PartA Analysis 2011Lectures

7/31/2019 PartA Analysis 2011Lectures

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Part A: Analysis Complex Analysis

Michaelmas Term, 2011Mathematical Institute, Oxford

November 25, 2011


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Contents

1 Topology of Euclidean spaces 11.1 Topology on R n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Topology on a subspace of R n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4 Bounded sets, compact subsets in R n . . . . . . . . . . . . . . . . . . . . . . . . . 61.5 Connected and path-connected sets . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2 Holomorphic functions 112.1 Power series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.2 Exponential and trigonometric functions . . . . . . . . . . . . . . . . . . . . . . . 142.3 The logarithm function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.4 Complex power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.5 Path integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.6 The Cauchy Theorem and The Cauchy Integral Theorem . . . . . . . . . . . . . . 23

2.6.1 The Cauchy Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.6.2 The Cauchy Theorem: other versions . . . . . . . . . . . . . . . . . . . . . 242.6.3 The Cauchy Integral Formula . . . . . . . . . . . . . . . . . . . . . . . . . 282.6.4 Some Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.7 Taylors Theorem and Laurent expansions . . . . . . . . . . . . . . . . . . . . . . 352.8 Isolated singularities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392.9 The Residue Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442.10 Contour integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

3 Conformal mappings 633.1 Some examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 643.2 Riemann mapping theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

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iv CONTENTS

Introduction The Cauchy-Riemann equations

These are lecture notes for Part A Analysis, based on the course and lecture notes designedby previous lecturers, in particular Glenys Luke, whose revised notes are posted on the course

web page for additional reading.In Mods Analysis, we have studied convergence of sequences and series, continuity and dif-ferentiability of real functions of one variable, and we have studied the theory of Riemannsintegration for functions of one real variable . In Part A Option Multi-Calculus, we will studycontinuity and differentiability of vector valued functions of several (real) variables .

In this part of Analysis, we will investigate carefully the meaning of differentiability of acomplex function of one complex variable. We are going to prove several surprising results . Forexample, if f is differentiable on the unit disk D of the complex plane, then its derivative f isalso differentiable, and all high order derivatives f (n ) exist. In particular, f is continuous.

It is necessary to develop a new machinery in order to prove the preceding simple statement.Indeed it is the goal of this course to develop a large array of mathematics which proves quiteuseful in both theoretical research and applied areas such as uid dynamics.

I will use the standard notions and notations in current literature. R and C denote the eldof real numbers and the eld of complex numbers. A complex number z = x + iy, where x, yare real numbers, x = Re z is called the real part of z, y = Im z is called the imaginary part of z. (x, y) is called the real coordinates of z, and z is called the complex coordinate of the point(x, y). z = x iy is the conjugate of z, and |z| = x2 + y2 is called the modulus which measuresthe distance between z and 0. We note the relations that x = 12 (z + z) and y = 12i (z z).A subset G C is open if for every point z G there is a small disk about z, B r (z) G(for some r > 0), where B r (z) = {w : |w z| < r }.In what follows, it is assumed that G is non-empty open subset of C . If f : G

C , then one

can writef (z) = f (x, y) = u(x, y) + iv(x, y) for z = x + iy G

where u(x, y) (resp. v(x, y)) the real (resp. imaginary part of the complex number f (z)), so thatu and v are real functions of two variables. The partial derivatives ux etc., if exist, are denoted byux etc. If u and v have partial derivatives [in x and in y respectively], then the partial derivativeof f exist, and

f x

= ux + ivx ,f y

= uy + ivy.

Now suppose that f (z) exists at a point z = x + iy G, that is, the limit

f (z) =d

dz f (z) = limh0f (z + h)

f (z)

h (0.0.1)

exists, i.e. f is [complex] differentiable at z, then we particularly have

f (z) = limh0h R

f (z + h) f (z)h

= limh0h R

u(x + h, y) u(x, y) + i (v(x + h, y) v(x, y))h

=f x

= ux + ivx


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CONTENTS v

exists. Similarly we also have

f (z) = limh0h R

f (z + ih) f (z)ih

=

1

i limh0h Ru(x, y + h)

u(x, y) + i (v(x, y + h)

v(x, y))

h

= if y

=1i

uy + vy = vy iu y.It follows that

ux + ivx = vy iu y.Proposition 0.0.1 If f has [complex] derivative at a point z = x + iy G, then

1) all partial derivatives ux , uy , vx and vy at (x, y) exist;2) the Cauchy-Riemann equations hold

ux = vy, uy = vx ; (0.0.2)3) we have

f (z) = f x

= if y

=12

f x i

f y

. (0.0.3)

Denition 0.0.2 If G C is an open set, then f is holomorphic on G if f (z) exists for every z G, if A C is a subset then when we say f is holomorphic on A if f is holomorphic on an open set G containing A. In particular, when we say f is holomorphic at a point a C , we

means that there is a disk B r (a) for some r > 0, such that f is holomorphic on B r (a).Let us look at the equality (0.0.3) from another angle. Let z = x + iy and z = x iy so thatx = 12 (z + z) and y = 12i (z z). Now let us forget the relations between z and its conjugate z,and consider z and z as independent variables in C , apply the chain rule formally to f , to dene

z and

z as the following

z

f =f x

xz

+yz

f y

=12

x i

y

f

and

z f = x z x f + y z y f = 12 x + i y f .

We then consider these equations as denitions of the partial differential operators of rst order:

z and

z . That is

z=

12

x i

y

,

z=

12

x

+ i

y. (0.0.4)

Example . 1) f (z) = zn and g(z) = zm (where n, m N). Then z f (z) = nzn1 and

z f (z) = 0, and

z g(z) = 0 and

z g(z) = mz

m1.2) If f (z) = zn zm then z f (z) = nz

n1zm and z f (z) = mz n zm1.


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vi CONTENTS

Proposition 0.0.3 If f has partial derivatives, then f z = 0 is equivalent to the Cauchy-Riemann equations. Hence, if f has a (complex) derivative, then f (z) = f z (z) and

f z (z) = 0 .

Proof. Work out

z f =

12

x + i

y f

=12

x

+ i

yu +

12

i

x+ i

y

v

=12

(ux vy) + i(uy + vx )so that z f = 0 if and only if the Cauchy-Riemann equations hold.

Conversely, we have

Theorem 0.0.4 If G is an open subset of C and f = u + iv such that ux , uy, vx , vy exist and continuous on G, and if the Cauchy-Riemann equations hold: ux = vy , uy = vx on G(equivalently z f = 0 on G), then f is holomorphic in G.

Proof. We show that, for every z G, the complex derivative f (z) exists. Let z G. Thenthere is > 0 such that z + h G for any |h| . Write h = h1 + ih 2 and z = x + iy. Supposez + h G then

f (z + h) f (z) = u(x + h1, y + h2) u(x, y)+ i(v(x + h1, y + h2) v(x, y)).

Consider g(t) = u(x + th 1, y + th 2) u(x, y) t [0, 1].Then g : [0, 1] R is continuous, and differentiable on (0 , 1) with

g(t) = ux |(x+ th 1 ,y+ th 2 ) h1 + uy|(x+ th 1 ,y+ th 2 ) h2 t (0, 1).Thus MVT (Analysis II)

g(1) g(0) = g(t1)for some t1 (0, 1). That is

u(x + h1, y + h2) = u(x, y) + h1ux (x + t1h1, y + t1h2)+ h2vy(x + t1h1, y + t1h2)

= u(x, y) + ux (x, y)h1 + uy(x, y)h2 + 1(h)

where

1(h1, h2) = ( ux (x + t1h1, y + t1h2) ux (x, y)) h1+

uy (x+ t 1 h1 ,y+ t 1 h2 )

uy (x,y )

h2.


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CONTENTS vii

Since ux anduy are continuous, so that

limh0

1(h)h

= limh0

ux (x+ t1 h1 ,y+ t1 h2 ) ux (x,y ) h1

h

+ limh0

uy (x+ t1 h1 ,y+ t1 h2 )

uy (x,y ) h2

h= 0

as h jh 1. Similarly there is a t1 (0, 1) such thatv(x + h1, y + h2) v(x, y) = vx (x, y)h1 + vy(x, y)h2 + 2(h)

and limh02 (h)

h = 0.Hence

f (z) = limh0

f (z + h) f (z)h

= limh0

uxh1 + uyh2 + ivxh1 + ivyh2h

= limh0

uxh1 + iu xh2 + ivxh1 vxh2h

= limh0

uxh + ivxhh

= ux + ivx

exists.Actually the continuity assumption on the partial derivatives is redundant, and only existence

of partials at any point of G and the Cauchy-Riemann equations are necessary.

Theorem 0.0.5 (Looman-Menchoff) If f = u + iv : G C is continuous, if all partial deriva-tives ux , vx , uy , vy exist on G, and if the Cauchy-Riemanns are satised: ux = vy , uy = vx ,then f is holomorphic in G.[This Theorem is not examinable] The proof is much more difficult, see pages 43, R. Narasimhan:

Complex Analysis in One Variable. See also Theorem 2.6.14.


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Chapter 1Topology of Euclidean spaces

[Week 1 Lecture 1]In order to study the limit

f (z) = limh0

f (z + h) f (z)h ,the derivative of a complex function f , we need to know some features of subsets in the complexplane which are signicantly complicated than those for the real line.

We begin with some common terminologies used by all mathematicians, such as open sets,closed sets, limiting points etc. If A R , a point x A is an interior point of A if (x, x + ) Afor some > 0. A subset U is open if all points of U are interior points of U . An interval(x , x + ) can be described as {y R : d(x, y) < }, where d(x, y) = |x y| is the distancebetween x and y. d is called a distance function (or called a metric ) on R . Thus, in order todene the concept of open sets, for example in a complex plane

C, we only need a distancefunction on C , nothing else such as the eld or the vector space structure is relevant.

For the benet of other options as well, let us study the topology of Euclidean spaces R n .

1.1 Topology on R n

If n is a natural number, R n is the set of all ordered n-tuples ( x1, , xn ) where all coordinatesx i are real numbers. R n is an example of vector spaces in Linear Algebra, but we will not use itsvector structure , we will only use the fact that each point x = ( x1, , xn ) in R n , x i are reals.Dene distance function d : R n

R n

[0,

) by

d(x, y) = (x1 y1)2 + + ( xn yn )2. (1.1.1)for x = ( x1, , xn ) and y = ( y1, , yn ), which is the distance between x and y. In terms of summation sign

d(x, y) = ni=1

(xi yi)2. (1.1.2)

d(x, y) is called the Euclidean distance between x and y, and is also denoted by |x y|.1


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2 CHAPTER 1. TOPOLOGY OF EUCLIDEAN SPACES

Example . As sets, C may be identied with the two dimensional Euclidean space R 2 byz = x + iy, i.e. z (x, y). If z1 = x1 + iy1 and z2 = x2 + iy2 then

d(z1, z2) = (x1 x2)2 + ( y1 y2)2= |z1 z2|which is the modulus (absolute value) of z1 z2.Proposition 1.1.1 The distance function d possesses the following properties.

1) d(x, y) 0, and d(x, y) = 0 if and only if x = y.2) d(x, y) = d(y, x) for any x, y R n .3) The triangle inequality holds:

d(x, z ) d(x, y) + d(y, z) x,y,z R n .Proof. 1) and 2) are obvious. 3) is of course the most important. Let us rst show Cauchy-

Schwartz inequality:n

i=1

a ibi ni=1

a2i ni=1

b2i . (1.1.3)

If both a2i and b2i are zero, then (1.1.3) is trivial, so assume thus that a2i > 0. Since forany real

(a i + bi)2 0,so that by working out the squares and rearranging terms, one has

( a2i )2 + 2( a ibi) + ( b2i ) 0 R .Therefore the discriminate

4( a ibi)2 4( a2i )( b2i ) 0which is equivalent to (1.1.3).

It follows from (1.1.3) that

n

i=1 (ai + bi)

2

n

i=1 a2i +

n

i=1 b2i

2

hence

ni=1

(a i + bi)2 ni=1

a2i + ni=1

b2i . (1.1.4)

Setting a i = x i yi and bi = yi zi (so that a i + bi = x i zi) we obtain the triangle inequality.Any function d on R n R n satisfying conditions 1-3) is called a distance function on R n .


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1.1. TOPOLOGY ON R N 3

Remark 1.1.2 As a vector space, R n may be equipped with a norm

||x||= x21 + + x2ncalled the Euclidean norm on R n . The distance d(x, y) = ||x y||. Clearly ||x|| 0 and ||x||= 0if and only if x = 0 . The triangle inequality may be stated as

||x + y| || |x||+ ||y|| .In addition, ||cx||= |c|||x|| for any scalar c and x R n .Remark 1.1.3 There are many other distance functions on R n one can use. Here are someexamples.

1) d(x, y) = ||x y|| where ||x|| = max {|x1|, , |xn |}is the L-norm.2) d1(x, y) = ||x y||1 where ||x||1 = |x1|+ + |xn | is the L1-norm.3) For p

1, d p(x, y) =

||x

y

|| p where

||x

|| p = p

|x1

| p +

+

|xn

| p called the L p-norm.

We can now dene the topology of R n : the collection of open subsets in R n . Let r > 0 anda R n . The open ball B r (a) centered at a with radius r is the subset {x R n : d(x, a ) < r }.Denition 1.1.4 A subset U R n is open (in R n ) if for every a U there is > 0 such that B(a) U .

Lemma 1.1.5 B r (a) is open, where a R n and r > 0.

Proof. If x B r (a) then d(x, a ) < r . Let = rd(x, a ). Then > 0 and for any y B(x)d(y, a ) d(y, x) + d(x, a ) [Triangle Ineq.]< + d(x, a )

= r ,

hence B(x) B r (a). B r (a) thus is open by denition.

Lemma 1.1.6 If U R n is open then U is a union of some open balls.

Proof. For each a U there is a > 0 such that Ba (a) U . Clearly we have

U = a U Ba (a).

Proposition 1.1.7 The collection of all open sets (together with the empty set) in R n is a topology in the following sense:

1) Empty set is open [you may considered it as part of the denition]. The whole space R nitself is open.

2) The union of some (any many) open sets is open.3) The intersection of nite many open sets is again open.


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Proof. 2) Suppose U = U where all U R n are open. For a U then a U forsome . Since U is open, there is > 0 such that B(a) U U , so that U is open.

3) Suppose U 1, , U n are open sets in R n , and U = ni=1 U i . If a U then a U i fori = 1 , , n . Since each U i is open, there is i > 0 such that B i (a) U i , i = 1 , , n . Let = min {1, , n}. Then > 0 and B(a) Bi (a) U i for all i. Hence B(a) ni=1 U i andthus U is open.You should think about, why the proof of 3) does not work for innite many open sets.

For an abstract space in place of R n , items 1) - 3) become the dening properties of a topology.It is the topology which is needed to describe the continuity of mappings between spaces.

Denition 1.1.8 A subset F R n is closed (in R n ) if its complement F c = R n \ F is open.Due to de Morgans law, we can translate the properties of topology for closed sets.

Proposition 1.1.9 The following properties hold:

1) Both empty set and the whole space Rn

are closed.2) The intersection of some closed subsets is closed.3) The union of nite many closed sets is again closed.

There are many subsets which are neither open nor closed. For example, [0 , 1](0, 1) is suchan example in R 2.Let A R n , and a R n . We say a is a limiting or accumulation point of A if for every > 0,

B(a) (A\{a}) = . Note that an accumulation point of A may not be in A.Proposition 1.1.10 A subset F R n is closed, if and only if all accumulation points of F

belong to F . [Exercise].

1.2 Topology on a subspace of R n

[Week 1, Lecture 2]The distance function d on R n restricted on a subset X R n is a distance function on X :

1. d(x, y) 0 for any x, y X , and d(x, y) = 0 if and only if x = y.2. d(x, y) = d(y, x) for x, y X .

3. d(x, y) d(x, z ) + d(z, y) for any x,y,z X .For r > 0 and a X , the ball (in X ) centered at a with radius r is

B Xr (a) = {x X : d(x, a ) < r }= X B r (a). (1.2.1)

The topology on X (under the distance d) can be constructed in terms of balls in X . Namely,a subset U X is open (in X ) if for each a U there is > 0 such that B X (a) X .


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1.3. CONTINUITY 5

Proposition 1.2.1 The collection of all open sets in X is called the sub-space topology on X R n , which satises the properties of topology:

1) The empty set and the whole space X are open (in X ).2) The union of some open sets is open.3) The intersection of nite many open sets is open.

A subset F X is closed (in X ) if its complement (relative to X ) F c = X \ F is open (inX ).Proposition 1.2.2 U X is open in X if and only if U = X V for some open set V in R n .Similarly F X is closed in X if and only if F = X B for some closed set B in R n .

Proof. If U X is open in X , then for each a U there is a > 0 such that

B Xa (a) = X Ba (a) X .Let V =

a U B

a(a). Then V is open in R n and

U = a U B Xa (a) = a U (X Ba (a))= X V .

Conversely, suppose U = X V for some open set V in R n , then for any a U = X V , sinceV is open, there is > 0 such that B(a) V , so thatB X (a) = X B(a) X V = U .

1.3 ContinuityLet X R n . X together with its distance function d is a metric space, written as ( X, d).

Denition 1.3.1 Let f : X R m be a mapping [such a mapping is called a function valued in R m with domain X R n ], and a X . We say f is continuous at a [along X ] if for every > 0there is > 0 such that

d(f (x), f (a)) < for x X and d(x, a ) < . (1.3.1)

f : X R m is continuous if it is continuous at any a X .Lemma 1.3.2 f : X R m is continuous [on X ] if and only if for any a X and > 0, thereis > 0 such that

B X (a) f 1(B(f (a))) . (1.3.2)

Proof. This is follows from (1.3.1) which says that f (x) B(f (a)) [which is equivalent tothat x f 1(B(f (a))) by denition] as long as x B X (a).

The following theorem shows the continuity of mappings may be formulated in terms of thetopology on R m (target space) and the subspace topology on X (the domain).


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Theorem 1.3.3 Let X R n and f : X R m . Then f is continuous if and only if for any open set U R m , the pre-image f 1(U ) is open [in X ], or equivalently, if and only if for any closed subset F R m , f 1(F ) is closed in [in X ].

Proof. Since f 1(R m \ F ) = X \ f 1(F ) so we only need to prove the open case. = part. Suppose f

1

(U ) is open in X for any open U Rm

. For any a X and > 0,B(f (a)) is open, so that f 1(B(f (a))) is open in X . Obviously a f 1(B(f (a))), hence thereis > 0 such that B X (a) f 1(B(f (a))). f is continuous at a.

= part. Suppose f is continuous. For open U R m , we show that f 1(U ) open set inX . If a f 1(U ), then f (a) U . Since U is open, there is a > 0 such that B(f (a)) U .Since f is continuous at a, there is > 0 such that (1.3.2) holds, so that

B X (a)) f 1(B(f (a))) f 1(U ).

Therefore f 1(U ) is open in X .

Remark 1.3.4 Let X Rn, f : X R

mand Y = f (X ). Then f is continuous if any only if for any open U in Y , f 1(U ) is open in X .

1.4 Bounded sets, compact subsets in R n

[Week 1, Lecture 3]A subset A R n is bounded if it is contained in a ball, that is, there is r > 0 and a R n

such that A B r (a). A is bounded if there is M > 0 such that ||x|| M for all x A.In Mods Analysis, we have proven the following important property of the real numbers.Bolzano-Weierstrass Theorem. Any bounded sequence of reals has a convergent subse-

quence.This theorem can be extended to sequences in R n . Namely

Theorem 1.4.1 Any bounded sequence in R n has a convergent subsequence.

Proof. We prove this for R 2. Let an = ( a1n , a2n ) be a bounded sequence in R 2. Then {a1n}isa bounded sequence of reals, so that there is convergent subsequence, denoted by {a1n k }. Then,{a2n k }is bounded in R , so that there is a convergent subsequence say {a2n k l }. Then {a1n k l }is asubsequence of the convergent sequence {a1n k }, so that both {a1n k l }and {a2n k l }are convergent, sois an k l = ( a

1n k l

, a2n k l ).

Theorem 1.4.2 (Cantors Intersection Theorem) Let F 1 F 2 F k be a decreasing sequence of non-empty , bounded and closed subsets in R n . Then k=1 F k is non-empty.Proof. Let F = k=1 F k . If there is N such that F k = F N for all k N , then clearlyF = N k=1 F k is non-empty. Otherwise, {F k}has strictly decreasing subsequence {F kl }, thatis {F kl } is decreasing and all F kl are different. Thus chose akl F kl but akl / F kl +1 . Since

{akl } F 1 is bounded, hence {akl }has a convergent subsequence. Without losing generality, wemay assume that akl a R n . Since for any N , akl F kl F kN for all l N , and F kN isclosed, so that a F kN , and therefore a N =1 F kN = F . F is non-empty.


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1.4. BOUNDED SETS, COMPACT SUBSETS IN R N 7

Theorem 1.4.3 Let A R n be bounded and closed. Suppose {U : }is a family of open subsets in R n such that A U [such a family {U : }is called an open cover of A ].Then there are nite many {U i : i = 1 , , k}such that A ki=1 U i .Proof. Since A is bounded, A B r (0) for some r > 0. We rst prove a weaker statement:

there is a countable many {U i : i = 1 , }such that A i=1 U i [i.e. there is a countablesub-cover]. Let U = U . If x U = U , then x U x for some index x . Since U x isopen and Q is dense in R , we can nd a cx U x with rational coordinates, and rational r x > 0such that

B r x (cx ) U x and x B r x (cx ).

HenceU = U = U B r x (cx ).

The collection of different balls B r x (cx ) (as x runs through U ) is at most countable, thereforecan be listed as

{B r i (ci) : i = 1 , 2,

}where for each i, B r i (ci) U i for some i . By construction,U = B r i (ci) i=1 U i

so that {U i : i = 1 , 2. }is a countable cover of A. Let V k = ki=1 U i . Then V 1 V 2 isan open cover of A. Suppose for any k, ki=1 U i A is not true, then F k = A \ V k are closed,bounded and non-empty, and F 1 F 2 , but k=1 F k is empty as {V k : k = 1 , 2, }is aopen cover of A, a contradiction.Denition 1.4.4 Let X R n . A family of open subsets

{U :

}is an open cover of X if

U X . {U : }is called a nite open cover if is a nite set. X is called compact,if any open cover of X possesses a nite sub-cover.Theorem 1.4.5 (Heine-Borel Theorem) A subset A R n is compact if and only if A is bounded and closed.

Proof. We have proven the If Part.Only if part. Since R n = i=1 B i(0), so that A i=1 B i(0). Thus there is nite manyi1, , ik N , such that A kl=1 B i l (0). Let M = max {i1, , ik}. Then A BM (0), A isbounded. We next prove that A is closed. Suppose A is not closed, then there is an accumulation

point a of A, such that a / A. Therefore, for every > 0, B(a) A = . Let U k =R n

\ F kwhere F k = {x : d(x, a ) 1k}for k = 1 , 2, . Thenk=1 U k = R n \ kF k = R n \ {a} A.

Since A is compact, so that there is N such that

N k=1 U k = U N = R

n \ F N A.so that {x : d(x, a ) 1N }A = which contradicts to the assumption that a is an accumulationpoint.


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Theorem 1.4.6 If A is compact and f : A R m is continuous, then f (A) is compact, hencef (A) is closed and bounded.This follows from the denition of compact sets and Theorem 1.3.3.A direct corollary is the following generalization of a result in Mod Analysis II: a continuous

function f : A Rm

is bounded if A is compact. In fact, by the previous theorem, f (A) iscompact in R m , and therefore by the Heine-Borel theorem, f (A) must be bounded, that is, f isbounded on A.

As an exercise, the reader may prove the following generalization of another result from ModAnalysis.

Proposition 1.4.7 If A is compact and f : A R m is continuous, then f is uniformly continuous on A.Proof. Let > 0. For any x A, by denition, there is x > 0 such that

B Ax (x) f 1(B/ 3(f (x))) .

Clearly {B Ax / 3(x) : x A}is an open covering of A, so that, as A is compact, there is x1, , xkA such thatk j =1 B

Ax j / 3

(x j ) = A .

Let = 13 min{x1 , , xk }. Let x, y A such that d(x, y) < . Then there are i and j suchthat x B Ax i / 3(x i) and y B Ax j / 3(x j ), andd(xi , x j ) d(xi , x) + d(x, y) + d(y, x j )

< x i3 + + x j3

< x i

so that x j B Ax i (x i), and thus

d(f (x i), f (x j )) 0 be any butxed, and set

V 1 = X [a, c) = X (a , c)and

V 2 = X (c, b] = X (c, b + ).Both V 1, V 2 are open in X [Proposition 1.2.2], non-empty (as c is neither the greatest lower boundnor least upper bound) and disjoint. Moreover X = V 1 V 2 so X is disconnected, a contradictionto the assumption.


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Denition 1.5.6 X R n is path-connected, if for any a, b R n there is a continuous mapping p : [0, 1] X (such a mapping called a path in X ) such that p(0) = a and p(1) = b.

By denition, it is obvious that any interval in R is path-connected, so the concepts of connectedness and path-connectedness for subspace of R are the same: X R is path-connected

if and only if X is an interval.Proposition 1.5.7 If X R n is path-connected, then it is connected.

Proof. If X were not connected, then X = U 1 U 2, U i are open, disjoint and non-empty.Let a U 1 and b U 2. There is a path p : [0, 1] X such that p(0) = a and p(1) = b. LetV 1 = p1(U 1) and V 2 = p1(U 2). Then V 1, V 2 are open in [0, 1], disjoint and non-empty. Moreover[0, 1] = p1(X ) = V 1 V 2, which contradicts to that [0 , 1] is connected.

However there are connected sets in R n which are not path-connected. For example X =

{(x, y) : x > 0, y = x sin 1x } {(0, 0)}is connected but not path-connected. On the other handwe haveProposition 1.5.8 If X R n is open and connected, then X is path-connected.

Proof. Actually we prove that any two points in X can be joined by a polygon which lies inX . Let a X . Let A be the subset of X whose points can be joined to a by a polygon in X .

1) A is non empty. In fact, since X is open there is > 0 such that B(a) X . Clearly anypoint in B(a) can be joined to a by a straight line in B(a). Thus B(a) A.

2) A is open in R n so is in X . Indeed, if x A, so that x is connected to a by a polygon inX . Since X is open, there is > 0 such that B(x) X . Any point in B(x) can be joined by apolygon to x hence to a. Therefore B(x) A.

Let B be the subset of X whose points can not be joined to a by a polygon. Similarly, wecan show that B is open in X [Exercise]. By denition X = A B and AB = , we thus musthave B = as A is not empty, open and closed.Denition 1.5.9 A connected open set in R n is called a region.

Therefore any region is path-connected. A region in C is also called a domain by someauthors.


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Chapter 2Holomorphic functions

[Week 2, Lecture 5]Let G stand for a non-empty open subset of C , unless otherwise specied.It is an easy exercise to verify the algebra of derivatives, as well as the chain rule for com-

position functions and the rule for inverse functions apply equally to the complex derivatives forcomplex functions:

(i) (cf )(z) = cf (z) for any c C ;(ii) (f + g)(z) = f (z) + g(z);(iii) (fg )(z) = f (z)g(z) + f (z)g(z);(iv) f g

(z) = f (z)g(z)f (z)g

(z)g(z)2 ;

(v) (f g)(z) = f (g(z))g(z).(vi) If G and D are open subsets in C , and f : G D is 1-1 and onto, so its inverse g = f 1exists. Suppose f has complex derivative on G and f (z) = 0 for all z G, then g has complexderivative on D, and

g(w) = 1f (g(w))

for w D.

By denition, f is complex differentiable at z if

f (z + h) f (z) = Ah + o(h) as h 0 (2.0.1)for some complex number A, which is equivalent to the existence of f (z), and A = f (z).In (2.0.1), the so-called complex structure of C has been used, and the differentiability incomplex context as required in (2.0.1) is much more demanding than those in the real case.

Indeed, f : G C may be identied with a mapping f : G R2

and write asf (x, y) = u(x, y) + iv(x, y)

= ( u(x, y), v(x, y))

for z = x + iy = ( x, y) G. Then f : G R 2 is differentiable in REAL SENSE, by denition, if u(x + h1, y + h2) u(x, y) = A11h1 + A12h2 + o( h21 + h22)v(x + h1, y + h2) v(x, y) = A21h1 + A22h2 + o(

h21 + h22)

11


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12 CHAPTER 2. HOLOMORPHIC FUNCTIONS

as h21 + h22 0 (where h1, h2 are reals). In this case one can identify A11 = ux , A12 = uyetc. It is a theorem (Option: Multi-Val Calculus) that if u, v have continuous partial derivativeson G, then f (x, y) = ( u(x, y), v(x, y)) is differentiable at any point in G. On the other hand,(0.0.1) implies much more: the partial derivatives of u and v must satisfy the Cauchy-Riemannequations (see below)!

The denition of f (z) looks no difference from the real case, but, as we allow h 0 incomplex plane C , and therefore z + h approaches to z (though restricted in the open set G) inarbitrary ways (in contrast with the real case in which there are only two sided limits!). As aconsequence, the very existence of the complex derivative (of rst order) in G has far reachingconsequences. For example, the existence of the rst derivative of f at every point z G impliesnot only the existence of its all higher derivatives, but also the analyticity of f at any z Gwhich says the Taylor theorem for f holds near z.

We will prove these claims by means of integration along paths .

2.1 Power series[Week 2 Lecture 6]

The basic examples of holomorphic functions are given by power series. Let us begin withthe following elementary identity.

Lemma 2.1.1 For any n 2 and w = zwn znw z

nz n1 =n1

k=1

zn1k wk zk . (2.1.1)

Proof. We have for z = 1

1 zn1 z

= 1 + z + z2 + + zn1 n 1. (2.1.2)Replace z by w/z or z/w in (2.1.2) to obtain

wn znw z

nz n1 = zn1 + zn2w + + zwn2 + wn1

zn1 zn1 zn1 zn1=

n

1

k=1

zn1kwk zn1

=n1

k=1

zn1k wk zk .

Let

f (z) =

n =0

an zn (2.1.3)


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2.2 Exponential and trigonometric functions[Week 3 Lecture 7]

The exponential function is dened by power series

exp(z) = ez

= 1 + z +z2

2 +z3

3! + +zn

n! + =

n =0

zn

n!

(with the convention 0! = 1) which converges everywhere in C , so ez is holomorphic on the entirecomplex plane C . Clearly ez = ez and e0 = 1, ex 1 for x 0.Remark 2.2.1 Holomorphic functions on C are called entire functions. Hence exp is an entire function.

The trigonometric functions are dened by

sin z = z z3

3!+ =

n =0

(1)nz2n +1

(2n + 1)!

and

cos z = 1 z2

2+

z4

4!+ =

n =0

(1)nz2n

(2n)!.

Both sin and cos are entire functions. Clearly sin( z) = sin z and cos(z) = cos z.The previous three power series look like close friends, indeed, after an inspection you maydiscover the following relations:

eiz = cos z + i sin zwhich is called the Euler formula and, by denition,

sin z = eiz eiz2i , cosz = e

iz + eiz2

.

Next, we prove the properties you expect ez , sin and cos should have.

Proposition 2.2.2 1) ez+ w = ezew for any z, w C . 2) ez = 0 and ez = 1ez for any z C .

Proof. 2) follows from 1). To show 1), we know that ez = n =0 zn

n ! converges absolutely inany disk, we can multiply ez and ew term by term

ezew =

n =0

zn

n!

m =0

wm

m!=

n,m =0

1n!m!

zn wm

=

k=0 n + m = k

1k!

k!n!(k n)!

zn wkn

=

k=0

1k!

k

n =0

k!n!(k n)!

zn wkn

=

k=0

1k!

(z + w)k = ez+ w .


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2.2. EXPONENTIAL AND TRIGONOMETRIC FUNCTIONS 15

Corollary 2.2.3 1) sin2 z + cos 2 z = 1 for all z C . 2) ez = ez . 3) |ei | = 1 for any R .Proof. Using Eulers formula

sin2 z + cos 2 z =

1

4eiz

eiz 2 + 1

4eiz + eiz 2

=12

eiz eiz + 12

eiz eiz = 1

which is 1). Since all coefficients in the power series of ez are real, so that

ez =

n =0

1n!

zn = ez

which is 2). Finally, if is real then

|ei |2 = ei ei = ei ei = 1.

In fact, ei maps the real line onto the circle S 1 = {z : |z| = 1}. That is, for eachcomplex z such that |z| = 1, there is R , such that ei = z. is called [the] argument of z.To this end, we need the following fact from Mods Analysis II.Exercise 2.2.4 In particular

sin x > x 1

x2

6x > 0 (2.2.1)

and

cos x < 1 x2

2+

x4

24x > 0. (2.2.2)

Proof. In fact, since x > 0 we have cosx 1 so thatsin x =

x

0cos xdx < x ,

integrating again to obtain

cos x 1 = x

0sin xdx >

12

x2

so that

sin x = x

0cos xdx >

x

01

12

x2 dx

= x 16

x3.


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Finally, integrating again to deduce that

cos x 1 = x

0sin xdx

0 and for k = 0 , 1, 2, .Lemma 2.2.5 There is a unique x0 (0, 3) such that cos x0 = 0 and sin x0 = 1 . Dene = 2 x0.

Proof. We have cos 0 = 1 and

cos3 < 1 32

2+

3424

= 1 32 + 924=

18

< 0.

Therefore, by IVT (Mods Analysis), there is a root x0 of cos in (0, 3). Sincecos x = sin x < x 1

x2

6

< 12

x x (0, 3)cos is strictly decreasing on (0 , 3), so that x0 is the unique root of cos on (0, 3). Since

sin x > x 1 x2

6>

12

x > 0, x (0, 3)so that sin x0 > 0, and therefore

sin x0 = 1 cos2 x0 = 1.


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w is called (the) logarithm of z.There are innite many solutions for z = 0, we must be clear which one of logz stands for.

In particular, if a is a positive number, the logarithm of a, is the collection of all loga + 2 kiwhere k runs through Z . However, as a rule, if a is a positive real number , then log a denotes theunique real number such that elog a = a, unless otherwise specied.

We aim to dene a logarithm function z log z which is an inverse of ez

and is holomorphic(on some domain), and therefore it has complex derivative 1z . Since z ez is not one to one andhas period 2i , so it is impossible to dene logz which is holomorphic on the whole plane. Wehave to restrict the domain of ez on which z ez is one to one, and choose a domain which isopen so that complex derivatives make sense.

A useful way to dene a branch of the logarithm function is to specify the region that thevalues of arg z are allowed to take, which gives innite many branches of the logarithm, theydiffer by 2ki for k Z .

We will show it is impossible to dene a holomorphic function inverse of ez on any simplyconnected domain which contains the origin. Any branch of the logarithm function can not bedened at 0 (so that 0 is called a branch point).

For any (xed), the exponential function z ez maps the stripS , 2 = {z = x + iy : < y < + 2 }

one to one and onto the region

D = C \ {re i : r 0}= {z C \ {0}: < arg z < + 2 }.

Notice that D may be obtained by cutting the complex plane C from 0 to innite and removingthe semi-line from 0 with direction . The inverse function of the exponential from S , 2 to Dis called a (holomorphic) branch of the logarithm, by abusing the notations, denoted by log z.That is

log z = log |z|+ i arg z (where < arg z < + 2 )for any z D .

Both S , 2 and D are open, and exp maps S , 2 one to one and onto D , so that its inverselog z is holomorphic on D and ddz log z =

1z for z D .

By varying (i.e. by cutting through different lines), we obtain innite many differentholomorphic branches. The reason we need to cut from the branch point 0 is that C \ {0}isnot simply connected : there is a hole at 0 in the region C \ {0}, thus a continuous closed curveabout 0 can not be contracted into a trivial one through C \ {0}. By removing a semi-line fromthe branch point to innity, for example G =

{z C

\ {0

}: 0 < arg z < 2

}, the region G is

opened, and becomes simply connected, that is there is no any hole in G. Yet, those are not allpossible holomorphic branches of the logarithm, and we will prove that it is possible to dene aholomorphic branch of the logarithm on any simply connected region which does not contain 0,see Corollary 2.6.22. This is another reason why 0 is called a branch point of the logarithm.

In applications, we should choose a branch (though never unique) so that it is holomorphicon the region which you are interested in.

Example 2.3.1 Taking D0 = {z C \ {0}: 0 < arg z < 2}, which is obtained by cutting out positive x-axis, and denelog z log |z|+ i arg z, (with 0 < arg z < 2)


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2.4. COMPLEX POWER 19

for z D0, which is holomorphic on D0 and ddz log z =1z . With this branch, log i =

2 i and

log(2) = log 2 + i etc.Example 2.3.2 Cutting the plane through 0 the negative part of the real line, so that D ={z C \ {0}: < arg z < }we obtain the principal branch of the logarithm

log z log |z|+ i arg z, arg z (, )which coincides with the real one for real variable, but not dened for negative reals. Again logis analytic on D and

ddz log z =

1z .

2.4 Complex powerThe denition of the complex power za (where a is a complex number) depends on the choice of a branch of logarithm function.

Denition 2.4.1 Choose a branch of the logarithm on C \{0}(for example by cutting a straight line from 0 to ), denoted by log, then z ea log z is called a branch of the power function za .Lemma 2.4.2 If a = n Z is an integer, then whatever the choice of a branch of log, z zais the same function on C \ {0}(except for those cutting lines).

2.5 Path integrals

[Week 3 Lecture 9]The main tool in the complex analysis is the theory of path integration (or called line inte-gration) integrals along curves in the complex plane, which will be developed in this section.What we need is a theory of integration for continuous (complex) functions along continuous,piece-wisely differentiable curves in regions of the complex plane. Such a theory can be estab-lished by using Riemann integration for real functions on intervals [Mods Analysis III, TrinityTerm], so there is nothing new here except we introduce some notations about paths and curvesetc., together with some elementary facts about path integration.

In Mods Integration, we have studied Riemann integrals for continuous functions:

b

a f (t)dt = lim| |0 l f (t l1)( t l t l1)here on the left-hand side the limiting process of Riemann sums runs over all nite partitions of [a, b]. This denition applies well to complex valued function f (t) = u(t) + iv(t) on [a, b],namely we dene

b

af (t)dt =

b

au(t)dt + i

b

av(t)dt.

We wish to extend this to integrals along a curve in the complex plane in place of theinterval [a, b].


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Denition 2.5.1 Let G C be a subspace. A [continuous] path or a parameterized curve in G is a continuous mapping from a closed interval [a, b] to G. The image { (t) : t [a, b]}iscalled a curve, denoted by or by [ ] and the path : [a, b] G is a parameterization of thecurve , also called a path in G from (a) to (b) (which indicates the orientation of ). iscalled closed if (a) = (b). is a simple path if (s) = (t) as long as s, t [a, b] and s = t

except possibly (a) = (b). A parameterized curve : [a, b] C which is simple and closed iscalled a (parameterized) Jordan curve.Suppose that : [a, b] G is a continuous path, which traces out a directed curve in G from (a) to (b), is differentiable on [a, b] with continuous derivative [such a path called a continuously

differentiable path in G], and suppose that f is a complex function on G, continuous along thepath , that is, t f ( (t)) is continuous on [a, b], then the path (line) integral of f along thepath is dened by

f (z)dz = b

af ( (t))d (t) =

b

af ( (t)) (t)dt (2.5.1)

[effectively, set z = (t) and dz = (t)dt] the last integral on the right-hand side is the Riemannintegral of g(t) = f ( (t)) (t) on [a, b].

A [continuous] path : [a, b] G is piece-wisely smooth , if there is a nite partition a = t0


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2.5. PATH INTEGRALS 21

Lemma 2.5.4 Let be a parameterized curve in an open set G. Suppose f has an anti-derivative[in complex sense] F , that is, if F is holomorphic on G and F = f , then

f (z)dz = F ( (b)) F ( (a)) . (2.5.5)Proof. Indeed f ( (t)) (t) = ddt F ( (t)) and (2.5.5) follows from the FT in calculus.Thus the Fundamental Theorem of Calculus holds for path integrals. In particular, in this

case, f (z)dz depends only on the end points of the path but independent of the path , aslong as it starts at (a) nishes at (b). In particular, if f has an anti-derivative on G: F = f ,and is a closed curve, then f (z)dz = 0.Example 2.5.5 f (z) = 1 constant function, and is any path from z1 to z2, i.e. : [a, b] Csuch that (a) = z1 and (b) = z2. Then

dz =

b

a (t)dt = (b) (a)

= z2 z1.That is, dz does not depend on the path joining z1 to z2.Example 2.5.6 : [a, b] C be a curve. Then

zdz = b

a (t) (t)dt

=12

(b)2 (a)2 .

Example 2.5.7 If : [0, 2] C by (t) = eit

, then

1z dz = i 2

0eit eit dt

= 2 i

and

zn dz = i 2

0e(n +1) it dt = 0 if n Z , n = 1.

Example 2.5.8 If : [0, 2] C by (t) = a + eit where a C and > 0. Then

dz(z a)n = 2i , n = 1 ,0, for n = 1 , n Z .Example 2.5.9 There is no function f holomorphic on a domain G which contains a circle

{z : |z| = r}for some r > 0, such that f (z) = 1z on G. That is, it is impossible to dene a holomorphic branch of log on such domain. Indeed, if there were such a holomorphic function,then

|z|= r1z

dz = |z|= r f (z)dz = f (end ) f (initial ) = 0on the other hand

|z|= r

1z dz = 2 i which is a contradiction.


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[Week 4, Lecture 10]

Lemma 2.5.10 (Max-Length Inequality) If f : G C is continuous along a path : [a, b] G,then

f (z)dz l( )max |f | (2.5.6)where l( ) =

ba | (t)|dt is the length of the curve .

Proof. Let I = f (z)dz and I = |I |ei where is the argument of I . Then

|I | = ei I = ei f (z)dz= ei f (z)dz =

b

aei f ( (t)) (t)dt

=

b

aRe ei f ( (t)) (t) dt

b

a |f ( (t))|| (t)|dt supt [a,b ] |f ( (t))| b

a | (t)|dtwhich completes the proof.

Theorem 2.5.11 If f n , f are continuous on G and f n f uniformly on along the path , then

f n (z)dz f (z)dz.Proof. Applying (2.5.6) to f n f to obtain

f n (z)dz f (z)dz l( )max |f

n f |0

as n .We may extend the concept of path integration of a function f : G C along chains. Achain in G is a formal sum of paths in G = n1 1 + n2 2 + + nk k

where n i are integers and i are curves G, so that the formal sum represents the union of severalpieces of parameterized curves. Dene

f (z)dz = n1 1 f (z)dz + + nk k f (z)dz (2.5.7)One of course needs to verify (2.5.7) is well dened: the right hand-side is independent of theway we represent a chain .

Proposition 2.5.12 1) f f (z)dz is linear for any chain .2) f (z)dz is additive: if 1, 2 two chains, then 1 + 2 f (z)dz = 1 f (z)dz + 2 f (z)dz.


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2.6. THE CAUCHY THEOREM AND THE CAUCHY INTEGRAL THEOREM 23

2.6 The Cauchy Theorem and The Cauchy Integral The-orem

In this section, we prove the most important results in Complex Analysis: The Cauchy Theoremand The Cauchy Integral Formula.

2.6.1 The Cauchy TheoremThe following is the rst version of The Cauchy Theorem.

Theorem 2.6.1 Suppose f is holomorphic on G, and G is a triangle whose boundary is a closed path in G oriented counter clockwise, then

f (z)dz = 0 .Proof. [Not examinable. The proof is due to E. Goursat (1900, Toulouse)]. If is a triangle

(with orientation counter clockwise) in G, then I () denotes the path integral f (z)dz.Equally dividing the sides of the triangle to form four similar triangles 1, 2, 3 and 4,so that the length l( i) = 12 l( ) and

I () = I ( 1) + I ( 2) + I ( 3) + I ( 4).

By Triangle Inequality

|I () | |I ( 1)|+ |I ( 2)|+ |I ( 3)|+ |I ( 4)|,so at least one of |I ( i)| exceeds 14 |I () |. Choose one of them, say (1) such that

|I ( (1) )| 14|I () |.

Applying the same procedure to the triangle (1) to construct (2) such that (2) (1) ,

l( (2) ) =12

l( (1) ) =122

l( )

and

|I ( (2) )

| 1

4|I ( (1) )

| 1

42 |I ()

|.

Repeating the previous procedure we may construct a sequence of triangles (n ) (with (0) = )so that the following properties hold:

1) (1) (n ) , andl( (n )) =

12

( (n1) ) = =12n

l( ). (2.6.1)

2)

|I ( (n ))| 14|I (

(n1) )| 14n |I () |. (2.6.2)


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According to Cantors intersection theorem, together with the fact that l( (n )) 0, thereis a unique point z0 n =1 (n ) . Since f (z0) exists, so that, for every > 0 there is > 0 suchthatf (z) f (z0)

z z0 f (z0) < for 0 < |z z0| <

hence|f (z) f (z0) f (z0) (z z0)| < |z z0| for 0 < |z z0| < (2.6.3)

Since (n ) {z0}and l( (n )) 0, there is N such that|z z0| l( (n )) < z (n ) for n N

Each (n ) is a closed path so that

( n ) dz = 0, ( n ) zdz = 0 (2.6.4)which implies that

I ( (n )) = ( n ) (f (z) f (z0) f (z0) (z z0)) dz.Applying the Max-Length inequality to obtain

I ( (n )) l( (n ))| maxz ( n ) |f (z) f (z0) f (z0) (z z0) | l( (n )) maxz ( n ) |z z0| n N [by (2.6.3)]

l( (n )

)2

= l( ) 2

4n n N .Combining this inequality with (2.6.2) we have

14n |I () | I (

(n )) l( ) 2

4nn N ,

multiplying 4 n both sides of the previous inequality to obtain

|I () | l( ) 2 > 0.Since is arbitrary, we must have

f (z)dz = 0 hence

f (z)dz = 0.

[Week 4, Lecture 11]

2.6.2 The Cauchy Theorem: other versionsNext we present a version of The Cauchy Theorem for simply connected domains .

By initiative, a region G (i.e. G is open and connected, so it is path connected) is simply connected , if there are no holes inside G, hence any simple closed curve in G can be deformedcontinuously and contracted in G to a single point of G. This can be made precise as thefollowing.


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Two closed curves 1, 2 : [0, 1] G are homotopic [in G] if there is a continuous mappingH : [0, 1][0, 1] G such that H (0, t) = 1(t), H (1, t) = 2(t) for any t [0, 1], and H (s, 0) =H (s, 1) for any s [0, 1]. The curves s (t) = H (s, t ) (t [0, 1]) are continuous deformations of 1 into 2. We say a closed curve : [0, 1] G can be deformed in G into a point, denoted by 0 [ is homotopic to zero], if is homotopic to a constant curve 0: 0(t) = p for any t, forsome p G. By denition, a connected open set G is simply connected, if any closed continuouscurve in G is homotopic [in G] to a constant curve, i.e. 0.Example 2.6.2 B r (a) = {z : |za| < r }is simply-connected, but B r (a)\{a}is not. An annulusAr,R (a) = {z : r < |z a| < R }( R > r 0) is connected but not simply-connected. On theother hand, the set obtained by removing a segment from the inner circle to the outside circle of Ar,R (a) is simply connected.

In order to prove the version of Cauchys Theorem for simply connected domains, we needthe following facts about the geometry of simply connected domains.

If A C , then we use A to denote the set of A together with its accumulation points. A =AAc is the boundary of A. We will deal with very simple domains whose boundaries are curves.For example, if G is the domain inside a simple and closed curve : [a, b] C , then G is theclosed curve i.e. { (t) : t [a, b]}. For example, B r (a) is the circle C r (a) = {z : |z a| = r}.Similarly, A r,R (a) is the union of two circles C r (a) and C R (a).Lemma 2.6.3 Let G be a region, and f be continuous on G. Suppose is a piecewise smooth path in G, then for every > 0 there is a polygonal path in G such that

f (z)dz

f (z)dz < .

Proof. [Outline of the proof] First we can choose a bounded and closed subset K of C , suchthat K G and the curve lies in K . Since f is continuous, so it is uniformly continuous in K ,hence there is > 0 such that

|f (z) f (w)| 0 there is a closed polygon in G such that

f (z)dz f (z)dz < ,and since f (z)dz = 0, thus f (z)dz < . Since > 0 is arbitrary, so we must have

f (z)dz = 0.


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Corollary 2.6.5 If G is a simply connected domain, f is holomorphic on G, then the path integral f (z)dz depends only on the end points of the path : [a, b] G.

Proof. Suppose : [c, d] G such that (a) = (a) and (b) = (b). The reversal path : [b, b+ d c] G by (t) = (b + d t)traces back from (b) to (a), + is a closed path in G, so that according to Cauchys theorem

+ f (z)dz = f (z)dz + f (z)dz = 0.It follows that

f (z)dz = f (z)dz = f (z)dz.

[Week 4, Lecture 12]A large class of simply connected domains in the complex plane C are given by Jordans curvetheorem.

Denition 2.6.6 A a simple and closed curve in C is called a Jordan curve .

The following famous theorem, its proof however is out of the syllabus of the Part A Analysis,and is not examinable. The interested reader may refer to a good text book on AlgebraicTopology. For example, Algebraic Topology, A First Course by W. Fulton contains much morethan we need here.

Theorem 2.6.7 A Jordan curve = { (t) : t [a, b]}divides the complex plane C into exactly two connected components, one of them is bounded and the other is unbounded. The bounded one G is simply connected, called the (simply connected) domain enclosed by the Jordan curve .The Jordan curve itself is denoted by G , called the boundary or the contour of G. Moreover G = G G is closed.

Let us state yet another version of Cauchys Theorem.

Theorem 2.6.8 Let G be a simply connected domain enclosed by a piece-wise smooth Jordan curve G . Suppose f is holomorphic on G = G G , then

G f (z)dz = 0 .

This follows from Theorem 2.6.4 with contour = G.In many applications, we will need to apply Cauchys theorem to more complicated domainsthan simply connected ones. To this end, let us make a convention that f is holomorphic on A(which is not necessary open) if it is holomorphic on an open set G such that A G.

Let D be a simply connected domain enclosed by a [piecewise smooth] Jordan curve D , andlet D j D be disjoint simply connected domains enclosed by Jordan curves D j for j = 1 , , k.G is called a multiply connected domain with k holes D1, , Dk . We want to apply Cauchystheorem to functions holomorphic on G = D \ k j =1 (D j D j ). The boundary of G is the unionof Jordan curves

G = D D 1 D k .


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G is oriented positively with respect to G: D is oriented counter clockwise and D j sclockwise, so denoted by D j . Suppose that f is holomorphic on G. We may open G into asimply connected one by cutting G through a path L j connecting a point on D j and a pointon D for each j , to obtain

G = G \ k j =1 L jwhich is simply connected. Applying Cauchys Theorem to f on G,

G f (z)dz +k

j =1 L j f (z)dz + L j f (z)dz = 0and therefore

G f (z)dz = 0.That is, Cauchys theorem is valid even for multiply connected domains.

As a consequence, we have the following homotopic version of The Cauchy Theorem.Suppose 1 and 2 are two Jordan curves in an open subset G C which are homotopic with

respect to G (so 2 is a continuous deformation of 1 in G preserving the orientation) and f isholomorphic in G, then

1 f (z)dz = 2 f (z)dz .2.6.3 The Cauchy Integral FormulaRecall that we say f is holomorphic on a subset A C if it is holomorphic on an open subset G A. If a C and r > 0, then C r (a) denotes the circle centered at a with radius r , orientedcounter clock-wise.

The following is the famous integral formula of Cauchy.

Theorem 2.6.9 Suppose f is holomorphic on a closed disk B r (a), then

f (a) =1

2i C r (a ) f ( ) a d . (2.6.6)Proof. By the homotopic version of the Cauchy Theorem,

C r (a )

f ( )

a

d =

C (a )

f ( )

a

d

for any 0 < < r . We show that

lim0 C (a ) f ( ) a d = 2 if (a) (2.6.7)

which then yields (2.6.6).Since f is continuous on B r (a) so is uniformly continuous, hence for any > 0 there is a

r > > 0 such that

|f ( ) f (a)| < B(a).


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Since

C (a ) f (a) a d = f (a) C (a )1

ad = 2 i

we have, for any (0, r )

C (a ) f ( ) a d 2if (a) = C (a ) f ( ) f (a) a d 2 max C (a ) |

f ( ) f (a)|| a|

= 2 max C

|f ( ) f (a)|

2for any (0, r ), so that (2.6.7).

The following theorem provides a general form of the Cauchy Integral Formula.

Theorem 2.6.10 Let G = D \ k j =1 D j be a multiply connected domain with holes D1, , Dkas described at the end of the last subsection, and f is holomorphic on G. Then

f (z) =1

2i G f ( ) z d z G,where G is oriented positively with respect to G. In particular, if f is holomorphic on a closed disk B r (a), then

f (z) =1

2i C r (a )

f ( )

zd z B r (a) .

Proof. In fact, if z G, then since G is open, there is r > 0 such that B2r (z) G. ApplyThe Cauchy Theorem to Gr = G \ B r (z) and g( ) = f ( ) z which is holomorphic on Gr . Then

G g( )d + C r (z) g( )d = 0and therefore

G

f ( )

z

d =

C r (z)

f ( )

z

d = 2 if (z).

In particular, a holomorphic function in a region G is determined by its values on the boundaryG.

[Week 5 Lecture 13]With the help of the Cauchy Integral Formula, we are now in a position to show that a

holomorphic function is innitely differentiable. What we need to show, which is the followingLemma, is that it is legitimate to differentiate the right-hand side of (2.6.6) under the integralsign which in turn gives the Cauchy Integral Formula for higher order derivatives of a holomorphicfunction.


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Lemma 2.6.11 Let : [0, 1] C be a [piece-wisely smooth] path, and continuous along .Let m N and deneF m (z) = ( )( z)m d for z C \ [ ]

where [ ] is the range of the path . Then F m is holomorphic and F m = mF m +1 on C

\[ ].

Proof. Let a D = C \ [ ] which is open, so we can choose > 0 such that B2(a) D.For z B(a), considerF m (z) F m (a) = 1( z)m

1( a)m

( )d .

For any positive integer n

1( z)n

1( a)n

=1

z 1

a i+ j = n

1

1( z)i

1( a) j

= ( z a)n1

j =0

1( z)n j

1( a) j +1

. (2.6.8)

It follows thus thatF m (z) F m (a)

z a= h(z, ) ( )d

where

h(z, ) m1

j =0

1( z)m j

1( a) j +1

.

HenceF m (z) F m (a)

z a mF m +1 (a) = h(z, ) m( a)m +1 ( )d .

By using the Max-Length estimate we obtain

F m (z) F m (a)z a

mF m +1 (a)

l( )max | |max h(z, ) m

( a)m +1.

Let us consider the difference

h(z, ) m

( a)m +1=

m1

j =0

1( z)m j

1( a) j +1

m( a)m +1

=m1

j =0

1( z)m j

1( a)m j

1( a) j +1

= ( z a)m1

j =0

m j1

i=0

1( z)m ji

1( a)i+ j +2

.


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Since and thus / B2(a), so that | z| and | a| , thus

h(z, ) m

( a)m +1 |z a|

m1

j =0

m j1

i=0

1m +2

m2

m +2 |z a|so that

F m (z) F m (a)z a

mF m +1 (a) l( )m2 max | |

m +2 |z a| 0as z a.Theorem 2.6.12 Let G be a simply connected domain enclosed by a simple closed curveG and f is holomorphic on G. Then f (n ) exists for any n = 1 , 2, , and

f (n )

(z) =

n!

2i Gf ( )

( z)n +1 d z G, (2.6.9)where G is oriented positively with respect to G.

Proof. According to The Cauchy Integral Formula

f (z) =1

2i G f ( ) z d z G.Lemma 2.6.11 implies that f exists on G and

f

(z) =1

2i Gf ( )

( z)2 d z G.Then we can use Lemma 2.6.11 again, to show that

f (z) = 22i G f ( )( z)3 d z G

and repeat the use of Lemma 2.6.11 again and again to obtain (2.6.9).

Remark 2.6.13 Let G = D \ k j =1 D j be a multiply connected domain with holes D1, , Dk :all D and D j s are Jordan domains, D j are disjoint, and D j D . Suppose f is holomorphic on G, then

f (n )(z) =n!

2i G f ( )( z)n+1 d z G,where G is oriented positively with respect to G.

We are now in a position to prove the result we have announced at the beginning of thecourse: a holomorphic function is innitely differentiable.

Theorem 2.6.14 If f is holomorphic on an open set G, then f is innitely differentiable on G.In particular, f is continuous.


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Proof. For any z G, choose a disk B r (z) G (it is possible as G is open). Apply theprevious theorem to f on B r (z). It follows from (2.6.9) that

f (n )(w) =n!

2i C r (z) f ( )( w)n +1 d w B r (z)so f has all higher derivatives at z.

It took us a long way to achieve the above theorem.

Corollary 2.6.15 f = u + iv is holomorphic on G if and only if 1) all partials ux , uy , vx and vy exist and continuous on G; 2) the Cauchy-Riemann equations hold: z f = 0 on G.

2.6.4 Some Applications[Week 5 Lecture 14] As applications, we prove several interesting results, including The Funda-mental Theorem in Algebra.

Example 2.6.16 (Cauchys Inequality, Cauchys Estimate) If f is holomorphic on the closed disk BR (a), where R > 0, then

|f (n )(a)| n!Rn

max|za |= R |f (z)|, n = 1 , 2, . (2.6.10)

Proof. Apply (2.6.9) to f on Ba (R) one has

f (n )(a) =n!

2i C R (a ) f ( )( a)n +1 d ,so that [using the Max-length inequality]

|f (n )(a)| n!2

2R max C R (a )

|f ( )|| a|n +1

=n!2

2R1

Rn+1max

C R (a ) |f ( )|.

Example 2.6.17 (Liouvilles Theorem) A bounded entire function is constant.

Proof. If f is holomorphic on C and |f (z)| M for all z C , then the Cauchy inequality(for n = 1, and f on BR (z)) implies that

|f (z)| 1R

M

for any R > 0 and z C . Letting R we obtain |f (z)| 0 so that f (z) = 0 for any z C .Thus f must be constant.Example 2.6.18 (The Fundamental Theorem in Algebra) If P (z) = an zn + + a0 is a poly-nomial of degree n 1 (so that an = 0 ), then P (z) = 0 has at least one solution in C .


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Proof. Suppose P (z) = 0 had no solution, so that P = 0, thus f (z) = 1P (z) is holomorphicon C . We claim that P (z) as z . Indeed

|P (z)| |z|n |an | |an1||z|

|a0||z|n

|z|n |an | |an1|+ + |a0||z| |z| > 1

Choose R > 1 such that

|an | |an1|+ + |a0|

R 12|an |.

Then

|P (z)| 12|an ||z|

n for any |z| > Rand therefore P (z) as z . Thus f (z) 0 as z , so that f is bounded [Exercise].According to Liouvilles Theorem, f is constant, so is P , a contradiction.

As another application, we prove the Maximum Modulus Theorem.Example 2.6.19 (Maximum Modulus Theorem) Suppose f is holomorphic on a domain G, and suppose there is a point a G such that |f (a)| |f (z)| for all z G, then f is constant.

Proof. We may assume that f (a) 0 instead, otherwise consider g(z) = ei f (z) insteadwhere is the argument of f (a). Consider A = {z G : f (z) = f (a)}. Then A is non-emptyand closed as f is continuous. We prove that A is open. Let w A G. Since G is open, thereis B r (w) G.

Applying the Cauchy Integral Formula to f on B(w) for any 0 < < r

f (w) = 12i C (w) f ( ) wd and using the parameterization = w + eit where t runs from 0 to 2, we have

f (w) =1

2i C f (w + eit )

eitie it dt

=1

2 2

0f (w + eit )dt.

[This formula is called the Mean Value Theorem]. Therefore

f (a) = f (w) =1

2 2

0f (w + eit )dt

so that

2

0(f (a) f (w + eit ))dt = 0.

Taking the real part we obtain

2

0(f (a) Ref (w + eit ))dt = 0 .


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Since f (a) Ref (w + eit ) 0 for all t, we must have Ref (w + eit ) = f (a) for all t for any < r . Therefore Ref (z) = f (a) for any z B r (z0), so that, according to the Cauchy-Riemannequations, Im f (z) = 0 for any z B r (z0), hence f (z) = f (a) for any z B r (z0). HenceB r (z0) A. Thus A is non-empty, closed and open, A = G as G is connected.

Finally we show the inverse of the Cauchy Theorem is also true, which is called the Moreras

theorem. There are various versions of Moreras theorem, we choose the triangle version.In the proof of the following theorem, for z1, z2 C , [z1, z2] denotes the parameterized linesegment from z1 to z2.

Theorem 2.6.20 (Moreras Theorem) If f is continuous in an open set G, and if for any triangle G

f (z)dz = 0 ,then f is holomorphic on G.

Proof. Clearly we only need to show f is holomorphic on any disk B r (a) G, thus without

losing generality, let as assume that G = B r (a). Dene

F (z) = [z0 ,z ] f ( )d for z B r (a). (2.6.11)We show that F is holomorphic on B r (a) and F = f .

For any z B r (a), there is > 0 such that B2(z) G. For w B(z) and w = z, byassumption, the line integral along the triangle {z0, z ,w}vanishes, so that

[z0 ,z ] f ( )d + [z,w ] f ( )d + [w,z 0 ] f ( )d = 0thus

F (w) = [z0 ,w] f ( )d = F (z) + [z,w ] f ( )d .

HenceF (w) F (z) = [z,w ] f ( )d ,

so thatF (w) F (z)w z

f (z) = [z,w ] (f ( ) f (z)) d .By the Max-Length inequality

F (w) F (z)w z

f (z) max B (z) |f ( ) f (z)| 0 as w z.Therefore F (z) = f (z) for any z G. F is holomorphic, so it is innitely differentiable. Inparticular F

= f exists in G. f is holomorphic on G by denition.From the proof of Moreras theorem, we in fact have proven the following


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2.7. TAYLORS THEOREM AND LAURENT EXPANSIONS 35

Theorem 2.6.21 Let G be an open set. If f is continuous on G, and for any path in G, thepath integral f (z)dz depends only on the end points of in G, then there is a holomorphic function F on G such that F = f on G. In particular, if f is holomorphic on a simply connected domain G, then f has an anti-derivative in G: there is a holomorphic function F on G such that F = f on G.

Example 2.6.22 If G is simply connected region and a / G, then there is a holomorphic func-tion f such that f (z) = 1za for any z G and exp(f (z)) = z a for any z G. If z0 Gand w0 C such that ew0 = z0 a, we may choose f such that f (z0) = w0. f is a holomorphicbranch of log(z a) on the simply connected domain G and log(z0 a) = w0.

Proof. Since a / G, 1za is holomorphic on G. According to Theorem 2.6.21, there is aholomorphic function g on G such that g(z) = 1za for any z G.Considerh(z) =

eg(z)

z

a

, z G.

Then h is holomorphic on G and

h(z) = eg(z)g(z) (z a) eg(z)

(z a)2= 0

so that h must be constant. Hence

eg(z)

z a=

eg(z0 )

z0 a= ew0 + g(z0 ) .

Let f (z) = g(z)

g(z0) + w0. Then

ef (z) = eg(z)g(z0 )+ w0 = z a z Gand f (z0) = w0.

2.7 Taylors Theorem and Laurent expansionsPower series are important examples of holomorphic functions, which are indeed the only holo-morphic functions.

Theorem 2.7.1 (Taylors Theorem) Let f be a holomorphic function on G, a G and R > 0the largest number such that BR (a) G. Then

f (z) =

n =0

an (z a)n z BR (a) (2.7.1)

wherean =

f (n )(a)n!

=1

2i C (a ) f ( )( a)n+1 d , n = 0 , 1, 2, (2.7.2) for any 0 < < R .


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[A holomorphic function is analytic .]Proof. Let z BR (a) and choose < R such that z B(a). Applying The Cauchy Integral

Formula to f on B(a))

f (z) =1

2i

C (a )

f ( )

z

d z BR (a). (2.7.3)

For C (a), |z a| < | a|, so thatf ( ) z

=f ( )

( a) (z a)=

f ( ) a

11 za a

=f ( ) a

n =0

z a a

n

so that Hence

=

n =0

(z

a)n

f ( )

( a)n +1 C (a). (2.7.4)

Since(z a)n

f ( )( a)n +1

1

max C (a ) |f ( )| |

z a|

n

,

and |za | < 1, by the M-test, the series (2.7.4) converges uniformly in C (a), thus

f (z) =1

2i C (a ) f ( ) z d =

n =0

(z

a)n

1

2i C (a )

f ( )

( a)n +1 d

=

n =0

an (z a)n .

Corollary 2.7.2 f is holomorphic in an open set G if and only if at each point a G, there isr > 0 such that f has a Taylor expansion about a on B r (a).

Corollary 2.7.3 Suppose 0 < R < is the convergence radius of a power series f (z) =n =0 an (z

a)n , then f has at least one singularity on the circle C R (a). In other words, thereis no function F holomorphic on BR (a) such that F = f on BR (a).

By Cauchys Theorem, the path integral C (a )f ( )

( a)n +1 d does not depend on as long as0 < < R .[Week 5 Lecture 15]As an application of Taylors Theorem, we prove the following

Theorem 2.7.4 (The Identity Theorem) Let f and g be two holomorphic functions on an open subset G. Suppose 1) G is connected (i.e. G is a domain); 2) there is a G and a sequencezn a but zn = a for all n , 3) f (zn ) = g(zn ) for n = 1 , 2, . Then f = g on G.


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2.7. TAYLORS THEOREM AND LAURENT EXPANSIONS 37

Proof. We may assume that g = 0 otherwise consider f g instead. Also we may assumethat a = 0 by considering translation z a. Thus zn 0, 0, zn G, zn = 0, and f (zn ) = 0 forall n. We may assume that zn = 0.Let

A = {z G : f (k)

(z) = 0 k = 0 , 1, 2, }= k=0 f (k)1({0}).Since f (k) are continuous, A is closed in G. We show that 0 A so that A is non-empty. Indeed,since f is holomorphic on G and 0 G, by Taylors Theorem

f (z) =

k=0

akzk in B r (0)

for some r > 0. Suppose not all ak = 0, and m N such that a0 = = am1 = 0 but am = 0,then, since zn 0, there is N such that zn B r (a) for n N . By assumptions

k= m

akzkn = f (zn ) = 0 n N .

Since zn = 0, we may divide both sides by zmn to obtain

k= m

akzkmn = 0 n N .

Letting n , to conclude that am = 0, a contradiction. Therefore ak = 0 for all k, so thatf (k)(0) = 2k! ak = 0. Hence 0 A.Next we show that A is open. Suppose b A G, so that f (k)(b) = 0 for all k. G is open,

there is r > 0, B r (b) G, and according to the Taylor Theorem,

f (z) =

k=0

ck(z b)k z B r (b)

where

ck =f (k)(b)

k!= 0 k = 0 , 1, 2, .

Therefore f = 0 on B r (b), hence f (k)(z) = 0 for any z B r (b), so that B r (b) A. Thus A isopen. Since G is connected, we must have A = G. The proof is complete.

Corollary 2.7.5 Let f be holomorphic on a region G, and suppose a is a root of f : that isf (a) = 0 . 1) If f is not identically zero, then there is r > 0 such that f (z) = 0 for any z B r (a) \{a}. That is, a is an isolated zero of f . 2) If is not identically zero near a, then thereis a positive integer m and a function g holomorphic on some disk B r (a) such that g(a) = 0 and f (z) = ( z a)m g(z) for z B r (a). In this case we say a is a zero of f of order (or multiplicity)m .


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Proof. If f (a) = 0 and f is not identically zero, then by the identity theorem, there is r > 0such that f (z) = 0 for any B r (a) \{a}. Since f is holomorphic, so that f (z) = n = m an (z a)nwhere am = 0. Then f (z) = ( z a)m g(z) with g(z) = n =0 an + m (z a)n which is holomorphicon B r (a).

The Taylor Theorem tells us that a holomorphic function on a disk has Taylor expansion.

We now consider functions holomorphic on annuli. For example functions holomorphic on a diskB r (a) \ {a}with its center removed (a punctured disk ).Theorem 2.7.6 (Laurent expansion) Suppose that f is holomorphic on an annulus Ar,R (a) =

{z : r < |z a| < R }for some 0 r < R [ Ar,R (a) is connected but not simply connected],then f (z) =

n = cn (z a)n

wherecn =

12i

C (a )

f ( )(

a)n +1

d

and r < < R .

Proof. According to the homotopic version of The Cauchy Theorem [applying to holomorphicfunction f ( )( a)n +1 on an annulus centered at a], cn dened along a circle C (a) does not dependon as long as C (a) is located inside Ar,R (a).

For z Ar,R (a), since r < |z| < R , so we can choose1 < 2 such thatr < 1 < |z| < 2 < R .

Applying The Cauchy Integral Formula to f on the domain A1 ,2 (a), we have

f (z) = 12i C 2 (a ) f ( ) z d 12i C 1 (a ) f ( ) z d

=1

2i C 2 (a ) f ( ) z d +1

2i C 1 (a ) f ( )z d .For C 2 (a), za a < 1, so that

1 z

=1

a1

1 za a=

n =0

(z a)n( a)n +1

and therefore f ( ) z

=

n =0

(z a)nf ( )

( a)n +1and the series converges uniformly on C 2 (a). Hence, by integrating term by term, we obtain

12i C 2 (a ) f ( ) z d =

n =0

(z a)n1

2i C 2 (a ) f ( )( a)n+1 d =

n =0

cn (z a)n .


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Denition 2.8.4 If a is an isolated singularity and f (z) = n = cn (z a)n near a, then thepart of series with negative powers1

n = cn (z a)n =

n =1

cn (z a)n

is called the principal part of f about the singularity a, and

c1 =1

2i C (a ) f ( )d is called the residue of f at the isolated singularity a, denoted by Res(f, a ).

Denition 2.8.5 If a is an isolated singularity of f with Laurent expansion about a

f (z) =cm

(z a)m+ +

c1z a

+

n = cn (z a)n z B r (a) \ {a}

so that its principal part (about a)cm

(z a)m+ +

c1z a

has only nite terms, then a is called a pole. If cm = 0 ( m 1) then a is a pole of order m . If all cn = 0 for any n = 1 , 2, , then a is removable .Example 2.8.6 Let

f (z) =1

(z 1)(z 2)=

11 z

12 z

which is holomorphic except two isolated points 1 and 2.

1. In the unit disk B1(0) [the largest possible disk about 0] Taylors theorem applies, and

11 z

=

n =0

zn |z| < 1 ,

12 z

=12

11 z2

=

n =0

12n +1

zn |z| < 2,so that

f (z) = 1 1

2n +1zn , |z| < 1.

2. On the annulus A1,2(0), Laurents expansion applies: for z A1,2(0), z2 < 1 so we stillhave the power series for 12z , but since |z| > 1 (i.e.

1z < 1)

11 z

= 1z

11 1z

=

n =0

1zn +1

thus

f (z) =

n =0

1zn +1

n =0

12n +1

zn 1 < |z| < 2.


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2.8. ISOLATED SINGULARITIES 41

3. Similarly, in the annulus A2,(0), the Laurent expansion is given by

f (z) =

n =0

2n 1zn +1 |z| > 2 .

4. f is holomorphic on B1(1) \{1}(and you may see that the radius 1 is the largest possible),and its Laurent expansion about 1 is given byf (z) =

11 z

11 (z 1)

= 1

z 1

n =0

(z 1)n , 0 < |z 1| < 1

so the principal part at 1 is 1z1 and Res( f, 1) = 1.5. Similarly, f is holomorphic on B1(2) \ {2}(again the radius 1 is the largest possible) and

has the Laurent expansion about 2f (z) =

11 z

+1

z 2=

11 + ( z 2)

+1

z 2=

n =0

(1)n1(z 2)n +1

z 2, 0 < |z 2| < 1

with its principal part 1z2 and Res( f, 2) = 1.

Example 2.8.7 f (z) = sin zz . 0 is an isolated singularity, with its Laurent expansion

f (z) =

n =0(1)

n

(2n + 1)! z2n |z| > 0

which has no principal part. In particular the right hand-side denes an entire function which coincides with f on C \ {0}. In this way the singularity is removed and f is extended holomor-phically to the singularity 0. 0 is a removable singularity.Example 2.8.8 f (z) = e

1z 2 has an isolated singularity 0 and its Laurent expansion

f (z) =

n=0

(1)nn!

z2n |z| > 0which has innite many negative power terms. 0 is an essential singularity.

Example 2.8.9 Consider

f (z) = sinz

4(z 1)= sin

4

+4

1z 1

=22

cos4

1z 1

+ sin4

1z 1


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has an isolated singularity at 1, so that it has Laurents expansion on C \ {1}. To work out itsexpansion, we use

f (z) =22

n =0

(1)n(2n)!

2n

42n(z 1)2n

+ 22n =0

(1)n(2n + 1)! 2n+1

42n +1(z 1)2n1

so there are innite many negative power terms. 1 is an essential singularity.

[Week 6 Lecture 17]In this lecture, we present the local behavior of f at an isolated singularity .

Theorem 2.8.10 Let a be an isolated singularity of f . Then the followings are equivalent.1) a is removable;2) limza f (z) = l exists ( l = );3) [Riemanns Removable Singularities Theorem] f is bounded in B r (a) \{a}for some r > 0.[ f is bounded near a ].Proof. 1)= 2) f has the Laurent expansion about a

f (z) =

n =0

an (z a)n

(with the principal part= 0), so that lim za f (z) = a1 exists.2)= 3) Trivial.3)= 1) According to Laurents expansion (applying to f on B r (a) \ {a}), the principal partof f about a is given by

P (z) =

n =1

cn1

(z a)nwhere

cn =1

2i C (a ) f ( )( a)n +1 d for any 0 < < r . By the Max-Length inequality

|cn | 1

22 max

C (a )|f ( )|

| a|n +1= n max

C (a ) |f ( )|.Thus if f is bounded on B r (a) \ {a}: say |f (z)| M for all z B r (a) \ {a}, then

|cn | lim0 n M = 0 for n = 1 , 2,

so that P (z) = 0. a is removable.


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By denition

f (z) =1

(z a)m

k=0

dk(z a)k

where d0 = 1(a) = 0, so that a is a pole of order m.

Corollary 2.8.12 If a is an isolated singularity of f , then a is essential if and only if limza f (z)doesnt exist nor limza f (z) = .Corollary 2.8.13 If a is an essential singularity of f , then a is also an essential singularity of 1f .

2.9 The Residue TheoremThe following theorem is an immediate consequence of The Cauchy Theorem.

Theorem 2.9.1 (The Residue Theorem) Let G be a simply connected domain enclosed by a simple and closed curve G . Suppose f is holomorphic on G \ {a1, , an}where a i G areisolated singularities of f . Then

G f (z)dz = 2 in

i=1

Res(f, a i) . (2.9.1)

Proof. Choose > 0 small enough such that B2(a i) G for i = 1 , , n. Apply TheCauchy Theorem to f on G

\k j =1 B(a i) so that

G f (z)dz +n

i=1 C (a i ) f (z)dz = 0which is equivalent to (2.9.1).

Remark 2.9.2 The Residue Theorem holds for a proper multiply connected domain G = D \k j =1 D j as well.

Of course, The Residue Theorem would be useless if we could not compute the residues

effectively by other means rather than path integrals.

[Week 6, Lecture 18]The following theorem provides a way of computing residues without using path integration.

Theorem 2.9.3 Let a be a pole of f with order m 1, and f (z) = (z)(za)m near a, where isholomorphic in B r (a) for some r > 0 and (a) = 0 .1) We have

Res(f, a ) =(m1) (a)

(m 1)!. (2.9.2)


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2.9. THE RESIDUE THEOREM 45

2) If a is a simple pole (so m = 1 ), then Res (f, a ) = (a), i.e.

Res(f, a ) = (a) = limza

(z a)f (z). (2.9.3)3) If a is a double pole (so m = 2 ) then

Res(f, a ) = (a). (2.9.4)Proof. Indeed, choose a small r > 0 so that

Res(f, a ) =1

2i C r (a ) (z)(z a)m dz=

(m1)(a)(m 1)!

the second equality follows from The Cauchy Integral Formula applying to .

Corollary 2.9.4 If a is a simple pole of f (z) =(z)

(z) , where , are holomorphic on some B r (a)(for some r > 0) (a) = 0 and (a) = 0 , (a) = 0 , then

Res(f, a ) =(a)

(a). (2.9.5)

Proof. By (2.9.3)

Res(f, a ) = limza

(z a)(z)

(z)

= limza

(z)(z)(a)za

=(a)

(a).

PLEASE REMEMBER THESE FORMULATE (2.9.2, 2.9.3, 2.9.4 and 2.9.5) AS WE WILLUSE THEM VERY OFTEN.

Example 2.9.5 Consider f (z) =

z 2(z 1)2 sin z

.

It is clear that 1 is a double pole [Theorem 2.8.11], and [using (2.9.4)]

Res(f, 1) = ddz z=1

z 2sin z=

sin z (z 2) cos zsin2 z z=1

=sin 1 + cos 1

sin2 1.

Since

sin z = z

n =0

(1)nz2n

(2n + 1)!


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k are zeros of sin z with order 1, so that k are simple poles of f , and

Res(f,k ) =k 2

(k 1)2 sin(k)= ( 1)k

k 2(k 1)2

.

Example 2.9.6 f (z) = tan z = sin zcos z has simple poles ak = k +12 for k Z . Using formula

(2.9.5)Res(f, a k) =

sin a kcos a k

= 1

.

Example 2.9.7 Let b be a pole of f with order m and holomorphic on B r (b) for some r > 0.Choose r > 0 small so that f = 0 on B r (b) \ {b}. b is a pole or a removable singularity of F = f f . We wish to calculate Res(F, b).

According to the Laurent expansion about b

f (z) = ( z

b)m g(z) on B r (b)

\ {b

}where g is holomorphic on B r (b) and g(b) = 0, ThereforeF (z) = (z)

f (z)f (z)

= m

z b(z) +

g(z)g(z)

(z)

on B r (b) \ {b}. Thus b is a simple pole of F (or removable) andRes(F, b) = lim

zb(z b)F (z) = m (b).

Similarly, if a is an isolated zero of f with order n, then a is a simple pole (or a removablesingularity) of F = f

f , and

Res(F, a ) = limza

(z a)F (z) = n (a).Example 2.9.8 [Not Examinable] (The Argument Theorem). Let G be a simple or multiply connected domain [A domain we can apply Cauchys Theorem is sufficient]. Let be holomorphicon G, f holomorphic on G except for nite many poles bi G of order m i ( i = 1 , , k) and f have zeros a j of order n j ( j = 1 , , l). Let F (z) = (z) f

(z)f (z) for

z G = G

\ {a1,

, a l , b1,

, bk

}.

The only possible singularities of F inside G are zeros and poles of f , so that, by applyingthe residue theorem to F and the previous example on residues of F at as and bs, we obtain

12i G (z) f (z)f (z) dz =

l

j =1

n j (a j ) k

i=1

m i (bi). (2.9.6)

In particular1

2i G f (z)f (z) dz = number of zeros number of poles


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2.9. THE RESIDUE THEOREM 47

both zeros and poles of f are counted with their multiplicities. This formula is called the argument principle .

If f is holomorphic, A C such that there is no solutions to f (z) = A on the contour G,then

12i

G

f (z)f (z)

A

dz = number of solutions to f (z) = A in G.

This formula can be used to estimate the number of solutions of certain equations.

Example 2.9.9 [Not Examinable] (Rouches Theorem). Let G be a simply connected domain en-closed by a simple, piecewise smooth and closed curveG . Suppose and are two holomorphic functions on G, and

(z) = 0 , |(z)| < | (z)| z G .Then the number of zeros of + in G equals the number of that of .

Proof. The proof is a good application of the existence of a holomorphic branch of thelogarithm in a simply connected region which does not contain the branch point 0, Corollary

2.6.22. According to the Argument Principle,1

2i G + + dzrepresents the number of solutions to + = 0 in G. Let

h = 1 +

.

Then1

2i G + + dz = 12i G dz + 12i G hh dz.Using a parameterization of G , say : [a, b] G . Since G is closed so that (a) = (b),and

12i G hh dz = 12i

b

a

h( (t))h( (t))

(t)dt

=1

2i b

a

ddt h( (t))h( (t))

dt.

Since < 1 so that h( (t)) B r (1) for some 0 < r < 1. Since 0 / B r (1), we can choosea holomorphic branch of the logarithm, denoted by log, which is holomorphic on B r (1), andd

dz log z =1z on B r (1). Thus, according to chain rule

ddt

log h( (t)) =ddt h( (t))h( (t))

t [a, b]

and therefore1

2i G hh dz = 12i b

a

ddt h( (t))h( (t))

dt

=1

2i b

a

ddt

log h( (t))dt

= 0.


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Hence1

2i G + + dz = 12i G dzwhich completes the proof.

2.10 Contour integration[Week 7, Lectures 19]

In this section we show by examples some interesting applications of The Residue Theorem.I. Evaluate an integral such as

20 Q(cos , sin )d, where Q is a rational function.

This kind of integrals can be converted into contour integrals along the unit circle, by usingsubstitution . On the unit circle with parameterization z = ei where goes from 0 to 2, sothat cos = 12 (z + z1) and sin =

12i (z z1) [on the unit circle z = z1] and d = 1iz dz. Theintegral is transformed into a contour integral

|z|=1 Q(12

(z + z1), 12i

(z z1))1iz

dz .

Example 2.10.1 Calculate integral

I = 2

0

(cos2)2 d1 2 pcos + p2

where 0 < p < 1.

Using z = ei , so that d = 1iz dz, cos = 12 (z + z1). Since, z2 = ei2

cos2 =12

(z2 + z2) =12

(z2 + z2)

so that

(cos2)2 =14

(z2 + z2)2 = 14

z4 + 2 +1z4

.

Hence

I = 14i |z|=1 z4 + 2 + 1z4 11 p(z + z1) + p2 1z dz

=14i |z|=1 z4 + 2 +

1z4

1(1 pz)

1(1 pz1)

1z

dz

=14i |z|=1 f (z)dz

wheref (z) = z8 + 2 z4 + 1

1z4

1z p

11 pz

.


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2.10. CONTOUR INTEGRATION 49

There are two singularities within the unit circle: a simple pole p with

Res(f, p ) = limz p

(z p)f (z)= p2 +

1 p2

2 11

p2

,

and 0 which is a pole of order 4. Now we compute Res(f, 0). Near zero, both 11 pz and1

z p areholomorphic and Taylors Theorem apply to obtain Taylors expansions

11 pz

=

n =0

pn zn | pz| < 1

and1

z p=

1 p

11 z p

= 1 p

n =0

pn zn | p1z| < 1.Hence

f (z) = 1 p

z4 + 2 +1z4

(1 + pz + p2z2 + p3z3 + ) 1 + p1z + p2z2 + p3z3 + .

The coefficient of 1z is the residue, so by multiplying out the products [keep in mind that we areinterested in the terms with z1]: if we use to denote two functions have the same term of 1z ,then

f (z) 1 p

1z4 (1 + pz + p

2

z2

+ p3

z3

)

1 + p1z + p2z2 + p3z3

1 p

p3 + p + p1 + p3 1z

.

which givesRes(f, 0

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