Solutions to Assignment 6

1. Find the Laurent series centered at z0 = 0 of the function f(z) = −1(z−1)(z−2)

in the following domains:

(a) D1 : |z| < 1; [6]

(b) D2 : 1 < |z| < 2; [8]

(c) D3 : 2 < |z|. [6]

Solution (a). Recall that the series

1

1− w=∑n≥0

wn,

works whenever |w| < 1. If |z| < 1 then |z/2| < 1 and we obtain

f(z) = −1

2

1

(1− z)(1− z/2)= −1

2

(∑n≥0

zn

)(∑m≥0

zm

2m

)

= −1

2

∑k≥0

(k∑

`=0

1/2`

)zk

=∑k≥0

(1/2k+1 − 1)zk

Solution (b). In this case is better to perform first a partial fractiondecomposition. This method can also be used to solve items (a) and (c) ifyou like. If 1 < |z| < 2 then |1/z| < 1 and |z/2| < 1 and we have

f(z) =1

2− z+

1

z − 1

=1

2

1

(1− z/2)+

1

z

1

(1− 1/z)

=∑n≥0

zn

2n+1+∑n≥1

1

zn

Solution (c). If 2 < |z| then we have to be treat both singularities z = 1and z = 2. Since |1/z| < |2/z| < 1 we obtain

f(z) =−1

z21

(1− 1/z)(1− 2/z)=−1

z2

∑n≥0

1

zn

∑m≥0

2m

zm

=−1

z2

∑k≥0

2k+1 − 1

zk

=∑k≥2

1− 2k−1

zk

2. Show that for any integer k ≥ 0 we have

1

(1− z)k+1=∞∑n=0

(n+ k

n

)zn,

and that it converges absolutely for any |z| < 1. [10]

Solution. We know that

f(z) =1

1− z=∑n≥0

zn,

is analytic if |z| < 1. Its k-derivative is

f (k)(z) =k!

(1− z)k+1,

and this is clearly an analytic function for |z| < 1, with a singularity atz = 1. Thus, the maximal radius of absolute convergence of the Taylorseries of this function is exactly 1. This is the nice part, the informationabout the singularities gives all the information you need about the radiusof absolute convergence of the Taylor series. Therefore

f (k)(z) =k!

(1− z)k+1=∑n≥0

f (n+k)(0)

n!zn

and this series converges absolutely for |z| < 1. We easily see that f (n+k)(0) =(n+ k)! and we deduce that

1

(1− z)k+1=f (k)(z)

k!=∑n≥0

(n+ k)!

k!n!zn =

∑n≥0

(n+ k

n

)zn.

3. Let f(z) =∑∞

n=0 an(z − z0)n, with a0 6= 0, where the series converges for|z − z0| < R. Comment why the following series should hold

1

f(z)=∞∑n=0

cn(z − z0)n

for |z−z0| < r, for some r > 0 and some coefficients cn. Show that c0 = 1/a0,c1 = −a1/a20, c2 = (a21 − a0a2)/a30 and c3 = −(a31 − 2a0a1a2 + a20a3)/a

40.

[10]

Solution. Since f(z0) = a0 6= 0, this implies that the f(z) is not zero closeto z0. Thus, there is some r > 0 such that the function g(z) = 1/f(z) iswell-defined, and therefore analytic, if |z − z0| < r. This implies that g(z)has a Taylor series

g(z) =∞∑n=0

cn(z − z0)n

in the region |z−z0| < r. It remains to compute the first coefficients. Whatwe know is

cn =g(n)(z0)

n!.

andn!an = f (n)(z0).

We have c0 = g(z0) = 1/f(z0) = 1/a0. We have

g′(z) = − f′(z)

f(z)2,

so c1 = g′(z0) = −a1/a20. We have

g′′(z) = −f′′(z)f(z)2 − 2f ′(z)2f(z)

f(z)4=

2f ′(z)2 − f ′′(z)f(z)

f(z)3

so c2 = g′′(z0)/2 = (a21 − a0a2)/a30. Omitting z we have

g′′′ =(4f ′f ′′ − f ′′′f − f ′′f ′)f 3 − 3(2f ′2 − f ′′f)f 2f ′

f 6

=6f ′f ′′f − f ′′′f 2 − 6f ′3

f 4.

We obtain

c3 =1

6g′′′(z0) =

2a1a2a0 − a3a20 − a31a40

.

4. (Special Case of the Proof of Laurent Series for z0 = 0). Let f(z) be ananalytic function in the region R between the circles C1 : |z| = r1 andC2 : |z| = r2, with r2 > r1, both positively oriented. Let C be any simpleclosed contour lying inside R (C could be C1 or C2 if you like).

(a) Comment why for any z ∈ R we have

f(z) =1

2πi

∫C2

f(w)

w − zdw − 1

2πi

∫C1

f(w)

w − zdw

[Extra 5]

(b) Define

an =1

2πi

∫C2

f(w)

wn+1dw

and

bn =1

2πi

∫C1

f(w)

w−n+1dw.

Assuming we can interchange summation with the integral sign∫

aswe please, show that for any z ∈ R we have

f(z) =∞∑n=0

anzn +

∞∑n=1

bnzn.

[Extra 5]

(c) Now comment why the contours C1 and C2 in the definition of an andbn may be replaced by C. Note that if we replace C1 and C2 by Calready in the beginning of the calculations to show item (b) we havea problem since in the summation process we simultaneously wouldneed |z/w| < 1 and |w/z| < 1, which is impossible. [Extra 5]

Solution (a). Consider the following Contour C∗. Since C∗ is a contouraround the point z, we deduce by Cauchy’s integral formula that

1

2πi

∫C∗

f(w)

w − zdw = f(z).

However C∗ = C2 ∪ (−C1) ∪ B1 ∪ B2, where B2 = −B1. Therefore, theabove integral is

1

2πi

∫C∗

f(w)

w − zdw =

1

2πi

∫C2

f(w)

w − zdw − 1

2πi

∫C1

f(w)

w − zdw.

Solution (b). Using the assumption that we can interchange summationwith the integral sign

∫as we please we obtain

∞∑n=0

anzn +

∞∑n=1

bnzn

=1

2πi

∞∑n=0

zn∫C2

f(w)

wn+1dw +

1

2πi

∞∑n=1

1

zn

∫C1

f(w)

w−n+1dw

=1

2πi

∫C2

∞∑n=0

znf(w)

wn+1dw +

1

2πi

∫C1

∞∑n=1

1

znf(w)

w−n+1dw

=1

2πi

∫C2

∞∑n=0

(z/w)nf(w)

wdw +

1

2πi

∫C1

∞∑n=1

(w/z)nf(w)

wdw.

Note that in the first integral above we have w ∈ C2, that is |w| = r2 andz ∈ R, which implies |z| < r2, thus |z/w| < 1. Analogously, in the secondintegral we have |w/z| < 1, thus we can say that the series in the firstintegral equals to 1/(1− z/w) and in the second integral (w/z)/(1−w/z).We obtain

∞∑n=0

anzn +

∞∑n=1

bnzn

=1

2πi

∫C2

1

1− z/wf(w)

wdw +

1

2πi

∫C1

w/z

1− w/zf(w)

wdw

=1

2πi

∫C2

f(w)

w − zdw − 1

2πi

∫C1

f(w)

w − zdw

= f(z).

Solution (c). By the principle of deformation, since the functions f(w)/wn+1

and f(w)/w−n+1 are analytic inside region R and we can continuously de-form C1 or C2 into C inside R, the contour integrals of these functions

over C1 and C2 respectively are the same if we replace any of them byC. Although this is true, in the actual proof we needed that paths to beparticularly chosen so that we can say that

∞∑n=0

(z/w)n = 1/(1− z/w)

in the first integral (that is, |z/w| < 1) and

∞∑n=1

(w/z)n = (w/z)/(1− w/z)

in the second integral (that is, |w/z| < 1). This is only achieved by choosingto use C2 in the definition of an and C1 in the definition of bn.