that the series∑∞

j=1 uj(z) is uniformly convergent on [D]. An elementary consequenceof the uniform convergence and the continuity of the terms of the series is that the sum,

f(z) =

∞∑

j=1

uj(z),

is continuous on [D].

Now give Γ a positive orientation. We wish to prove that f(z) is analytic in D and the statementsabout term-by-term differentiation of the series. According to Cauchy’s integral formula for closedJordan regions,

uj(z) =1

2πi

∫

Γ

uj(ζ)

ζ − zdζ,

for all z ∈ D. Before we proceed, we point out that we can prove the remaining parts of theWeierstrass limit theorem without calling upon this strong form of Cauchy’s integral formula, which

54

we have only proved for starlike closed regions. The role of Γ can be replaced by any positivelyoriented contour Γ1 that is interior to D and arbitrarily close to Γ. Then the ordinary form ofCauchy’s integral formula will suffice. For example, analyticity inside Γ1 implies analyticity insideΓ because any point inside Γ, together with a neighbourhood of that point, can be enclosed withina suitable Γ1. Any compact subset of the interior of Γ is also interior to a suitable Γ1. The uniformconvergence of the series on [D] implies the uniform convergence on Γ1. Knowing that this optionis available, we will continue to work with Γ.

The uniform convergence of the series on Γ implies that the series together with any continuousmultiplier can be integrated term-by-term along Γ. The multipliers we have in mind are 1/(ζ − z)and powers thereof, where ζ is a moving point along Γ and z is a fixed point in D, so that |ζ − z|has a positive lower bound. It follows that

f(z) =

∞∑

j=1

uj(z) =

∞∑

j=1

1

2πi

∫

Γ

uj(ζ)

ζ − zdζ =

1

2πi

∫

Γ

( ∞∑

j=1

uj(ζ)

ζ − z

)

dζ =1

2πi

∫

Γ

f(ζ)

ζ − zdζ,

for all z ∈ D. Thus f(z) is defined by a Cauchy integral, which immediately implies its analyticityin D according to Theorem 4.4. Furthermore, f(z), which is already known to be continuous on [D],satisfies Cauchy’s integral formula in D with the integral taken around Γ.

We can differentiate the Cauchy integral under the integral sign. The nth derivative of f(z), forz ∈ D, is given by,

f (n)(z) =n!

2πi

∫

Γ

f(ζ)

(ζ − z)n+1dζ.

Similarly, the derivatives of the uj(z) are given by

u(n)j (z) =

n!

2πi

∫

Γ

uj(ζ)

(ζ − z)n+1dζ.

Again, the uniform convergence of the original series on Γ allows the following interchange ofsummation and integration:

∞∑

j=1

u(n)j (z) =

∞∑

j=1

n!

2πi

∫

Γ

uj(ζ)

(ζ − z)n+1dζ

=n!

2πi

∫

Γ

( ∞∑

j=1

uj(ζ)

(ζ − z)n+1

)

dζ

=n!

2πi

∫

Γ

f(ζ)

(ζ − z)n+1dζ

= f (n)(z).

Hence, the series f(z) =∑∞

j=1 uj(z) is differentiable term by term to all orders in D. In otherwords,

f (n)(z) =

∞∑

j=1

u(n)j (z),

n = 0, 1, 2, 3, . . . , for all z ∈ D.

55

We now turn to the proof of the final statement of the theorem, that each derived series is uniformlyconvergent on all closed subsets of D. Let E be any closed subset of D. Since D is open, E isproperly interior to D. By a Heine-Borel argument previously used in the proof of Cauchy’stheorem, we can construct a positive lower bound δ to the distance between points of Γ and pointsof E. (Alternatively, we can fit a contour Γ1 properly between E and Γ and let δ be a positivelower bound to the distance between points of Γ1 and points of E.)

Recall that the uniform convergence of the original series on Γ implies that, given arbitrary ǫ > 0,there exists N depending only on ǫ such that

∣

∣

∣

∣

M∑

j=N

uj(ζ)

∣

∣

∣

∣

< ǫ,

for every M > N and every ζ ∈ Γ. Suppose z ∈ E. The uniform convergence of the original serieson Γ, the triangle inequality for contour integrals, and the lower bound δ for |ζ − z|, ζ ∈ Γ, z ∈ E,allows us to construct the following bound:

∣

∣

∣

∣

M∑

j=N

u(n)j (z)

∣

∣

∣

∣

=

∣

∣

∣

∣

M∑

j=N

n!

2πi

∫

Γ

uj(ζ)

(ζ − z)n+1dζ

∣

∣

∣

∣

=

∣

∣

∣

∣

n!

2πi

∫

Γ

( M∑

j=N

uj(ζ)

(ζ − z)n+1

)

dζ

∣

∣

∣

∣

≤ n!

2π

∫

Γ

∣

∣

∣

∣

M∑

j=N

uj(ζ)

(ζ − z)n+1

∣

∣

∣

∣

|dζ|

≤ n!

2π

∫

Γ

ǫ

δn+1|dζ|

=n!L ǫ

2π δn+1,

where L is the finite arc length of the rectifiable curve Γ. This bound holds for all M > N and allz ∈ E. It proves that the nth derived series,

f (n)(z) =

∞∑

j=1

u(n)j (z),

is uniformly convergent on E, and therefore on all closed subsets of D. (Note that the case n = 0is included here, but is superseded by the stronger statement in the first part of the theorem.)

There are alternative statements of the Weierstrass limit theorem that can be derived as corol-laries of the theorem just proved. Some are less general. For example, Γ may be interior to asimply connected region of analyticity, and the uniform convergence may be assumed to hold onall compact subsets, not just a single Jordan curve. Some appear to be more general. For example,the region of analyticity may be multiply connected, but the gain in generality is minimal sincethe compact sets of uniform convergence include those that can be enclosed by Jordan curves onsimply connected regions of analyticity. The version appearing in Ahlfors’ Complex Analysis, whichis intended to cover a situation that commonly occurs, states:

56

Theorem 4.15. Let Ωn, n = 1, 2, 3, . . . , be a nested sequence of open regions,

Ω1 ⊆ Ω2 ⊆ Ω3 ⊆ . . . ,

and let Ω be the union of all the Ωn. Suppose that fn(z) is analytic in the region Ωn and that thesequence fn(z) converges uniformly to a limit function f(z) on all compact subsets of Ω. Thenf(z) is analytic in Ω. Moreover, f ′n(z) converges uniformly to f ′(z) on every compact subset of Ω.

Proof. Let E be any compact subset of Ω. For some n0, E will be properly interior to Ωn0, and

therefore interior to Ωn for every n ≥ n0. A positively oriented piecewise smooth Jordan curve Ccan be drawn to enclose E and be properly interior to Ωn0

. Then the fn(z), n ≥ n0, convergeuniformly to f(z) on C. This C satisfies the conditions on Γ in the main theorem. (Of course,the theorem can be written for series rather than sequences, and the last statement in the theoremextends to higher-order derivatives.)

Let C∗ denote the extended complex plane C ∪ ∞. The extended complex plane ishomeomorphic to a 2-sphere, called the Riemann sphere, by the following construction. Let theordinary complex plane C be represented in the usual way by the real xy-plane, and let the latterbe the xy-plane of a three-dimensional Euclidean xyu-space. The 2-sphere with centre (0, 0, 1

2)and radius 1

2 touches the xy-plane at the origin, which we will call the south pole of the sphere.The north pole is at the point (0, 0, 1) on the u-axis. Draw a line segment from the north poleto any point in the xy-plane. It intersects the sphere at exactly one point other than the northpole. Conversely, every point of the sphere except the north pole occurs on exactly one such linesegment. So we have a one-to-one map, called a stereographic projection, from either C orR2 to a sphere with one point (the north pole) missing. Now since large complex numbers in C

map to points in the neighbourhood of the north pole, we can close up the missing point by lettingthe north pole correspond to a single point at infinity in the extended complex plane C∗. Thusthe stereographic projection extends to a one-to-one map from the extended complex plane to thecomplete Riemann sphere.

Using this map, we can impose the Riemann sphere topology on the extended complex plane bydefining open and closed sets in C∗ to be the inverse images of open and closed sets on the Riemannsphere. In this topology, all closed sets are compact, including C∗ itself. Bounded closed sets inthe R2 topology (that is, the standard topology for C that we have been using) remain closed setsin the Riemann sphere topology. Unbounded closed sets need to have the point at infinity attachedto qualify as closed in the Riemann sphere topology.

A neighbourhood of infinity can be understood in both the R2 topology and the Riemannsphere topology. In the former, it is the exterior of any Jordan curve in C, which is often taken tobe a circle. The same set would be regarded as a deleted neighbourhood of infinity in the Riemannsphere topology. By appending the point at infinity, it then becomes an ordinary neighbourhoodof infinity.

We can put a metric on C∗ by defining distances between points or arc lengths of curves in C∗ tobe their corresponding values determined from the first fundamental form on the Riemann sphere.Then the metric distance d(z1, z2) between two points z1, z2 ∈ C∗, which always takes a value inthe interval [0, π/2], is given by

d(z1, z2) = tan−1

∣

∣

∣

∣

z1 − z21 + z1z2

∣

∣

∣

∣

, d(z1,∞) = cot−1 |z1|,

57

where the bar denotes the complex conjugate. Similarly, the metric arc length of a locally rectifiablecontour C ⊂ C∗ is given by the line integral,

d(C) =

∫

C

|dz|1 + |z|2 .

We could just as easily use a sphere of unit radius with centre at the origin. The north pole is againat (0, 0, 1), but the south pole moves to (0, 0,−1). The effect on the metric is simply to doubledistances and arc lengths.

Partial classification of singularities. The removable singularities theorem makes the followingdefinitions well defined. The definitions of “pole” and “isolated essential singularity” given here arenot the standard ones, which will be given in the next chapter, but they are equivalent to them.First, recall that a singularity of a locally analytic function is a point to which the function cannotbe analytically continued.

Notes.

• When referring to analyticity of a function in the neighbourhood of a singularity, it will alwaysbe understood that the function is single-valued near the singularity unless we explicitly statethat the function is or may be multi-valued or that the singularity is or may be a branchpoint.

• All removable singularities will be treated as ordinary points of analyticity on account of theremovable singularities theorem, Theorem 4.11.

Definition 4.16. Analyticity at infinity. Suppose f(z) is analytic on an unbounded set anddefine g(z) = f(1/z). The point ∞ ∈ C∗ is a point of analyticity of f(z) if z = 0 is a point ofanalyticity of g(z). Similarly, z = ∞ is a singularity of a particular type of f(z) if g(z) has asingularity of that same type at z = 0.

If f(z) is analytic in a neighbourhood of ∞ and f(z)/z → 0 as z → ∞, then the removablesingularities theorem implies that f(z) is analytic at ∞.

Examples. The functions, g(z) = (az + b)/(cz + d), c 6= 0, h(z) = z sin(1/z), and k(z) = (1 +1/z)z are analytic at z = ∞. Their values at infinity are, respectively, g(∞) = a/c, h(∞) = 1, andk(∞) = e.

Definition 4.17. An analytic function f(z) has a zero at z = z0 if f(z0) = 0. It is a zero of

order k if f(z)/(z − z0)k is analytic and nonzero at z0, where k must be a positive integer according

to the removable singularities theorem. A zero of order one is a simple zero, a zero of order twois a double zero, and so on. (Of course, a zero is a point of analyticity, not a singularity.)

Examples. The functions sin z and tan z have simple zeros and sin2 z and tan2 z have double zerosat z = nπ, where n ∈ Z. Jn(z) has a zero of order n at z = 0 if n is a positive integer, and Jν(z)has infinitely many simple zeros on the real axis, and only on the real axis, when ν ≥ −1. Thefundamental theorem of algebra states that a polynomial of degree n with real or complexcoefficients has exactly n zeros, if multiplicities or orders are counted.

Definition 4.18. A function f(z) which has a singularity at z0 and is analytic in a deletedneighbourhood N ′(z0, δ) of z0 has a pole at z0, or, more specifically, a pole of order k, if

58

(z − z0)kf(z) is analytic and nonzero at z0, where k must be a positive integer according to the

removable singularities theorem. A pole of order one is a simple pole, a pole of order two is adouble pole, and so on. (See also Definition 6.5 below.)

Examples. The function cot z has simple poles and cot2 z has double poles at z = nπ, where n isan integer. Γ(z) has simple poles at zero and the negative integers. ζ(z) has a simple pole at z = 1.(sin z)/z2 has a simple pole and (cos z)/z2 has a double pole at z = 0. A polynomial of positivedegree n has a pole of order n at ∞.

If f(z) and g(z) are analytic in a region D, f(z)/g(z) has a pole at each zero of g(z) in D,provided it is not cancelled by a zero in f(z) of the same or higher order.

Definition 4.19. A function is meromorphic (sometimes called globally meromorphic foremphasis) if it is analytic in the whole complex plane C except for poles. (An entire function isthe special case having no poles.) A function is meromorphic in a region D ⊂ C∗ if it is analyticin D except for poles. If the number of poles is infinite in the latter case, all their limit points mustoccur on the boundary of D in the Riemann sphere topology.

A theorem of Weierstrass states that every meromorphic function is a ratio of entire functions.Examples are rational functions as ratios of polynomials, tan z = sin z/ cos z, and elliptic functionsas ratios of Jacobi theta functions or Weierstrass sigma functions. Similarly, functions meromorphicin a region are ratios of functions analytic in that region. Such constructions are never unique.

Entire functions admit infinite product formulae over their zeros, which carry over to meromor-phic functions written as ratios of entire functions. Examples of entire functions are

sinπz = πz

∞∏

k=1

(

1 − z2

k2

)

,1

Γ(z)= z eγz

∞∏

k=1

e−z/k

(

1 +z

k

)

,

where γ is Euler’s constant (value 0.577215664901532...). These products converge uniformly oncompact sets that avoid the zeros of the functions on the left. From these two examples, onecan read off the identity, Γ(z) Γ(−z) = −π/(z sinπz), which rearranges to the more familiar form,Γ(z) Γ(1 − z) = π/ sinπz.

Meromorphic functions also admit infinite partial fraction expansions, as in the example,

cot z =1

z+

∞∑

k=1

2z

z2 − k2π2.

This series converges uniformly on all compact sets that avoid the poles of cot z (multiples of π).

Definition 4.20. A singularity z0 of f(z) is an isolated essential singularity of f(z) if f(z)is analytic in a deleted neighbourhood N ′(z0, δ) of z0 and z0 is not a pole of f(z). (See alsoDefinition 6.6 below.)

Examples. The functions e1/z , sin(1/z), cos(z−1/2) and J0(1/z3) each have an isolated essential

singularity at z = 0, which is a limit point of zeros in the last three examples. All entire functionsexcept polynomials have an isolated essential singularity at ∞, examples being ez, sin z, J0(z),1/Γ(z) and (z − 1)ζ(z). (A few theorems on isolated essential singlarities are given in Chapter 6.)

All other singularities of single-valued locally analytic functions are also called essential sin-

gularities, or nonisolated essential singularities for emphasis. These come in a variety of

59

shapes. The simplest cases are limit points of poles. They can also be limit points of other essentialsingularities or of poles and essential singularities together. They can combine to form singularcurves called natural barriers or natural boundaries. They can also form nondense closed setslike one- or two-dimensional Cantor sets.

Examples. The functions tan(1/z) and tan(tan(1/z)) have a nonisolated essential singularity atz = 0. They are limit points of poles, and, in the second example, also limit points of other essentialsingularities. All meromorphic functions except rational functions have an essential singularity atinfinity, which may be either isolated (as in the case of (cos z)/z and ζ(z)) or nonisolated (as inthe case of tan z and 1/J0(z)). The circle of convergence |z| = 1 of the power series

∑∞n=0 z

n! isa natural barrier of the function defined within. Natural barriers often occur in the solutions ofanalytic differential equations. They can be smooth or fractal in character.

Definition 4.21. A point in the neighbourhood of which a locally analytic function f(z) analyt-ically continues to a multi-valued function is a branch point of f(z). Branch points join two ormore sheets of a Riemann surface. An algebraic branch point joins a finite number of sheets.A quadratic branch point joins two sheets, a cubic branch point joins three sheets, and so on. Alogarithmic branch point joins an infinite number of sheets.

If f(z) has an algebraic branch point of degree n at z0, the n-fold neighbourhood of z0 maps toa single-valued neighbourhood of the origin ζ = 0 under the map z = z0 + ζn.

The functions√z and cot

√z both have quadratic branch points at z = 0 and z = ∞. In

the latter case, the singularity at infinity is a quadratically branched essential singularity. Thefunction z−1/3 has cubic branch points at zero and infinity. The functions log z and z

√2 both

have logarithmic branch points at zero and infinity. (The multi-valued functions log z and zα are

discussed briefly in Chapter 1.) The Bessel functions Yn(z) and H(1)n (z), where n is an integer, also

have logarithmic branch points at zero and infinity.

The branching in these examples can be understood by putting z in polar form,

z = reiθ = r(cos θ + i sin θ),

and letting θ (which is arg z) run outside its principal range (−π, π].

In the cases with quadratic branching at zero and infinity, the functions are single-valued ona two-sheeted Riemann surface parametrised by r > 0 and 0 ≤ θ < 4π or −π < θ ≤ 3π (or anyhalf-open 4π-interval). In the cases with cubic branching at zero and infinity, the Riemann surfacehas three sheets and θ runs over a half-open 6π-interval. In the cases with logarithmic branchingat zero and infinity, the Riemann surface has infinitely many sheets and θ runs from −∞ to ∞.

The inverse sine function sin−1 z has quadratic branch points at z = ±1 and a logarithmicbranch point at z = ∞. The quadratic branch points are infinite in number because the fullRiemann surface contains infinitely many copies of the points 1 and −1. Each such point joins twoRiemann sheets. The logarithmic branch points at infinity are just two in number and each join acountably infinite number of Riemann sheets.

Branch points are not necessarily isolated. They can be limit points of unbranched singularities(poles and essential singularities) or of other branch points. Also natural barriers can be branched.For example, the function, f(z) = log z

∑∞n=0 z

−n!, is analytic outside the unit circle, which is anatural barrier across which f(z) cannot be analytically continued inwards, but the factor log z

60

requires the exterior of the circle to be unwound into a Riemann surface with infinitely manysheets.

Generally, complex analysis textbooks do not attempt a detailed classification of nonisolatedsingularities or logarithmic branch points. However, in the analytic theory of differential equations,one needs to be able to distinguish singularities somewhat more carefully than in complex analysisgenerally. There are simple tests for singular behaviour in differential equations that do not requirea knowledge of exact solutions. A lot can be said about the nature of the integrability of adifferential equation based on the results of such tests. This is one of the main themes of Painleve

analysis, named after the French mathematician and former French prime minister Paul Painleve(1863–1933).

A differential equation in the complex domain is said to have the Painleve property if itsgeneral solution is free of branch points whose position depends on one or more integration con-stants. A stronger statement of the Painleve property also rules out essential singularities thatdepend on integration constants, regardless of whether or not they are branched, so that only polesare accepted. Either way, differential equations possessing the Painleve property have been foundto be integrable, either by getting themselves explicitly solved or linearised, or by being mappedinto another space where Riemann-Hilbert theory comes into play. The functions appearing in thesolutions of Painleve-type differential equations include elliptic functions, hyperelliptic functions,abelian functions, automorphic functions and, of course, Painleve transcendents. The first two (ofsix) classical Painleve transcendents satisfy the respective differential equations,

d2w

dz2= 6w2 + z,

d2w

dz2= 2w3 + zw + α, α ∈ C.

A construction analogous to the Riemann sphere can be used to extend the sheets of a Riemannsurface. A separate point at infinity can occur on each sheet, as in z(z2 + 1)−1/2, or a point atinfinity can be a branch point that joins two or more sheets, as in z1/3 and

√z3 + 1 . In the case

of sin−1 z, two distinct branch points at infinity occur, as already mentioned.

Algebraic functions are roots w = f(z) of polynomial equations P (w, z) = 0, P being a poly-nomial in both variables. The extended Riemann surfaces of algebraic functions are compact,and are all homeomorphic to a 2-sphere with a finite number of handles (or a torus with a finitenumber of holes, including the 2-sphere itself). The number of handles or holes is a topologicalinvariant called the genus of the algebraic function or of the Riemann surface. There is an equiva-lence relation on algebraic functions that preserves genus. These are the invertible birational maps(z,w) 7→ (z1, w1) modulo P (z,w). They induce a polynomial relation P1(w1, z1) = 0 on z1 and w1

and so define a new algebraic function w1 = f1(z1). Even though degrees and numbers of Riemannsheets may change, the new algebraic function has the same genus as the original.

The global domains including poles and branch points of the algebraic functions,

w = zp/q, w =

(

az + b

cz + d

)p/q

, w =√

az2 + bz + c ,

where p/q is a rational number, are each homeomorphic to an ordinary 2-sphere, which has genuszero. These functions have two branch points except in the special cases where they are rational.

61

Under a suitable birational transformation, they can each be mapped to w1 = 0. Every rationalfunction is also an algebraic function of genus zero.

The zero set of an arbitrary cubic polynomial in two variables,

a1w3 + a2w

2z + a3wz2 + a4z

3 + a5w2 + a6wz + a7z

2 + a8w + a9z + a10 = 0,

is known as an elliptic curve. The algebraic function w = f(z) solving this cubic has genus one,unless it degenerates to genus zero. Under a birational map, the non-degenerate case is equivalentto the standard Weierstrass elliptic curve,

w2 = 4z3 − g2z − g3, g 32 6= 27g 2

3 .

The quartic w2 = b0z4 + b1z

3 + b2z2 + b3z + b4 is also birationally equivalent to the Weierstrass

elliptic curve, and has genus one, provided the right-hand side has distinct roots. (Otherwise it hasgenus zero.) Inequivalent elliptic curves are indexed by a single complex paramater.

If Pk(z) denotes a polynomial in one variable of degree k with no square factors, then thealgebraic functions,

w =√

P2n+1(z) , w =√

P2n+2(z) ,

each have 2n + 2 quadratic branch points, one being at infinity in the former example. TheirRiemann surfaces have two sheets and, with appropriate points at infinity attached, they arehomeomorphic to a sphere with n handles (or, equivalently, a torus with n holes), and so havegenus n. The corresponding algebraic curves are known as hyperelliptic curves. The even-degreecase is equivalent to the odd-degree case. Elliptic curves are the subcases with n = 1. All algebraiccurves of genuses 0, 1 and 2 are birationally equivalent to hyperelliptic curves of the same genus.This is not true for genus three and higher. For example, the algebraic function P6(z)1/3 hasgenus four and is not equivalent to a hyperelliptic curve, in general. But if P6(z) has two simpleroots and two double roots, the genus drops to two, and it becomes birationally equivalent to thesquare root of a quintic polynomial.

62

5 The Cauchy-Taylor theorem and analytic continuation

This chapter will explore further consequences of Cauchy’s integral formula, which evaluates ananalytic function inside a simple closed curve in terms of its boundary values.

We have already seen that Cauchy’s integral can be differentiated under the integral sign. Hence,an analytic function is differentiable to all orders. This, of course, is enough to form the formalTaylor series of f(z) about any point z0 where f(z) is analytic:

∞∑

k=0

f (k)(z0)

k!(z − z0)

k.

Note that we did not write “f(z) =” on the left-hand side, which requires proof. The Cauchy-

Taylor theorem, which is the version of Taylor’s theorem applicable to complex analytic functions,will be seen to be a best possible result. We will show that the Taylor series does converge to f(z)in its disc of convergence, and the radius of convergence R is the distance from z0 to the nearestsingularity of f(z). More generally, a Laurent series is a power series in both positive and negativepowers of z − z0 which converges in an annulus, and we will prove a similar best-possible result forthese series as well.

In the real domain, a Taylor series does not always behave in an expected fashion. First, weshould not be surprised that the radius of convergence of the logarithm series,

log(1 + x) = x− x2

2+x3

3− x4

4+ . . . , −1 < x ≤ 1,

is R = 1 because the left-hand side is unbounded as x→ −1 from the right. However, consider

1

1 + x2= 1 − x2 + x4 − x6 + . . . , −1 < x < 1,

tan−1 x = x− x3

3+x5

5− x7

7+ . . . , −1 ≤ x ≤ 1.

In these examples, which also have R = 1, the left-hand sides are real-analytic for all real x anddo not undergo any glitch at x = 1 or x = −1. Here, the limitation on the radius of convergencebecomes transparent if we just change x to z and examine their complex analytic extensions. Thefunction 1/(1 + z2) has simple poles at z = ±i and tan−1 z has logarithmic branch points at z = ±i.These singular points occur at a distance R = 1 from the origin.

A more pathological example is the following real function:

f(x) = e−1/x2

, f(0) = 0.

This function is differentiable to all orders on the real axis, including at x = 0. However, allderivatives at x = 0 are zero, which follows from the limit,

limx→0

∣

∣x−ne−1/x2∣

∣ = limy→∞

yn/2

ey=

limy→∞

y

e2y/n

n/2

=

limy→∞

1

(2/n)e2y/n

n/2

= 0,

n > 0, where we applied l’Hopital’s rule. Thus f(x) has the Taylor series,

0 + 0x+ 0x2 + 0x3 + . . . ,

63

about x = 0, which is certainly convergent with R = ∞, but it converges to the wrong function.Again, replacing x by z reveals the problem. The function,

f(z) = e−1/z2

, f(0) = 0,

is unbounded near z = 0 if approached from either of the sectors π/4 < arg z < 3π/4 or −3π/4 <arg z < −π/4. The point z = 0 is an isolated essential singularity of f(z), and f(z) does not havea derivative at z = 0, much less a Taylor series about z = 0.

A second type of pathology in the real domain is a formal Taylor series with a zero radius ofconvergence. The real function,

g(x) :=

∫ ∞

0

e−t

1 + x2tdt,

is differentiable to all orders everywhere. At the origin, g(0) = 1, g(2n)(0) = (−1)nn!(2n)! andg(2n+1)(0) = 0. Its Taylor series about x = 0 is

1 − x2 + 2!x4 − 3!x6 + 4!x8 − . . . ,

which is divergent for all x except, trivially, x = 0. In other words, it has a zero radius of conver-gence.

To see what sort of singularity is lurking at z = 0 in the complex domain, change x to z andanalytically extend g(z) off the positive real axis. The substitution t = 1/u2 − 1/z2 gives

g(z) =2

z2e1/z2

∫ z

0

1

ue−1/u2

du,

where the path of integration approaches the origin inside the sector −π/4 < arg u < π/4. Withthe assistance of the formula,

γ := limn→∞

φ(n) − log n

=

∫ 1

0

1 − e−t

tdt −

∫ ∞

1

e−t

tdt,

for Euler’s constant γ, where φ(n) :=∑n

k=1 (1/k), we can rearrange our integral for g(z) to

g(z) = g1(z)(

2 log z − γ)

+ g2(z),

where

g1(z) =1

z2e1/z2

=

∞∑

k=0

1

k! z2k+2,

g2(z) =2

z2e1/z2

∫ ∞

z

1 − e−1/u2

udu =

∞∑

k=1

φ(k)

k! z2k+2.

The power series in 1/z (Laurent series) for g1(z) and g2(z) are everywhere convergent exceptat z = 0, where these functions have an isolated essential singularity according to the standarddefinition below. On the other hand, the factor log z shows that z = 0 is also a logarithmic branchpoint for g(z). In fact, we can give a simple connection formula,

g(

zemπi)

= g(z) +2mπi

z2e1/z2

,

64

valid for m ∈ Z and all nonzero z on the Riemann surface.

Notice that g(xeπi) and g(xe−πi) do not coincide with the real-variable definition of g(−x) forreal positive x. In particular, g(z) is not an even function, whereas g(x) was. Notice also thatg(z) → ∞ as z → 0 through real values on every sheet of the Riemann surface except the positivereal axis of the initial sheet. Thus z = 0 is both a branch point and an essential singularity of g(z)and we now have a reasonably good picture of it. This example illustrates what might happen inthe complex domain when a Taylor series in the real domain has a zero radius of convergence.

Lemma 5.1. Taylor’s theorem with remainder. Suppose that f(z) is analytic in a simplyconnected region D. Let C be a simple closed positively oriented rectifiable contour in D. Then,for every z and z0 inside C,

f(z) = f(z0) + f ′(z0)(z − z0) +f ′′(z0)

2!(z − z0)

2 + . . . +f (n)(z0)

n!(z − z0)

n +Rn(z, z0),

where

Rn(z, z0) =(z − z0)

n+1

2πi

∫

C

f(ζ)

(ζ − z0)n+1(ζ − z)dζ.

Proof. In the identity,

a+ ar + ar2 + ar3 + . . . + arn =a(1 − rn+1)

1 − r, r 6= 1,

put a = 1/(ζ − z0) and r = (z − z0)/(ζ − z0). Then

1

ζ − z=

n∑

k=0

(z − z0)k

(ζ − z0)k+1+

(z − z0)n+1

(ζ − z0)n+1(ζ − z).

Multiply both sides by f(ζ)/(2πi) and integrate with respect to ζ along C. Then Cauchy’s integralformula and its derivatives give the desired result.

Notation. We have already introduced the notation D(z0, r) for the open disc |z − z0| < r,[D](z0, r) for its closure, and C(z0, r) for the circular boundary |z − z0| = r. Let the latter bepositively oriented if an orientation is required. Use the simpler notations D(r), [D](r) and C(r),respectively, if the centre is understood from the context. Then D(∞) can be an alternative namefor the whole complex plane C. Let M(r) denote the maximum of |f(z)| on C(r). If the centrez0 needs to be specified, the notation can be expanded to M(z0, r). If the function f needs to bespecified, the notation can be expanded to M(f, z0, r).

Theorem 5.2. The Cauchy-Taylor theorem. Suppose that f(z) is analytic at z0 and the disc,

D(R) =

z : |z − z0| < R

, 0 < R ≤ ∞,

is the largest open disc, centre z0, on which f(z) is analytic. Then the Taylor series,

f(z) =

∞∑

k=0

f (k)(z0)

k!(z − z0)

k,

converges absolutely to f(z) on D(R) and uniformly on all compact subsets.

65

Proof. Choose two fixed radii R1 and R2, 0 < R2 < R1 < R. Then Taylor’s theorem with re-mainder holds for the contour C = C(R1). The remainder term is

Rn(z, z0) =(z − z0)

n+1

2πi

∫

C(R1)

f(ζ)

(ζ − z0)n+1(ζ − z)dζ.

Let z be inside or on C(R2) and write r := |z − z0| ≤ R2. Then |ζ − z| ≥ R1 − r for ζ ∈ C(R1).The ML formula gives

|Rn(z, z0)| ≤rn+1

2π

M(R1)

(R1)n+1(R1 − r)2πR1

=Mr

R1 − r

(

r

R1

)n

≤ MR2

R1 −R2

(

R2

R1

)n

.

Since R2/R1 < 1,lim

n→∞Rn(z, z0) = 0,

uniformly for all z inside or on C(R2). Hence the Taylor series of f(z) about z0 converges tof(z) uniformly for all z inside or on C(R2). Since R2 can be arbitrarily close to R, the Taylorseries converges to f(z) everywhere in the open disc D(R) and uniformly on all compact subsets.Convergence on D(R) is absolute according to a standard elementary theorem on power series.

Remarks.

• If the function f(z) is known on a domain of analyticity that may or may not be maximal,then the theorem applies to the largest open disc D(R) with given centre z0 that can be fitin the known domain of analyticity. If the actual radius of convergence of the Taylor seriesexceeds R, then the sum of the Taylor series provides an analytic continuation of f(z) to alarger domain.

• All the standard power series theorems apply to a Taylor series. For example, an explicitformula for the radius of convergence is

1

R= lim sup

n→∞|an|1/n, an :=

f (n)(z0)

n!.

If R is finite and the Taylor series converges absolutely at one point of the circle of convergenceC(R), then it converges absolutely and uniformly on the closed disc [D](R). More generaltheorems governing the convergence behaviour on or near C(R) are Abel’s theorem and theTauber-Littlewood theorem (Theorems 1.9 and 1.10 in Chapter 1).

• If the radius of convergence R is finite, the domain of analyticity of f(z) may be larger thanthe disc D(R). Alternatively, if an analytic function f(z) is defined by a power series, it maybe possible to analytically continue f(z) outside of its disc of convergence. Either way, theCauchy-Taylor theorem guarantees at least one singularity of f(z) on the circle of convergenceC(R) according to the following lemma.

66

Lemma 5.3. Let the power series f(z) =∑∞

k=0 ak(z − z0)k have a finite positive radius of con-

vergence R, and suppose that f(z) can be analytically continued beyond its disc of convergenceD(R). Then f(z) has at least one singularity on the circle of convergence C(R). In other words,the radius of convergence of a power series is the distance from the centre to the nearest singularityof its sum function f(z).

Proof. Suppose, on the contrary, that f(z) is analytic everywhere on C(R). Then every pointof C(R) is an interior point of a neighbourhood of analyticity. Because C(R) is compact, a finitenumber of these neighbourhoods is able to cover C(R). Then there exists R1 > R such that D(R1)is a subset of the union of D(R) and the latter neighbourhoods. Then f(z) is analytic in D(R1) andthe Cauchy-Taylor theorem guarantees that the Taylor series for f(z) about z0 converges in D(R1).But the Taylor series is identical to the original power series, and we have found a larger disc ofconvergence. This contradicts the hypothesis that R was the radius of convergence. Hence, f(z)has at least one singularity on C(R). Since f(z) has no singularities inside C(R), the value of R isthe distance from z0 to the nearest singularity of f(z). (Of course, there may be several singularitiesat the same distance.)

Examples. Lemma 5.3, together with the removable singularities theorem, makes it possible tocalculate the radius of convergence of a power series when the coefficients are unknown. In the caseof compound expressions built out of elementary functions, it is generally easy to calculate a smallnumber of terms of their Taylor series about a point by algebraic operations on power series (forexample, long division, Cauchy product), but one is much less likely to be able to find an explicitformula for the nth term. Nevertheless, finding the radius of convergence R is relatively easy.

• In the Taylor series,

1

2 cos z − 1= 1 + z2 + 11

12z4 + 301

360z6 + 15371

20160z8 + . . . , |z| < π/3,

there is no simple formula for the nth term. Nevertheless, we can read off its radius ofconvergence R = π/3 because the left-hand side has simple poles at z = ±π/3, and these arethe nearest to the origin.

• The Bernoulli numbers Bn, n = 0, 1, 2, . . . , are defined by the generating function,

z

ez − 1=

∞∑

n=0

Bn

n!zn, |z| < 2π.

The left-hand side has simple poles at the nonzero integer multiples of 2πi. Hence, the radiusof convergence of the power series on the right-hand side is R = 2π. Adding z/2 to theleft-hand side converts it into the even function 1

2z coth 12z. Hence B1 = −1/2 and all other

Bernoulli numbers of odd index are zero. We deduce the following Taylor series:

z cot z =

∞∑

n=0

(−1)n22nB2n

(2n)!z2n, |z| < π,

tan z = cot z − 2 cot 2z =

∞∑

n=1

(−1)n−1 22n(22n − 1)B2n

(2n)!z2n−1, |z| < π/2,

z cosec z = z cot(12z) + z cot z =

∞∑

n=0

(−1)n2(22n−1 + 1)B2n

(2n)!z2n, |z| < π.

67

Since all the odd-order derivatives of tan z are positive at the origin, the even-index Bernoullinumbers alternate in sign, beginning at B2. The first dozen nonzero Bernoulli numbers are

B0 = 1, B1 = − 12 , B2 = 1

6 , B4 = − 130 , B6 = 1

42 ,

B8 = − 130 , B10 = 5

66 , B12 = − 6912730 , B14 = 7

6 ,

B16 = − 3617510 , B18 = 43867

798 , B20 = − 174611330 .

Notice that they stay small until about B14 and then grow rapidly with alternating signs.The actual rate of growth is exhibited by Euler’s formula for the Riemann zeta function atthe positive even integers:

ζ(2n) =(−1)n−1 22n−1B2n

(2n)!π2n.

Stirling’s asymptotic formula,

n! ∼ nne−n√

2πn ,

as n→ ∞ and ζ(n) ∼ 1 together give the asymptotic formula,

B2n ∼ 4(−1)n−1( n

πe

)2n √πn ,

as n → ∞. These results show directly that the power series for z/(ez − 1) has a radius ofconvergence R = 2π. There is a simple algorithm for calculating the exact fractional part ofa large Bernoulli number, known as the Clausen-von Staudt theorem. Euler’s formula andasymptotic methods allow B2n to be approximated closely enough to determine the integerpart. Together, these two methods allow large Bernoulli numbers to be calculated exactlywithout relying on recurrence relations.

• The Taylor series,

πz cot πz =∞∑

k=0

(−1)k(2π)2kB2k

(2k)!z2k = 1 − 2

∞∑

k=1

ζ(2k)z2k, |z| < 1,

has a radius of convergence R = 1 because the nearest poles to the origin are z = ±1. So alsodoes the geometric series,

z2

1 − z2=

∞∑

k=1

z2k, |z| < 1.

Adding twice the second series to the first cancels the poles at z = ±1. Hence the series,

πz cot πz +2z2

1 − z2= 1 − 2

∞∑

k=1

ζ(2k) − 1

z2k, |z| < 2,

has a radius of convergence R = 2. Similarly, the power series on the right-hand side of

πz cot πz +

m∑

j=1

2z2

j2 − z2= 1 − 2

∞∑

k=1

ζ(2k) −m

∑

j=1

1

j2k

z2k, |z| < m+ 1,

68

has a radius of convergence R = m + 1. Of course, if we follow this line of reasoning to itslogical conclusion, we arrive at the partial fraction expansion,

π cot πz =1

z+

∞∑

k=1

2z

z2 − k2, z ∈ C − Z.

This series converges uniformly on compact sets that avoid the integers. Integrating bothsides and exponentiating gives the infinite product expansion,

sinπz = πz

∞∏

k=1

(

1 − z2

k2

)

, z ∈ C.

• Just after the removable singularities theorem (Theorem 4.11), we gave the first, second andtenth derivatives at the origin of the three functions,

f(z) =sin z

z, g(z) = cos

√z , h(z) = (1 + z)1/z ,

each of which has a removable singularity at the origin. In the first case,

f(z) =

∞∑

k=0

(−1)k

(2k + 1)!z2k, z ∈ C.

Hence f (2n)(0) = (−1)n/(2n + 1) and f (2n+1)(0) = 0. In the second case,

g(z) =∞∑

k=0

(−1)k

(2k)!zk, z ∈ C.

Hence g(n)(0) = (−1)nn!/(2n)!. In the third case,

h(z) = (1 + z)1/z

= exp

1

zlog(1 + z)

= exp

1 − z

∞∑

k=0

(−1)k

k + 2zk

(|z| ≤ 1, z 6= −1)

= e

∞∑

m=0

(−1)m

m!zm

∞∑

k=0

(−1)k

k + 2zk

m

(|z| < 1)

= e

1 − 12z + 11

24z2 − 7

16z3 + 2447

5760z4 − 959

2304z5

+ 238043580608z

6 − 67223165888z

7 + 5594401991393459200 z

8

− 123377159309657600 z

9 + 2912885739173574645760 z

10 − . . .

.

Multiplying the coefficient of z10 by 10! gives h(10)(0) = 145644286955e/101376.

69

Cauchy’s integral formula provides a bound on the derivatives of an analytic function, and henceon the coefficients of a convergent power series.

Theorem 5.4. Suppose that f(z) is analytic in an open disc D(z0, ρ) with centre z0 and continuousonto its boundary C(z0, ρ) from within. Then, for z ∈ D(z0, ρ),

|f(z)| ≤ M(ρ),

|f (n)(z)| ≤ n! ρM(ρ)

(ρ+ r)(ρ− r)n, n ≥ 1,

where r := |z − z0| < ρ and M(ρ) is the maximum of |f(z)| on C(z0, ρ).

Proof. Apply the ML formula to Cauchy’s integral formula. Let ζ = z0 + ρeiθ on C(z0, ρ) andz = z0 + reiφ, 0 ≤ r < ρ. Then,

|ζ − z| = |ρeiθ − reiφ| =√

ρ2 + r2 − 2ρr cos(θ − φ) ≥ ρ− r.

Then,

|f(z)| =

∣

∣

∣

∣

1

2πi

∫

C(z0,ρ)

f(ζ)

ζ − zdζ

∣

∣

∣

∣

≤ 1

2π

M(ρ)

ρ− r2πρ =

ρM(ρ)

ρ− r.

The same argument works equally well on f(z)k, where k is a positive integer, and we get

|f(z)|k ≤ ρM(ρ)k

ρ− r, |f(z)| ≤

(

ρ

ρ− r

)1/k

M(ρ),

for z ∈ D(z0, ρ). Taking the limit k → ∞ yields |f(z)| ≤M(ρ), as required. (Later, the maximummodulus theorem will imply strict inequality when f(z) is nonconstant.)

For n ≥ 1, apply the triangle inequality to the derivatives of Cauchy’s integral formula:

∣

∣f (n)(z)∣

∣ =

∣

∣

∣

∣

n!

2πi

∫

C(z0,ρ)

f(ζ)

(ζ − z)n+1dζ

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

n!

2πi

∫ 2π

0

f(z0 + ρeiθ) ρi eiθ

ρeiθ − reiφn+1 dθ

∣

∣

∣

∣

∣

≤ n!

2π

∫ 2π

0

Mρ∣

∣ρeiθ − reiφ∣

∣

n+1 dθ

=n! ρM

2π

∫ 2π

0

dθ∣

∣ρeiθ − r∣

∣

n+1 (translation in θ, period 2π)

=n! ρM

π

∫ π

0

dθ

ρ2 + r2 − 2ρr cos θ(n+1)/2

≤ n! ρM

π(ρ− r)n−1

∫ π

0

dθ

ρ2 + r2 − 2ρr cos θ

=n! ρM

π(ρ− r)n−1

π

ρ2 − r2

=n! ρM

(ρ+ r)(ρ− r)n,

70

where M denotes M(ρ). (Note that ρ can be the maximum radius R provided f(z) is continuousonto C(z0, R) from within, which is compatible with some types of singularities on C(z0, R).)

Corollary 5.5. Cauchy’s inequality. Suppose that f(z) is analytic in an open disc D(z0, ρ)and continuous onto its boundary C(z0, ρ) from within. Then

∣

∣f (n)(z0)∣

∣ ≤ n!M(ρ)

ρn, n ≥ 0,

where M(ρ) is the maximum of |f(z)| on C(z0, ρ). If a power series∑∞

k=0 ak(z− z0)k converges on

a closed disc [D](z0, ρ) to f(z), then

|an| ≤M(ρ)

ρn, n ≥ 0.

Proof. The first statement is just the special case z = z0, which implies r = 0, of the Theorem 5.4.It can also be derived directly from Theorem 4.7 applied to a circle:

f (n)(z0) =n!

2πi

∫

C(z0,ρ)

f(ζ)

(ζ − z0)n+1dζ

=n!

2πρn

∫ 2π

0e−inθ f(z0 + ρeiθ) dθ.

The triangle inequality (or ML formula) gives |f (n)(z0)| ≤ n!M(ρ)/ρn. The second statementfollows from an = f (n)(z0)/n!.

Corollary 5.6. Mean-value property. Suppose that f(z) is analytic in an open disc D(z0, ρ)and continuous onto its boundary C(z0, ρ) from within. Then

f(z0) =1

2π

∫ 2π

0f(z0 + reiθ) dθ, 0 ≤ r ≤ ρ.

Thus the value of an analytic function at a point is the average of the values of that function oncircles centred at that point. The same is true for its real and imaginary parts in the real xy-plane.

Proof. Apply Cauchy’s integral formula to the centre of the circle C(z0, r) and parametrise thecircle according to ζ = z0 + reiθ. Then,

f(z0) =1

2πi

∫

C(z0,r)

f(ζ)

ζ − z0dζ

=1

2π

∫ 2π

0f(z0 + reiθ) dθ,

as required. If z = x+ iy and f(z) = u(x, y) + iv(x, y), then taking real and imaginary parts gives

u(x0, y0) =1

2π

∫ 2π

0u(x0 + r cos θ, y0 + r sin θ) dθ,

v(x0, y0) =1

2π

∫ 2π

0v(x0 + r cos θ, y0 + r sin θ) dθ.

71

Thus u(x, y) and v(x, y) also obey the mean-value property separately.

Cauchy’s inequality severely restricts the types of entire functions that can have power lawgrowth.

Lemma 5.7. Let f(z) be an entire function whose growth is limited by the inequality,

|f(z)| < A|z|α,

uniformly for all sufficiently large z and some real A and α. If α ≥ 0, then f(z) is a polynomialwhose degree does not exceed α or is identically zero. If α < 0, then f(z) is identically zero.

Proof. Because f(z) is entire, it has a power series expansion,

f(z) =

∞∑

k=0

akzk,

with an infinite radius of convergence. Choose a circle C(R), with centre at the origin, with radiusR large enough that the given inequality applies to f(z) on this circle. Using our previous notation,we get

M(r) < Arα,

for all r ≥ R. Cauchy’s inequality gives

|an| ≤M(r)

rn< Arα−n.

Consider n > α and let r → ∞. For all such n, an = 0. This proves the lemma.

Corollary 5.8. Liouville’s theorem. If an entire function f(z) is bounded, or if it possiblygrows at a rate such that f(z)/z → 0 uniformly as z → ∞, then f(z) is constant.

Proof. Since f(z)/z is bounded for large z, the previous lemma applies with α = 1. Thus f(z) isa polynomial of degree one at most. But if the degree was actually one, it would fail to meet thegiven growth constraint. Hence, f(z) is a constant.

Liouville’s theorem plays a vital role in Riemann-Hilbert and Weiner-Hopf problems. In prob-lems of these types, two analytic functions in different domains are required such that they satisfygiven junction conditions across a common boundary. The strategy is to transform the problem sothat the junction conditions simplify to a continuous join, and then invoke Liouville’s theorem orone of its variants to restrict the class of analytic functions that satisfy any remaining conditionsat infinity or elsewhere.

Liouville’s theorem also provides a quick proof of:

Theorem 5.9. Fundamental theorem of algebra. A polynomial,

P (z) = anzn + an−1z

n−1 + . . . + a1z + a0 ,

aj ∈ C, j = 0, 1, 2, . . . , n, an 6= 0,

of degree n, n ≥ 1, has exactly n zeros (or roots), if a zero of order k (or multiplicity k) is countedas k zeros. (The theorem also holds trivially for n = 0 since a polynomial of degree zero is just anonzero constant.)

72

Proof. Suppose a polynomial P (z) of degree n ≥ 1 has no zeros at all. Then |P (z)|, being real-valued and continuous, attains its minimum on every compact set. Also since P (z) is uniformlylarge for large z, the modulus |P (z)| is bounded away from zero. This implies that the reciprocal1/P (z) is a bounded entire function. According to Liouville’s theorem, it must be constant. Thiscontradiction forces us to conclude that P (z) must have at least one zero, say, z1. The proof thatP (z) has exactly n zeros if multiplicities are counted is an elementary consequence of having onezero, by induction on the degree. So P (z) has a factorization, unique up to ordering of factors:

P (z) = an(z − z1)(z − z2) · · · (z − zn).

When applied to real-variable polynomials with real coefficients, this theorem implies that everysuch polynomial factors into real linear and quadratic factors because the roots are either real orform complex conjugate pairs.

Examples.

x4 + 1 =(

x2 +√

2x+ 1)(

x2 −√

2x+ 1)

,

x5 − 1 =(

x− 1)(

x2 + 12 (1 +

√5 )x+ 1

)(

x2 + 12 (1 −

√5 )x+ 1

)

.

The next two theorems on uniqueness of analytic continuation are remarkably useful. Wehave already mentioned that analytic functions are often presented in the form of infinite series ordefinite integrals whose domains of validity are smaller than the maximal domains of analyticity ofthe functions being represented. These theorems imply, for example, that if an analytic functionis defined in the neighbourhood of a particular point, or is defined on an interval of the real axis,then it analytically extends to a unique global analytic function, whose domain may be a Riemannsurface. All of its singularities and the shape of its Riemann surface are uniquely determined.Similarly, if an identity involving analytic functions is known to be true in the neighbourhood of aparticular point, or on an interval of the real axis, then it is guaranteed to be true globally.

Consider, for example, the gamma function,

Γ(z) =

∫ ∞

0tz−1e−t dt, Re z > 0.

The identity,

Γ(z)Γ(1 − z) =π

sinπz,

is easiest to prove when z is real and in the open interval (0, 1). The Weierstrass identity theo-

rem guarantees that the identity extends from the real interval to the vertical strip 0 < Re z < 1,which is the common domain of Γ(z) and Γ(1 − z) according to the original definition. But theidentity provides a definition of Γ(z) throughout the domain of Γ(1−z) except for the obvious polesat zero and the negative integers. Hence, Γ(z) is globally defined as a meromorphic function andits reciprocal 1/Γ(z) is entire. The Weierstrass identity theorem also guarantees that this extensioninto the left half-plane is unique. The second theorem proves uniqueness under more primitivehypotheses.

Theorem 5.10. Uniqueness of analytic continuation 1. Weierstrass identity theorem.Suppose that f(z) and g(z) are functions analytic in a common open region D. Let H be a subset

73

of D that contains a convergent sequence zk, k = 1, 2, 3, . . . , whose limit a is in D. (Allsubregions, arcs and line segments in D contain such a sequence.) If

f(z) = g(z), z ∈ H,

then f(z) ≡ g(z) everywhere in D.

Proof. First note that the limit point a must be in D itself, and not just on its boundary. Forf(z) = sin(π/z) and g(z) = 0 agree on the sequence 1/n with limit point zero, but f(z) is notanalytic at this limit point. Observe that the continuity of f(z) and g(z) at z = a forces f(a) = g(a).So the limit point a can be appended to the sequence zk (and to the set H that contains it) if itis not already a member. In fact, without loss of generality, we can let H be just the sequence zktogether with its limit point a.

By considering the difference of the two analytic functions, we can set g(z) ≡ 0, without loss ofgenerality. Thus

f(zk) = 0, k = 1, 2, 3, . . . , f(a) = 0,

where zk → a as k → ∞. We want to prove that f(z) ≡ 0 in D under these hypotheses. Since f(z)is analytic at a, it admits a convergent Taylor series without constant term,

f(z) =

∞∑

n=1

f (n)(a)

n!(z − a)n,

valid at least in the largest open disc D(a, r) that fits inside D.

Now, define

f1(z) =f(z) − f(a)

z − a=

∞∑

n=1

f (n)(a)

n!(z − a)n−1.

Because f(z) is analytic at z = a, f1(z) → f ′(a) as z → a on all paths. In particular, we can let zrun through the zk. But f1(zk) = 0 for all k. Hence f ′(a) = 0. According to Lemma 4.12, f1(z) isanalytic at a. We have just proved that f1(a) = 0 = f ′(a).

Next, define

f2(z) =f1(z) − f1(a)

z − a=

∞∑

n=2

f (n)(a)

n!(z − a)n−2.

Precisely the same argument yields f2(a) = 0 = f ′′(a). The argument can be repeated indefinitely.The result is that all derivatives vanish at a:

f(a) = f ′(a) = f ′′(a) = . . . = f (n)(a) = . . . = 0.

Hence, the Taylor series for f(z) about a converges to the zero function in D(a, r).

So far, we have proved that f(z) ≡ 0 in a circular disc D(a, r) ⊂ D. If this is D itself, then we aredone. Otherwise, choose any point b ∈ D −D(a, r) and join a to b by an arc Γ in D (optionally asmooth Jordan arc). Now every point ζ ∈ Γ is a point of analyticity of f(z) and is therefore thecentre of an open neighbourhood N(ζ, δ(ζ)) ⊂ D of analyticity. Since Γ is a compact set, we cancover Γ with a finite number of these open neighbourhoods,

N(b0, δ0), N(b1, δ1), N(b2, δ2), . . . , N(bm, δm),

74

where b0 = a, bm = b and δ0 = R. In addition, by adding a finite number of discs, if necessary, wecan arrange the finite subcover so that

bj ∈ N(bj−1, δj−1), j = 1, 2, 3, . . . , m.

Since f(z) ≡ 0 in N(b0, δ0) and b1 ∈ N(b0, δ0), f(z) and all its derivatives vanish at b1. Hencethe Taylor series for f(z) about b1 converges to the zero function at least in the disc N(b1, δ1).In other words, f(z) ≡ 0 in N(b1, δ1). Next, b2 is in this disc, and the same argument shows thatf(z) ≡ 0 in N(b2, δ2). Continuing in this fashion, we find that f(z) ≡ 0 in all of the discs N(bj, δj).In particular, f(b) = 0. Since b was any point in D −D(a,R), we have proved that

f(z) ≡ 0, z ∈ D.

This completes the proof that f(z) ≡ g(z) in D under the original hypotheses.

The preceding theorem introduced the important concept of analytic continuation along an

arc, first studied by Weierstrass. This method can be used to analytically continue a function f(z)outside a known domain of analyticity into an area where the existence or analytic character of f(z)is unknown and to be determined. Of course, if f(z) has been analytically continued into a largerdomain by any method, that continuation will be unique by the Weierstrass identity theorem. Thearc is to be covered with a finite number of suitable discs in order, with the centre of each disc onthe arc and in the interior of the previous disc. Then f(z) can be constructed in the union of thediscs by convergent Taylor series, one disc at a time. A singularity of f(z) is a point that cannotbe reached in this manner.

Theorem 5.11. Uniqueness of analytic continuation 2. Monodromy theorem. Supposethat D is a simply connected region and that f(z) is analytic at a particular point a ∈ D, andtherefore in a neighbourhood of a. If f(z) can be analytically continued along any arc in Dthrough a, then f(z) analytically extends uniquely to D.

The proof uses elementary ideas from homotopy theory. See Ahlfors, Complex Analysis, Chap-ter 8.

In the monodromy theorem, the hypothesis of simple connectivity is essential. If f(z) is ana-lytically continued along arcs into a multiply connected region, it is possible that the values of thefunction will not agree when two paths meet after passing on opposite sides of a hole or singularpoint. In that case, as we have already mentioned, the way to proceed is to allow the function tocontinue uniquely onto a Riemann surface. If the multi-valuedness of f(z) can be localised tothe neighbourhood of a particular point, then that point is a branch point. Some examples offunctions having branch points are given in Chapter 4.

75

6 Laurent’s theorem and the residue theorem

Laurent’s theorem provides an extension of the Cauchy-Taylor theorem to annular regions and theneighbourhoods of isolated singular points (poles, isolated essential singularities), including thepoint at infinity.

Lemma 6.1. Cauchy’s integral formula for annular regions. Suppose that f(z) is analyticbetween and on two positively oriented rectifiable Jordan curves C1 and C2, with C1 properlyinside C2 as in Theorem 3.12. Then for all z in the annular region between the contours,

f(z) =1

2πi

∫

C2

f(ζ)

ζ − zdζ − 1

2πi

∫

C1

f(ζ)

ζ − zdζ.

Proof. In the proof of Theorem 3.12, it was shown how to find two bridges or cuts from C1 to C2

in the complex ζ-plane that were interior to the annular region between C1 and C2 except for theirendpoints. Choose one of these cuts, say K, and indent it, if necessary, so that it does not crossthe point ζ = z. Then the contour, C2 + (−K) + (−C1) + K, is in the domain of analyticity off(ζ) and forms the positively oriented boundary of a simply connected region of analyticity havingthe point z inside. Then the lemma is a consequence of Cauchy’s integral formula applied to thiscontour.

Lemma 6.2. Exterior Cauchy’s integral formula. If f(z) is analytic outside and on a positivelyoriented Jordan curve C and if f(z) → 0 uniformly as z → ∞, then, for z outside C,

f(z) = − 1

2πi

∫

C

f(ζ)

ζ − zdζ.

Proof. The method is the same as for Corollary 3.13. In the previous lemma, let C1 = C andC2 = C(0, R), where the radius R is large and the centre is the origin. By the ML formula, themodulus of the integral over C(0, R) is bounded above by 2πRM(0, R)/(R − |z|). This can bemade arbitrarily small by choosing R sufficiently large. This proves the lemma.

Theorem 6.3. Laurent’s theorem. Suppose that f(z) is analytic in the open circular annulus(including limiting cases),

D :=

z : R1 < |z − z0| < R2

, R1 ≥ 0, R2 ≤ +∞.

Then f(z) admits a power series expansion in both positive and negative powers (a Laurent

series),

f(z) =∞

∑

n=−∞an(z − z0)

n,

which is absolutely convergent in D and uniformly convergent on compact subsets. A formula forthe coefficient an is

an =1

2πi

∫

C(r)

f(ζ)

(ζ − z0)n+1dζ, −∞ < n <∞,

where C(r) is any positively oriented circle |ζ − z0| = r with R1 < r < R2.

76

Proof. Since the circles C(R1) and C(R2) may pass through singularities of f(z), or may beextreme cases (R1 = 0 or R2 = ∞, in which case let C(0) = z0 and C(∞) = ∞), we chooseauxiliary radii R3 and R4 near R1 and R2, respectively, such that

R1 < R3 < r < R4 < R2 .

Let E denote the open annulus in the complex ζ-plane between C(R3) and C(R4). According toLemma 6.1, for z ∈ E,

f(z) = f1(z) + f2(z),

where, for z in the indicated larger domains,

f1(z) := − 1

2πi

∫

C(R3)

f(ζ)

ζ − zdζ, z outside C(R3), including ∞,

f2(z) :=1

2πi

∫

C(R4)

f(ζ)

ζ − zdζ, z inside C(R4).

Note that f1(z) and f2(z) inherit the larger domains of analyticity according to Theorem 4.4. Fur-ther, by moving the contours C(R3) and C(R4) arbitrarily close to C(R1) and C(R2), respectively,with the extreme cases suitably interpreted, the domain of analyticity of f1(z) extends to all zoutside C(R1) and the domain of analyticity of f2(z) extends to all z inside C(R2).

Now f2(z) is analytic inside C(R2) and so has a convergent Taylor series expansion,

f2(z) =∞

∑

n=0

an(z − z0)n, |z − z0| < R2,

where

an =f

(n)2 (z0)

n!=

1

2πi

∫

C(r)

f2(ζ)

(ζ − z0)n+1dζ, n ≥ 0.

Since (ζ−z0)−n−1f1(ζ) is analytic outside and on C(r) and has a zero of order n+ 2 ≥ 2 at ζ = ∞,the exterior Cauchy theorem (Corollary 3.13) implies

1

2πi

∫

C(r)

f1(ζ)

(ζ − z0)n+1dζ = 0, n ≥ 0.

Hence,

an =1

2πi

∫

C(r)

f(ζ)

(ζ − z0)n+1dζ, n ≥ 0.

So far, one half of the Laurent series has been accounted for.

Next, f1(z) is analytic outside C(R1) and has a zero at z = ∞. Change variable,

z =1

z − z0, g2(z) = f1(z), g1(z) = f2(z).

In the complex z-plane, g2(z) is analytic inside the circle C(0, 1/R1) and has a zero at the origin,and g1(z) is analytic outside the circle C(0, 1/R2). Hence, g2(z) has a Taylor series without constant

77

term about the origin:

g2(z) =∞∑

n=1

bnzn, |z| < 1/R1,

bn =1

2πi

∫

C(0,1/r)

g2(ζ)

ζn+1dζ, n ≥ 1.

In terms of the original variables, we get

f1(z) =

∞∑

n=1

bn(z − z0)−n, |z − z0| > R1,

bn =1

2πi

∫

C(r)(ζ − z0)

n−1f1(ζ) dζ, n ≥ 1,

where we used the integration by substitution, ζ = 1/(ζ − z0), dζ = −dζ/(ζ − z0)2. The mi-

nus sign is cancelled by the reversal of orientation of the circle under this substitution. Because(ζ − z0)

n−1f2(ζ) is analytic inside C(r), we can replace f1(ζ) in the integrand by f(ζ) withoutaffecting the integral. Hence, with a−n := bn, we get the other half of the Laurent series:

f1(z) =∞∑

n=1

a−n(z − z0)−n, |z − z0| > R1,

a−n =1

2πi

∫

C(r)(ζ − z0)

n−1f(ζ) dζ, n ≥ 1.

So, putting the two halves of the series together, we get the full Laurent series:

f(z) = f1(z) + f2(z) =∞∑

n=−∞an(z − z0)

n, R1 < |z − z0| < R2,

an =1

2πi

∫

C(r)(ζ − z0)

−n−1f(ζ) dζ, −∞ < n <∞.

The absolute convergence of the series in the annulus between C(R1) and C(R2) and the uniformconvergence on compact subsets are inherited from the Taylor series for f2(z) and g2(z).

Remarks. Of course, all the elementary properties of power series carry over to each half of theLaurent series. In particular, if f1(z) is not identically zero and the inner radius R1 is minimised,then f(z) has a singularity on C(R1) (including the limiting case C(0) = z0). If f2(z) is neitherzero nor an entire function and the outer radius R2 is maximised, then f(z) has a singularity onC(R2).

Additional notes on singularities. At the end of Chapter 4, we gave a partial classificationof singularities. The definitions given there of a pole and an isolated essential singularity are notstandard, although they are easily seen to be equivalent to the standard definitions, which we nowpresent. In the following definitions, f(z) is analytic in a deleted neighbourhood N ′(z0, R2) of z0,which implies R1 = 0.

78

Definition 6.4. If R1 = 0, the part f1(z) of f(z) having the negative powers in the Laurent seriesabout z0 is called the principal part or singular part of f(z) at z0.

Definition 6.5. If the principal part terminates with leading term a−k(z − z0)−k, k ≥ 1, which

necessarily implies R1 = 0, then f(z) has a pole of order k at z0. The leading coefficient a−k iscalled the strength of the pole when k ≥ 2. (In the case of a simple pole k = 1, the term “strength”is superseded by “residue” below.)

Definition 6.6. If R1 = 0 and the principal part is an infinite series, then f(z) has an isolated

essential singularity at z0.

Definition 6.7. If z0 is a pole or isolated essential singularity of f(z), then the residue of f(z)at z0 is the coefficient a−1 of (z − z0)

−1 in the Laurent expansion f(z) =∑∞

k=−∞ ak(z − z0)k of

f(z) about z0. The notation for residue is

Resz=z0

f(z) = residue of f(z) at z0.

Lemma 6.8. Analytic functions are uniformly unbounded in the neighbourhood of a pole.

Proof. Write f(z) = g(z)/(z − z0)k, k ≥ 1, where g(z) is analytic and nonzero at z = z0. By

continuity, there exists a neighbourhood of z0 on which |g(z)| > A := |g(z0)|/2. On that neighbour-hood, with z0 punched out, |f(z)| > A/|z − z0|k, which tends to +∞ uniformly as z → z0 on allpaths.

Definition 6.9. An a-point of an analytic function f(z) is a root of the equation f(z) = a.Equivalently, it is a zero of the function f(z) − a.

Lemma 6.10. A limit point of zeros or a-points of a nonconstant locally analytic function f(z) isan essential singularity.

Proof. Let z0 be a limit point of a-points of f(z). If z0 were a point of analyticity of f(z), thenf(z0) = a and f(z) would be constant by the Weierstrass identity theorem (just the first part ofthe proof being needed). If f(z) were bounded near z0, then z0 would be a point of analyticityby the removable singularities theorem. Given that f(z) is nonconstant, we conclude that f(z)must be unbounded near z0. If z0 were a pole, f(z) would be uniformly unbounded near z0, andso could not be a limit point of a-points with finite a. The only possibility remaining is that z0 isan essential singularity of f(z), either isolated or nonisolated.

Theorem 6.11. Casorati-Weierstrass theorem. In any neighbourhood of an isolated essentialsingularity, an analytic function f(z) either takes the value a or approaches a arbitrarily closely,for every a ∈ C.

Proof. Let z0 be the isolated essential singularity of f(z) and let a ∈ C be given. If z0 is a limitpoint of a-points of f(z), there is nothing to prove. So suppose that the equation f(z) = a hasno roots in N ′(z0, δ). Then g(z) := 1/(f(z) − a) is analytic in this region. If g(z) were bounded,then z0 would be a point of analyticity of g(z) by the removable singularities theorem. But thenz0 would either be a point of analyticity or a pole of f(z), contradicting the hypothesis that z0 wasan isolated essential singularity. Thus g(z) must be unbounded in N ′(z0, δ). This implies that f(z)approaches a arbitrarily closely in a deleted neighbourhood of z0.

Two stronger theorems of this type are

79

Theorem 6.12. Picard’s first theorem. A nonconstant entire function takes every value a ∈ C,with at most one possible exception, at least once.

Theorem 6.13. Picard’s second theorem or the grand Picard theorem (1880). In anyneighbourhood of an isolated essential singularity z0 ∈ C∗, an analytic function f(z) takes everyvalue a ∈ C, with at most one possible exception, an infinite number of times.

A quick proof of the first theorem using elliptic modular functions can be found in Ahlfors, Com-

plex Analysis, Chapter 8. The second theorem is more difficult. Its proof is given in Titchmarsh,Theory of Functions, Chapter 8. According to the second theorem, the phrase “at least once” inthe first theorem can be upgraded to “an infinite number of times” when the entire function is nota polynomial.

Examples. The entire function ez never takes the value zero, because eze−z = 1. The entirefunctions sin z, J0(z) and 1/Γ(z) do not miss any values. In the neighbourhood of a nonisolatedessential singularity, an analytic function can miss two values. For example, the meromorphicfunction tan z never takes the values i or −i.

We now turn our attention to properties of residues of analytic functions at poles and isolatedessential singularities. Residues are of considerable importance in complex analysis. Consider theformal term-by-term integral of a Laurent series around one of the circles C(r) in the annulus ofconvergence. All the integrals vanish except the integral of the minus-one power, which is 2πi timesthe residue. The justification of the term-by-term integration will be provided in the proof of theresidue theorem below.

If f(z) is built out of elementary functions, a quick way to calculate the residue at z0 is tojust to let z = z0 + ǫ and expand the elementary functions in a sufficient number of powers of ǫ.However, in general, residues can be calculated without looking at the Laurent expansion. Thefollowing quick methods of calculation are straightforward consequences of the definition:

Lemma 6.14. If f(z) has a simple pole at z0, then

Resz=z0

f(z) = limz→z0

(z − z0)f(z) =

[

(z − z0)f(z)

]

z=z0

.

Proof. Put z = z0 in (z − z0)f(z) = a−1 + a0(z − z0) + . . . .

Lemma 6.15. If f(z) has a pole of order k at z0, then

Resz=z0

f(z) =1

(k − 1)!

[

dk−1

dzk−1

(

(z − z0)kf(z)

)

]

z=z0

.

Proof. Differentiate (z− z0)kf(z) = a−k + . . . +a−1(z− z0)k−1 + . . . k− 1 times and put z = z0.

Lemma 6.16. If g(z) and h(z) are analytic at z0 and h(z) has a simple zero there, then

Resz=z0

g(z)

h(z)=

g(z0)

h′(z0).

If g(z) also has a zero at z0, then the pole is absent and this formula yields a zero result.

80

Proof. This is an application of Lemma 6.14.

Remarks. A function always has a nonzero residue at a simple pole. It may have a zero residueat a higher-order pole or isolated essential singularity because the minus-one power may be absentfrom the Laurent series. The definition of residue is consistent with a function having a zero residueat each point of analyticity. Residues are not defined, in general, at other types of singularities(nonisolated singularities, branch points).

Theorem 6.17. The residue theorem. Suppose that f(z) is analytic inside and on a posi-tively oriented rectifiable Jordan curve C except for a finite number of poles or isolated essentialsingularities, z1, z2, . . . , zk, inside C. Then

∫

Cf(z) dz = 2πi

k∑

j=1

Resz=zj

f(z)

= 2πi[

sum of residues of f(z) inside C]

.

Proof. According to Corollary 3.14, the theorem will be true when C encloses k singular points ifit is true when C encloses any one singular point, say, z1. Then we can let C be the circle C(z1, r),0 < r < R, where R is the distance from z1 to the nearest other singularity (of any character)of f(z). Laurent’s theorem applies to the deleted neighbourhood N ′(z1, R). Let

I =

∫

Cf(z) dz.

As in the proof of Laurent’s theorem, let

f(z) = f1(z) + f2(z), I1 :=

∫

Cf1(z) dz, I2 :=

∫

Cf2(z) dz,

where

f1(z) =−1∑

n=−∞an(z − z1)

n, z ∈ C∗ − z1,

and f2(z) is analytic inside the circle C(z1, R). By Cauchy’s theorem,

I2 = 0, I = I1.

Also, as in the proof of Laurent’s theorem, we change variables:

z =1

z − z1, g(z) = f1(z).

In the complex z-plane, g(z) is the entire function,

g(z) =

∞∑

n=1

a−nzn.

(If the original singularity of f(z) at z1 was a pole, then g(z) is a polynomial; if the originalsingularity was an isolated essential singularity, then this power series has an infinite radius ofconvergence.) Integration by substitution gives

I =

∫

C(0,1/r)

g(z)

z2dz,

81

where the circle C(0, 1/r) is positively oriented. Now,

g(z)

z2=

a−1

z+ h(z),

where h(z) is entire. The integral of h(z) over C(0, 1/r) vanishes by Cauchy’s theorem. Finally,

I =

∫

C(z1,r)f(z) dz =

∫

C(0,1/r)

a−1

zdz = 2πi a−1.

This completes the proof of the residue theorem.

The residue theorem is the central pillar of a branch of complex analysis that deals with theevaluation of definite integrals and summation of series. An immediate extension based on Corol-lary 3.17 is:

Corollary 6.18. Suppose that f(z) is analytic in a simply connected region D except for a finitenumber of poles or isolated essential singularities, z1, z2, . . . , zk, interior to D. Let C be anyrectifiable closed contour or cycle in D that does not cross any of the zj and let nj be the windingnumber of C about zj . Then

∫

Cf(z) dz = 2πi

k∑

j=1

nj Resz=zj

f(z).

As in the case of Cauchy’s theorem and Cauchy’s integral formula, there is an exterior versionof the residue theorem and a version for the annular region between two Jordan curves. The latteris straightforward. For the exterior version, we need:

Definition 6.19. Residue at infinity. Suppose that f(z) is analytic everywhere outside abounded region, in which case it admits a convergent Laurent series,

f(z) =

∞∑

n=−∞anz

n, R < |z| <∞.

Then the residue of f(z) at z = ∞ is minus the coefficient a−1:

Resz=∞

f(z) = − a−1.

Theorem 6.20. Exterior residue theorem. Suppose that f(z) is analytic everywhere outsideand on a positively oriented rectifiable Jordan curve C except for a finite number of poles or isolatedessential singularities, z1, z2, . . . , zk. Then

∫

Cf(z) dz = − 2πi

k∑

j=1

Resz=zj

f(z) − 2πi Resz=∞

f(z).

Proof. Let C(R) be a circle, centre z = 0, large enough to enclose C and all the singularitiesoutside C. Then, by the residue theorem, for r > R,

∫

C(r)f(z) dz −

∫

Cf(z) dz = 2πi

k∑

j=1

Resz=zj

f(z).

82

From the proof of Laurent’s theorem, we know that we can integrate the Laurent series for f(z)for R < |z| <∞ term by term on C(r). Only the a−1 term contributes, and we get

∫

C(r)f(z) dz = 2πi a−1 = − 2πi Res

z=∞f(z).

Theorem 6.21. The argument principle. Suppose that a function f(z) is analytic or mero-morphic inside and on a positively oriented rectifiable Jordan curve C and that f(z) has no zerosor poles on C itself. Let N be the number of zeros and P the number of poles of f(z) inside C,counting multiplicities. Then

N − P =1

2π

[

arg f(z)]

C.

Proof. The values of N and P must be finite because limit points of zeros or poles are essentialsingularities, and none are allowed either inside or on C by the hypotheses of the theorem. Considerthe contour integral,

I =1

2πi

∫

C

f ′(z)

f(z)dz.

The integrand is analytic on C (and therefore also on an open set that covers C) and has simplepoles at the zeros and poles of f(z) inside C. It has no other singularities inside C. Hence I equalsthe sum of the residues of f ′(z)/f(z) inside C.

If z1 is a zero of order m of f(z) inside C, then

f(z) = (z − z1)mg(z),

where g(z) is analytic and nonzero at z1. Then

f ′(z)

f(z)=

m

z − z1+g′(z)

g(z), Res

z=z1

f ′(z)

f(z)= m.

Hence the sum of the residues of this type is the sum of the orders of all the zeros of f(z) inside C,which is N .

If z2 is a pole of order n of f(z) inside C, then

f(z) = (z − z2)−nh(z),

where h(z) is analytic and nonzero at z2. Then

f ′(z)

f(z)= − n

z − z2+h′(z)

h(z), Res

z=z2

f ′(z)

f(z)= −n.

Hence the sum of the residues of this type is minus the sum of the orders of all the poles of f(z)inside C, which is −P .

So far, the residue theorem gives

N − P = I =1

2πi

∫

C

f ′(z)

f(z)dz.

83

Now f ′(z)/f(z) has the multi-valued primitive log f(z) in its non-simply connected domain ofanalyticity. By choosing any branch of log f(z) at a particular point of C and then varying log f(z)continuously as one follows C once around in the direction of its arrow, we get

N − P =1

2πi

[

log f(z)]

C

=1

2πi

[

log |f(z)| + i arg f(z)]

C

=1

2π

[

arg f(z)]

C,

where we used the fact that log |f(z)| is single-valued.

Theorem 6.22. Rouche’s theorem. Suppose that f(z) and g(z) are analytic inside and on apositively oriented rectifiable Jordan curve C and that

|f(z) − g(z)| < |f(z)|

on C. Then f(z) and g(z) have the same number of zeros inside C, counting multiplicities.

Proof. First, the inequality implies that neither f(z) nor g(z) can have a zero on C. Define themeromorphic function h(z) := g(z)/f(z). Then |h(z) − 1| < 1 on C. This means that the values ofh(z) on C are confined to an open disc of centre 1 and radius 1. Then arg h(z) returns to its startingvalue after a circuit of C. According to the argument principle, N − P = 0 for h(z) inside C. Thisis equivalent to the statement that f(z) and g(z) have the same number of zeros inside C, countingmultiplicities.

Rouche’s theorem provides the standard proof of the fundamental theorem of algebra,Theorem 5.9. One alternative proof has already been given. An elementary proof not requiring thecomplex integral calculus will be given in the next chapter.

Standard proof of Theorem 5.9. Suppose that a nonconstant polynomial P (z) of degree nhas leading term anz

n. Then P (z) − anzn is a polynomial of lower degree (or identically zero).

On a sufficiently large circle, the conditions of Rouche’s theorem are satisfied by f(z) = anzn and

g(z) = P (z). Inside the circle, f(z) has a zero of order n at z = 0 and no other zeros. Hence P (z)has exactly n zeros, counting multiplicities.

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7 Maximum principles and harmonic functions

We begin this chapter with a theorem that is of central importance in analytic function theory.It is a consequence of the mean-value property of an analytic function, but can be proved in anelementary way that does not involve the complex integral calculus. We give two proofs.

Theorem 7.1. The maximum modulus theorem or maximum principle. Suppose that f(z)is analytic in a region D in the extended complex plane C∗ and continuous on its closure [D] (in theRiemann sphere topology). Then the maximum of |f(z)| on [D] is attained on the boundary ∂D.If this maximum is also attained at an interior point of D, then f(z) is constant on [D].

Corollary 7.2. The minimum modulus theorem. Suppose that f(z) is analytic in a regionD ⊂ C∗ and continuous on its closure [D]. Then the minimum of |f(z)| on [D] is either zero or isattained on the boundary ∂D. If the minimum is nonzero and is also attained at an interior pointof D, then f(z) is constant on [D].

Proof. In both theorems, f(z) is bounded and uniformly continuous on [D]. If f(z) is nonzeroon [D], then 1/f(z) is analytic in D and continuous on [D]. So the maximum modulus theorem for1/f(z) implies the minimum modulus theorem for f(z).

We can restrict attention to bounded regions D ⊂ C by the following argument. If the conclusionof one of the theorems is denied for an unbounded region D, then a nonconstant locally analyticfunction will take its maximum modulus or its nonzero minimum modulus at an interior point of D.If this point is finite, then we can cover it with a bounded region of analyticity E ⊂ D and get animmediate contradiction. If the point is infinity, just map it to the origin by ζ = 1/z. The imageof D will cover the origin and we again have a contradiction.

Suppose z0 is a finite interior point of D and |f(z)| ≤ |f(z0)| in a neighbourhood of z0 within D.Then a circle C(z0, r) together with its interior D(z0, r) can be drawn within this neighbourhood.The mean-value property states that

f(z0) =1

2πi

∫

C(z0,r)

f(ζ)

ζ − z0dζ

=1

2π

∫ 2π

0f(z0 + reiθ) dθ.

The triangle inequality for Riemann integrals gives

|f(z0)| ≤1

2π

∫ 2π

0|f(z0 + reiθ)| dθ.

But, by hypothesis, |f(z0 + reiθ)| ≤ |f(z0)| on [0, 2π], since the left-hand side is |f(z)| evaluatedon C(z0, r). If strict inequality occurs at any point, it must also occur on a subinterval of positivelength on [0, 2π], understood to be periodically extended, on account of the continuity of |f(z)|on C(z0, r). But in that case the mean value of |f(z)| on C(z0, r) would be strictly less than|f(z0)|. This would force |f(z0)| < |f(z0)|, which is a contradiction. So the hypothesis that|f(z)| ≤ |f(z0)| in a neighbourhood of z0 forces |f(z)| to be constant on C(z0, r) and take thevalue |f(z0)|. This result is independent of r for all sufficiently small r, and so |f(z)| is constanton an open set covering z0. It is a simple consequence of the Cauchy-Riemann equations that theconstancy of |f(z)| implies the constancy of f(z) itself on the open set covering z0. Then f(z) is

85

constant throughout D. So a nonconstant analytic function cannot take its maximum modulus ata finite interior point of D. (An infinite interior point is handled by the map z 7→ 1/z as alreadymentioned.)

The preceding proof ultimately relied on Cauchy’s integral formula, which was behind the mean-value property used. However, the maximum modulus theorem does not require the apparatus ofthe complex integral calculus. It is an elementary consequence of the properties of power series, and,without the benefit of the Cauchy-Taylor theorem, it can be understood to apply to any functionsof a complex variable that have local convergent power series representations. Such functions arelocally analytic on account of the term-by-term differentiability of power series.

Elementary proof. Suppose f(z) has a convergent power series representation,

f(z) = a0 + a1(z − z0) + a2(z − z0)2 + . . . ,

in a disc |z − z0| < R, R > 0. When a0 = 0, there is nothing to prove, because zero is a permittedminimum value or a trivial maximum value for |f(z)|. So assume that a0 6= 0. Then f(z0) = a0 6= 0.Also assume that f(z) is nonconstant, so that the power series has at least one other term beyond a0.We wish to prove that the positive number |a0| is neither a local maximum nor a local minimum ofthe modulus |f(z)| in a neighbourhood of z0. Let an, n ≥ 1, be the first nonzero coefficient after a0.Then

f(z) = a0 + an(z − z0)n + (z − z0)

n+1ψ(z),

where ψ(z) is bounded in a neighbourhood of z0.

Certain simplifications are available without losing any generality. By scaling (that is, multiplyingf(z) by a constant), we can make a0 = 1. By translating (that is, adding a constant to z), wecan make z0 = 0. After that, by rotating (that is, replacing z by eiθz), we can make an real andpositive.

So far,f(z) = 1 +Azn + zn+1ψ1(z), A > 0,

where ψ1(z) is bounded near z = 0. By restricting z to a sufficiently small circular disc centred atz = 0, we can ensure that

∣

∣zψ1(z)∣

∣ < A/2.

Then the triangle inequality implies

|f(z)| ≤ |1 +Azn| + 12A|z|

n.

|f(z)| ≥ |1 +Azn| − 12A|z|

n.

In the first inequality, let z = ρeiπ/n, where ρ is small and positive. For such z, |f(z)| ≤ 1−Aρn/2,which is strictly less than 1. So 1 cannot be a local minimum for |f(z)|. In the second inequality,let z be small and positive. For such z, |f(z)| ≥ 1 + Aρn/2, which is strictly greater than 1. So1 cannot be a local maximum for |f(z)|. So the modulus of the original function f(z) cannotattain a local maximum or a local minimum at an interior point of D unless the minimum valueis zero (obviously allowed) or the function is constant. This completes the elementary proof of themaximum and minimum modulus theorems.

The method can be refined a little more. Let w0 = f(z0). Observe that f(z)−w0 has a zero oforder n, n ≥ 1, at z0. By adding a sufficiently small complex constant λ, we find that f(z)−w0 +λ

86

has exactly n simple zeros near z0 arranged in a small approximate regular n-gon enclosing z0.(There may, of course, be unrelated zeros elsewhere in the complex plane, which have no bearingon this discussion.) So the image of a sufficiently small open disc centred at z0 under the mapw = f(z) contains an open neighbourhood of w0. This proves the following statement:

Corollary 7.3. A nonconstant analytic function f : A → C, A ⊂ C, maps open sets to opensets.

A corresponding statement for real-analytic functions is false. For example, x 7→ x2 maps theopen interval (−1, 1) to the half-open interval [0, 1). A similar statement for continuous functionsapplies to inverse images. This property is used to define continuity in general topological spaces.Under a continuous map f : A→ B, where A and B are subsets of the same or different topologicalspaces, an open set in B pulls back to an open set in A.

The minimum modulus theorem provides a short elementary proof of the fundamental theo-

rem of algebra, Theorem 5.9. Two proofs relying on the complex integral calculus have alreadybeen given.

Elementary proof of Theorem 5.9. Consider a closed disc of radius R in the real xy-plane,where R is assumed to be large, and let z = x + iy. On this disc, being a closed bounded setin R2, a real-valued continuous function of x and y will attain its maximum and minimum values.The modulus of a polynomial P (z) is such a real-valued continuous function. Now a nonconstantpolynomial is uniformly large on a large circle because it is dominated by its leading term. It followsthat its modulus will not attain its minimum on the boundary of the disc. It must therefore attainits minimum at an interior point. The minimum modulus theorem forces that minimum value tobe zero. So P (z) has at least one zero (or root) in any sufficiently large disc. The proof thatone zero implies exactly n zeros, counting multiplicities, where n is the degree of P (z), follows thestandard division algorithm. (The minimum principle in this proof can be rewritten to eliminateany reference to analytic functions or power series.)

We now turn our attention to harmonic functions of two real variables. We already know thatreal and imaginary parts of complex analytic functions are harmonic, in fact conjugate harmonic,and that such harmonic functions possess the mean-value property according to Corollary 5.6. Itfollows that such harmonic functions also obey a maximum principle and a minimum principle onaccount of the mean-value property. However, we will not restrict attention to harmonic functionsthat arise in this manner.

Consider the example,U(x, y) = log(x2 + y2),

which is easily verified to be harmonic in R2\0, 0. In particular, it is harmonic in the annulus1 < r < 2, where (r, θ) are polar coordinates. This U(x, y) is the real part of 2 log z, z = x+ iy,which has a multi-valued imaginary part. So the conjugate harmonic function,

V (x, y) = 2θ +K = 2arg(x+ iy) +K,

where K is a real constant, is not harmonic in the same domain as U(x, y). It is harmonic on theinfinitely many sheeted Riemann surface for log z. The polar coordinate θ is not restricted to ahalf-open 2π-interval. Instead, it runs through all the real numbers.

Let us define harmonic and related terms more precisely.

87

Definition 7.4. A real-valued function U(x, y) of two real variables is harmonic in a regionD ⊂ R2 if it is continuous in D, possesses pointwise partial derivatives Ux, Uy, Uxx and Uyy in D,and satisfies Laplace’s equation,

∇2U = Uxx + Uyy = 0,

in D. (A complex-valued function of two real variables is harmonic in D if its real and imaginaryparts are harmonic separately.)

Definition 7.5. If D is a simply connected region in R2 and U(x, y) is a twice-differentiableharmonic function in D, then V (x, y), defined up to an additive constant by the Cauchy-Riemannequations,

Vx = −Uy, Vy = Ux,

is also twice-differentiable and harmonic in D and is called the conjugate harmonic functionof U . (Then −U is the conjugate harmonic function of V .)

Definition 7.6. A real-valued function of two real variables U(x, y) is subharmonic in a regionD ⊂ R2 if it is continuous and satisfies

U(x0, y0) ≤ 1

2π

∫ 2π

0U(x0 + r cos θ, y0 + r sin θ) dθ,

whenever (x0, y0) ∈ D and the circle with equation, x = x0 + r cos θ, y = y0 + r sin θ, and itsinterior belong to D.

Lemma 7.7. In terms of the two real variables x and y, where z = x+ iy, an analytic functionf(z) and its real and imaginary parts are harmonic and its modulus |f(z)| is subharmonic.

Proof. We use the fact that an analytic function is at least twice continuously differentiable. Letz = x+ iy and f(z) = u(x, y) + iv(x, y). Then u and v satisfy the Cauchy-Riemann equations,

ux = vy, uy = − vx .

Since we can take another derivative and the second derivatives are continuous, we get

∇2u := uxx + uyy = vyx − vxy = 0,

and, similarly, ∇2v = 0. Thus, in a region D of the xy-plane corresponding to a domain ofanalyticity of f(z), u(x, y) and v(x, y) are twice continuously differentiable (in fact, differentiableto all orders) and satisfy Laplace’s equation. In other words, they are conjugate harmonic functionsin D. Also f(z), regarded as a complex-valued function of x and y, satisfies Laplace’s equation andis therefore harmonic.

We have proved that f(z) obeys the mean-value property:

f(z0) =1

2π

∫ 2π

0f(z0 + reiθ) dθ, 0 ≤ r ≤ ρ.

Apply the triangle inequality for Riemann integrals:

|f(z0)| ≤1

2π

∫ 2π

0

∣

∣f(z0 +Reiθ)∣

∣ dθ.

This expresses the fact that |f(z)| is subharmonic in D.

88

Our definition of harmonic does not require U(x, y) to be differentiable once or to have direc-tional derivatives in directions not parallel to the axes. However, we intend to show that harmonicfunctions are necessarily real analytic, and so they are differentiable to all orders in D and arelocally the sums of their own convergent Taylor series. This occurs because harmonic functionshave an integral formula, namely, Poisson’s integral formula, analogous to Cauchy’s integralformula for analytic functions. They also possess the mean-value property, which we have shownholds at least for harmonic functions that occur as real or imaginary parts of analytic functions.Conversely, we will see that the mean-value property implies harmonic. Hence, harmonic is thecase of equality in the definition of subharmonic.

In the definition of conjugate harmonic, we imposed two additional hypotheses. We neededU(x, y) to be twice differentiable to guarantee the existence and equality of the mixed derivativesUyx = Uxy. We needed D to be simply connected to avoid the possibility of V being multi-valued.

Cauchy’s integral formula can be adapted to harmonic functions of two real variables, at leastfor certain simple contours such as circles and rectangles, and any Jordan curves whose insides canbe conformally mapped explicitly to a disc. (Even rectangles are not so simple in this context,since the conformal map from the inside of a rectangle to a disc involves elliptic functions.) Itprovides a solution of the Dirichlet problem for Laplace’s equation inside a circle or a Jordancurve. The Dirichlet problem for an elliptic partial differential equation for u(x), x ∈ Rn, is to solvethe equation for u(x) in a given bounded solid region in Rn having a piecewise smooth boundaryhypersurface (not necessarily connected) when u(x) is specified on the boundary. (Unboundedregions can be handled by imposing suitable conditions at infinity.) In the corresponding Neumann

problem, the normal derivative of u is specified on the boundary.

Let the domain of interest be the closed disc x2 + y2 ≤ R2. According to the bijection, C → R2,z 7→ x+ iy, we use symbols for regions and curves in C and R2 interchangeably. So the interiorof the disc can be denoted D(R), the closed disc [D](R), and the bounding circle C(R), accordingto our earlier notation, with the centre understood to be the origin. Let C denote C(R) with thepositive orientation.

Suppose u(x, y) is harmonic inside D(R) and continuous onto C from within. Restrict attentionat first to harmonic functions that occur as real parts of analytic functions, so that

f(z) = U(x, y) + iV (x, y), z = x+ iy,

where f(z) is analytic in D(R) and continuous onto C from within. Cauchy’s integral formula gives

f(z) =1

2πi

∫

C

f(ζ)

ζ − zdζ =

1

2π

∫ 2π

0

Reiθ

Reiθ − zf(Reiθ) dθ,

for z inside C. If z is outside C, the Cauchy integral is zero. Suppose z is inside C. Then R2/z isoutside C. So also is R2/z, where the bar denotes complex conjugation. Hence,

0 =1

2π

∫ 2π

0

Reiθ

Reiθ −R2/zf(Reiθ) dθ =

1

2π

∫ 2π

0

(

1 − Re−iθ

Rr−iθ − z

)

f(Reiθ) dθ.

Subtracting the last two equations and rearranging, we get

f(z) =1

2π

∫ 2π

0Re

Reiθ + z

Reiθ − z

f(Reiθ) dθ.

89

We have obtained a real kernel. Taking real and imaginary parts, we get

U(x, y) =1

2π

∫ 2π

0Re

Reiθ + x+ iy

Reiθ − x− iy

U(R cos θ,R sin θ) dθ,

whenever x2 + y2 < R2, and a similar equation for V (x, y). This is Poisson’s integral formula

for U(x, y) in the disc D(R). It expresses U(x, y) in the disc in terms of its boundary values andplays the same role for harmonic functions as Cauchy’s integral formula on a circle does for analyticfunctions. A simple translation z → z− z0 will move the disc to any desired location in the domainwhere f(z) is analytic. The special case (x, y) = (0, 0) states that U(x, y) obeys the mean-valueproperty.

So far, Poisson’s integral formula has been derived for harmonic functions that arise as realparts of analytic functions. This restriction will be lifted shortly. Before doing so, we observe thatV (x, y) in the disc can also be expressed in terms of U(x, y) on the circle. If we drop the realpart on the kernel, we again have a complex analytic function of z, being a Cauchy integral plus aconstant. It has the same real part as f(z). Hence,

f(z) =1

2π

∫ 2π

0

Reiθ + z

Reiθ − zU(R cos θ,R sin θ) dθ + iK,

for |z| < R and some real constant K. Putting z = 0 gives f(0) = U(0, 0) + iK, so K = V (0, 0).Taking imaginary parts gives

V (x, y) =1

2π

∫ 2π

0Im

Reiθ + x+ iy

Reiθ − x− iy

U(R cos θ,R sin θ) dθ + V (0, 0),

whenever x2 + y2 < R2.

Theorem 7.8. Harmonicity of Poisson integrals. Let G(θ) be a Riemann-integrable functionof the real variable θ on the interval [0, 2π] with G(2π) = G(0). Let z denote both x+ iy ∈ C and(x, y) ∈ R2, and similarly for z0, and let D(z0, R) denote the open disc |z − z0| < R in both C

and R2 and let C(z0, R) denote its circular boundary. For (x, y) ∈ D(z0, R), form the Poisson

integral:

U(x, y) :=1

2π

∫ 2π

0Re

Reiθ + z − z0Reiθ − z + z0

G(θ) dθ.

Then U(x, y) is real-analytic and harmonic in the disc D(z0, R), and is the real part of a complexanalytic function in the disc. If, in addition, G(θ) is continuous at a point θ0 ∈ [0, 2π] (where rightcontinuity at 0 implies left continuity at 2π and conversely), then U(x, y) tends to the limit G(θ0)as (x, y) → (x0 + R cos θ0, y0 + R sin θ0) along any path in the open disc D(z0, R). In particular,if G(θ) is continuous on [0, 2π], then U(x, y) extends to a continuous function on the closed disc[D](z0, R) and G(θ) is its boundary value at the point (x0 +R cos θ, y0 +R sin θ).

Proof. The theory of Poisson integrals is similar to the theory of Cauchy integrals. We candifferentiate them under the integral sign to all orders and form two-variable Taylor series. Soany U(x, y) formed by a Poisson integral possesses continuous partial derivatives to all orders inD(z0, R).

In particular, we can form the first and second derivatives of U(x, y) and verify directly that

Uxx + Uyy = 0, Uyx = Uxy,

90

for (x, y) ∈ D(z0, R). So U(x, y) is harmonic. In fact, the harmonicity of U(x, y) follows from theharmonicity of the kernel,

Re

Reiθ + z − z0Reiθ − z + z0

=R2 − (x− x0)

2 − (y − y0)2

R2 + (x− x0)2 + (y − y0)2 − 2R((x− x0) cos θ + (y − y0) sin θ),

for each fixed θ, which can be verified by direct substitution into Laplace’s equation. Also, theproof that U(x, y) is the real part of a complex analytic function follows by erasing the operation“Re ” in the kernel.

In addition h(z) := Ux−iUy is a complex analytic function in D(z0, R), because Ux and −Uy satisfythe Cauchy-Riemann equations in D(z0, R). This shows that Ux and Uy are real-analytic, and soU(x, y) itself is real-analytic because D(z0, R) is simply connected. In fact, a direct calculationshows that U(x, y) is the real part of a complex analytic function f(z) = U + iV , where V is givenby

V (x, y) :=1

2π

∫ 2π

0Im

Reiθ + z − z0Reiθ − z + z0

G(θ) dθ + K,

where K is constant.

The proof that G(θ0) is the boundary value of U(x, y) at a point of continuity θ0 of G(θ) is lessstraightforward. The method shown here is universal and applicable to many solutions of boundaryvalue problems given by integral formulas, such as the solution of the initial-value problem for theheat equation.

In D(z0, R), introduce polar coordinates x = x0 + r cosφ, y = y0 + r sinφ. Let G(θ) be periodicallyextended with period 2π. Then the Poisson integral reads

U(x0 + r cosφ, y0 + r sinφ) =1

2π

∫ 2π

0

R2 − r2

R2 + r2 − 2Rr cos(θ − φ)G(θ) dθ,

U(x0 + r cosφ, y0 + r sinφ) −G(θ0) =1

2π

∫ θ0+π

θ0−π

R2 − r2

R2 + r2 − 2Rr cos(θ − φ)

(

G(θ) −G(θ0))

dθ.

Let arbitrary ǫ > 0 be given. The continuity of G(θ) at θ0 implies the existence of δ > 0, dependingon ǫ, such that

∣

∣G(θ) −G(θ0)∣

∣ < ǫ/3

whenever |θ − θ0| < δ. Split the interval of integration into

∫ θ0+π

θ0−π=

∫ θ0−δ

θ0−π+

∫ θ0+δ

θ0−δ+

∫ θ0+π

θ0+δ.

The triangle inequality for Riemann integrals gives, for 0 ≤ r < R,

1

2π

∣

∣

∣

∣

∫ θ0+δ

θ0−δ

∣

∣

∣

∣

<1

2π

∫ θ0+δ

θ0−δ

R2 − r2

R2 + r2 − 2Rr cos(θ − φ)

ǫ

3dθ

<ǫ

6π

∫ 2π

0

R2 − r2

R2 + r2 − 2Rr cos(θ − φ)dθ

=ǫ

3.

91

This bound is uniform throughout D(z0, R).

Next, consider the interval [θ0 + δ, θ0 +π]. The hypothesis that G(θ) is Riemann integrable impliesthat it is bounded. Hence, there is a uniform bound M such that

0 ≤∣

∣G(θ) −G(θ0)∣

∣ ≤M,

for all θ ∈ [0, 2π]. As the point (r, φ) in polar coordinates approaches the boundary point (R, θ0)along any path in the open disc D(z0, R), it must eventually enter the region Aǫ given by

R− δ1 < r < R, θ0 − δ/2 < φ < θ0 + δ/2,

where δ1 is yet to be determined, and stay in that region until it arrives at the boundary point(R, θ0). Hence, in the region Aǫ,

−1 ≤ cos(θ − φ) < cos(δ/2),

R2 + r2 − 2Rr cos(θ − φ) ≥ R2 sin2(δ/2),

uniformly for θ ∈ [θ0 + δ, θ0 + π]. The triangle inequality gives the bound,

1

2π

∣

∣

∣

∣

∫ θ0+π

θ0+δ

∣

∣

∣

∣

< MR2 − r2

R2 sin2(δ/2).

The bound depends on r and ǫ and contains the factor R2 − r2. We may now choose δ1 > 0,depending on ǫ, such that

MR2 − r2

R2 sin2(δ/2)<

ǫ

3,

whenever R− δ1 < r < R. So the second integral is bounded by ǫ/3 whenever the point (r, φ) is inthe region Aǫ.

The third integral is handled similarly. The triangle inequality for real numbers gives

∣

∣U(r, φ) −G(θ0)∣

∣ <ǫ

3+ǫ

3+ǫ

3= ǫ,

whenever (r, φ) lies in the region Aǫ having the boundary point (R, θ0). This completes the proofthat U(x0 + r cosφ, y0 + r sinφ) → G(θ0) as (r, φ) → (R, θ0) along any path in D(z0, R). The finalstatement in the theorem is a straightforward consequence of the uniform continuity of G(θ) on[0, 2π].

Theorem 7.9. Real analyticity of harmonic functions. All harmonic functions are real-analytic on their domains of harmonicity and obey Poisson’s integral formula and the mean-valueproperty on closed circular discs interior to their domains. In particular, harmonic functions areuniquely determined in a disc by their values on the boundary.

Proof. Let u(x, y) be harmonic in a region E according to the minimalist hypotheses given inDefinition 7.4. Let D(z0, ρ) be a circular disc whose closure is interior to E, z0 denoting the point(x0, y0). Let w(x, y) be the real-analytic harmonic function determined by the Poisson integral overthe values of u(x, y) on the circle C(z0, ρ). So u(x, y) and w(x, y) are both harmonic in D(z0, ρ) andcontinuous onto the boundary from within, and are equal to each other on the boundary. Considerthe continuous function,

f(x, y) = u(x, y) − w(x, y) + ǫ(x− x0)2,

92

on the closed disc [D](z0, ρ), where ǫ > 0. This function takes its maximum somewhere on the closeddisc. If it takes its maximum at an interior point (x1, y1), then fxx(x1, y1) ≤ 0 and fyy(x1, y1) ≤ 0,which would imply ∇2f ≤ 0 at (x1, y1). But a direct calculation gives ∇2f = 2ǫ > 0. This contra-diction proves that f(x, y) takes its maximum on the boundary C(z0, ρ), along which f(x, y) ≤ ǫρ2.Since ǫ can be taken arbitrarily small, we conclude that u(x, y) ≤ w(x, y) on the closed disc. Byconsidering instead g(x, y) = w(x, y) − u(x, y) + ǫ(x− x0)

2, we reach the opposite conclusion thatu(x, y) ≥ w(x, y). This completes the proof that u(x, y) ≡ w(x, y) on the closed disc. So harmonicfunctions are uniquely determined by their boundary values on a circular disc and are given byPoisson’s integral formula. The mean-value property is a special case of Poisson’s integral formula.

Theorem 7.10. Maximum principle for harmonic and subharmonic functions. Suppose thatU(x, y) is harmonic or subharmonic in a bounded region D ⊂ R2 and continuous on its closure [D].Then the maximum of U(x, y) on [D] is attained on the boundary ∂D. If the maximum is alsoattained at an interior point of D, then U(x, y) is constant on [D]. In the case of harmonic functions,both the maximum and minimum are attained on the boundary.

Proof. In the case where U(x, y) is subharmonic, we follow the corresponding argument of themaximum modulus theorem for f(z), based on the subharmonic property of |f(z)|. The argumentis identical. In the case of harmonic functions, the mean-value property implies that harmonicfunctions in discs are subharmonic (equality instead of non-strict inequality). This proves themaximum principle for harmonic and subharmonic functions. But, also in the harmonic case, wecan apply the maximum principle to −U(x, y), thereby proving the minimum principle as well.

Theorem 7.11. Converse of mean-value property. Any continuous function of two realvariables on a region D ⊂ R2 that satisfies the mean-value property,

U(x0, y0) =1

2π

∫ 2π

0U(x0 + r cos θ, y0 + r sin θ) dθ,

for every (x0, y0) ∈ D and any particular positive r (depending on (x0, y0)) such that the closeddisc of radius r and centre (x0, y0) is interior to D, is harmonic in D.

Proof. Name the closed disc [D](z0, r), where z0 denotes (x0, y0). The mean-value propertyimplies that U and −U both obey the maximum principle on this disc. Use the Poisson integralto construct a real-analytic harmonic function V (x, y) having the same boundary value on C(z0, r)as U(x, y). Then V (x, y) is continuous onto C(z0, r) from within, and so is U − V . The functionU − V also has the mean-value property and therefore obeys both the maximum and minimumprinciples on the disc. But U − V is zero on the boundary C(z0, r). Hence, U − V is identicallyzero in the closed disc [D](z0, r). So U(x, y) is harmonic in the open disc D(z0, r) and continuousonto its closure. Since every interior point of D is the centre of such a disc, it follows that U(x, y)is harmonic in D.

Some textbooks prefer to use the mean-value property as their definition of harmonic. Fromthere, they deduce theorems that harmonic functions are real-analytic, satisfy Laplace’s equationand Poisson’s integral formula, and obey maximum and minimum principles.

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