Complex Analysis for Applications, Math 132/1, Home Work Solutions …mt/132.1.01w/sol-2.pdf ·...

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Complex Analysis for Applications, Math 132/1, Home Work Solutions-II Masamichi Takesaki Page 148, Problem 1. Γ Γ Γ 0 Page 129, Problem 2. If two contours Γ 0 and Γ 1 are respectively shrunkable to single points in a domain D, then they are continuously deformable to each other. Proof. Γ 1 Γ 0 Γ 0 Γ 1 Γ 1 Γ 0 Γ 1 Γ 1 Γ 0 Γ 0 Formal Solution. Two curves γ 0 : t [0, 1] z 0 (t) G and γ 1 : t [0, 1] z 1 (t) G in adomain G is called homotopic if the curve γ 0 can be continuously demormable to the second curve γ 1 , which means that there exists a continuous mapping z :(s, t) [0, 1] × [0, 1] z (s, t) G such that 1

Transcript of Complex Analysis for Applications, Math 132/1, Home Work Solutions …mt/132.1.01w/sol-2.pdf ·...

Page 1: Complex Analysis for Applications, Math 132/1, Home Work Solutions …mt/132.1.01w/sol-2.pdf · Complex Analysis for Applications, Math 132/1, Home Work Solutions-II Masamichi Takesaki

Complex Analysis for Applications, Math 132/1,Home Work Solutions-II

Masamichi Takesaki

Page 148, Problem 1.

Γ Γ

Γ0

Page 129, Problem 2. If two contours Γ0 and Γ1 are respectively shrunkable tosingle points in a domain D, then they are continuously deformable to each other.

Proof.

Γ1Γ0

Γ0

Γ1

Γ1

Γ0

Γ1Γ1

Γ0

Γ0

Formal Solution. Two curves γ0 : t ∈ [0, 1] 7→ z0(t) ∈ G and γ1 : t ∈ [0, 1] 7→z1(t) ∈ G in a domain G is called homotopic if the curve γ0 can be continuouslydemormable to the second curve γ1, which means that there exists a continuousmapping z : (s, t) ∈ [0, 1] × [0, 1] 7→ z(s, t) ∈ G such that

1

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{z(0, t) = z0(t),

z(1, t) = z1(t).

(0,0 ) (1,0 )

(1,1 )(0,1 )

(s,t ) z(s,t )z1(t)

z2(t)

z

We write this fact γ0 ∼ γ1. Now we claim the following facts about this relation oncurves:

γ ∼ γ for every curve γ in G;

γ0 ∼ γ1 ⇒ γ1 ∼ γ0;

γ0 ∼ γ1, γ1 ∼ γ2 ⇒ γ0 ∼ γ2.

The first fact can be seen easily by setting: z(s, t) = z(t), 0 ≤ s, t ≤ 1, withγ : t ∈ [0, 1] 7→ z(t) ∈ G. The second fact also easily check by setting z′(s, t) =z(1 − s, t), 0 ≤ s, t ≤ 1, where z : (s, t) ∈ [0, 1] × [0, 1] 7→ z(s, t) ∈ G gives γ0 ∼ γ1,i.e., z(0, t) = z0(t) and z(1, t) = z1(t). The third fact appears to be complicated.But it is not hard either. Let z1 : (s, t) ∈ [0, 1] × [0, 1] 7→ z1(s, t) ∈ G give thehomotopy: γ0 ∼ γ1 and z2 : (s, t) ∈ [0, 1] × [0, 1] 7→ z2(s, t) ∈ G give the homotopyγ1 ∼ γ2. Then set

z(s, t) =

{z1(2s, t), 0 ≤ s ≤ 1

2, 0 ≤ t ≤ 1;

z2(2s − 1, t), 12≤ s ≤ 1, 0 ≤ t ≤ 1.

Then the function z of (s, t) is continuous and z(0, t) = z0(t), z(1/2, t) = z1(t) andz(1, t) = z2(t).

Now suppose curves γ0 is homotopic to a point curve α0: t ∈ [0, 1] 7→ w0(t) =α0 ∈ G and γ1 is homotopic to another point curve α1: t ∈ [0, 1] 7→ w1(t) = α1 ∈ G.So, our assumption means that γ0 ∼ α0 and γ1 ∼ α1. Connect α0 and α1 by acontinuous curve: z : t ∈ [0, 1] 7→ z(t) ∈ G so that z(0) = α0 and z(1) = α1. Setz̄(s, t) = z(s), 0 ≤ s, t ≤ 1. Then z̄(0, t) = w0(t) = α0 and z̄(1, t) = z(1) = α1 =w1(t). Thus we get α0 ∼ α1 and

γ0 ∼ α0; α0 ∼ α1; α1 ∼ γ1 ⇒ γ0 ∼ γ1.

Page 149, Problem 3. In the domain D = {z ∈ C : 1 < |z| < 5}, which of thefollowing curves are homotopic to the curve Γ, z(θ) = 3 + eiθ, 0 ≤ θ ≤ 2π:

a) Γa : z(θ) = 3 + ei( π2 +θ), 0 ≤ θ ≤ 2π;

b) Γb : z(t) = 3i, 0 ≤ t ≤ 1;c) Γc : z(t) = 2eit, 0 ≤ t ≤ 2π;d) Γd : z(t) = −3 + eit, 0 ≤ t ≤ 2π;e) Γe : z(t) = 3 + e−it, 0 ≤ t ≤ 4π.

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Answer.Γ ∼ Γa ∼ Γb ∼ Γd ∼ Γe. 0 4

Except the curve Γc, all curves are homotopic to a point curve. As Γc encloses thehole |z| < 1 of the domain D, it cannot be shrunk to a point, consequently it cannotbe homotopic to any curve which is shrinkable to a point in D. ♥

Page 160, Problem 3. Evaluate the integrals over the positively oriented circleC, |z| = 2 :

a)Ia =

C

sin 3z

z − π2

dz; b)Ib =

C

zez

2z − 3dz; c)Ic =

C

cos z

z3 + 9zdz;

d)Id =

C

5z2 + 2z + 1

(z − i)3dz; e)Ie =

C

e−z

(z + 1)2dz; f)Ie =

C

sin z

z2(z − 4)dz.

Answer. a) Set f(z) = sin 3z. As π/2 falls inside the circle C, we have

Ia = 2πi f(π

2

)= 2πi sin

(3π

2

)= −2πi.

b) Set f (z) = 12zez . Then 3/2 is inside of C. So we have

Ib =

C

1

2

zez

z − 32

dz = 2πif

(3

2

)= 2πi

1

2

3

2e

32 =

3πi

2e

32 .

c) First observe that

cos z

z3 + 9z=

cos z

z(z2 + 9)=

cos z

z(z − 3i)(z + 3i),

and that the singularities z = ±3i lie outside of C and the singularity z = 0 sitsinside of C. So let

f (z) =cos z

z2 + 9

and conclude

Ic = 2πif (0) =2πi

9.

d) The singularity z = i sits inside of C. So we get with f (z) = 5z2 + 2z + 1:

Id =1

2!2πif ′′(i) = πi · 10 = 10πi.

e) The singularity z = −1 falls inside of C, so that with f (z) = e−z we have

Ie = 2πif ′(−1) = −2πie.

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f) The point z = 0 is the only singularity of the integrand inside of C. So with

f (z) =sin z

z − 4,

obtain

f ′(z) =(z − 4) cos z − sin z

(z − 4)2;

Ie = 2πif ′(0) = 2πi−4

16= −

πi

2♥

Page 160, Problem 4. Compute the contour integrals:∮

C

z + i

z3 + 2z2dz

along the following counterclockwise traveled circles once:

a) Ca : |z| = 1.b) Cb : |z + 2 − i| = 2.c) Cc : |z − 2i| = 1.

Ca

CbCc

z=-2

z=2i

z=-2+i

z=0

Answer. First locate the singularities of the integrand:

z + i

z3 + 2z2=

z + i

z2(z + 2).

From the above inspection, we conclude that z = 0 and z = −2 are the singularitiesand that z = 0 falls inside of Ca and outside of both Cb and Cc and z = −2 fallsinside of Cb and outside of both Ca and Cc. So Cc encloses no singularity of theintegrand, so that ∮

Cc

z + i

z3 + 2z2dz = 0.

Now with

f(z) =z + i

z2; g(z) =

z + i

z + 2, g′(z) =

(z + 2) − (z + i)

(z + 2)2=

2 − i

(z + 2)2

Ca

z + i

z3 + 2z2dz = 2πig′(0) =

πi(2 − i)

2= π

(1

2− i

);

Cb

z + i

z3 + 2z2dz = 2πif(−2) = 2πi

i − 2

4= −π

(1

2+ i

).

Page 5: Complex Analysis for Applications, Math 132/1, Home Work Solutions …mt/132.1.01w/sol-2.pdf · Complex Analysis for Applications, Math 132/1, Home Work Solutions-II Masamichi Takesaki

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Page 160, Problem 5. Let C be the ellipse

x2

4+

y2

9= 1

traversed once on the positive direction and set

G(z) :=

C

ζ2 − ζ + 2

ζ − zdz.

Evaluated G(1), G′(i) and G′′(−i).

z=1

z=-i

z=iC

Answer. Let

g(z) = z2 − z + 2, z ∈ C.

By the Cauchy Integral Formula, 2πig and G agree inside the ellipse C. Therefore,

G(1) = 2πig(1) = 2πi(12 − 1 + 2) = 4πi;

G′(i) = 2πig′(i) = 2πi(2i − 1) = −2π(2 + i);

G′′(−i) = 2πig′′(−i) = 4πi.

Page 160, Problem 6. Evaluate

|z|=3

eiz

(z2 + 1)2dz.

z=-i

z=i

Answer. We first prepare the factorization:

eiz

(z2 + 1)2=

eiz

(z + i)2(z − i)2

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and split the integration along the circle Γ : |z| = 3 to the sum of the integrationsover the smaller two circles Γ0 : |z − i| = 1

2and Γ1 : |z + i| = 1

2. With

f0(z) =eiz

(z + i)2; f1(z) =

eiz

(z − i)2;

f ′0(z) =

ieiz(z + i)2 − 2eiz(z + i)

(z + i)4=

ieiz(z + i) − 2eiz

(z + i)3

=eiz(iz − 3)

(z + i)3;

f ′1(z) =

ieiz(z − i)2 − 2eiz(z − i)

(z − i)4=

ieiz(z − i) − 2eiz

(z − i)3

=eiz(iz − 1)

(z − i)3

we compute

Γ

eiz

(z2 + 1)2dz =

Γ0

f0(z)

(z − i)2dz +

Γ1

f1(z)

(z + i)2dz

= 2πi(f ′0(i) + f ′

1(−i)) = 2πi

(e−1(−1 − 3)

−8i+

e(i · i − 1)

8i

)

= π

(1

e− e

2

).

Page 161, Problem 7. Compute the contour integral:

Γ

cos z

z2(z − 3)dz

along the contour Γ shown in the figure on the right.

Γ

0 3

Proof. The contour Γ is not a simple closed curve. Butthe loop inside does not enclose any singilarity of theintegrand, so that it does not contribute to the inte-gral by the Cauchy Fundamental Theorem. So the inte-gral along Γ is the same as the integral along the curveΓ′ obtained by removing the inner loop from Γ. Thus

Γ’

0 3

Page 7: Complex Analysis for Applications, Math 132/1, Home Work Solutions …mt/132.1.01w/sol-2.pdf · Complex Analysis for Applications, Math 132/1, Home Work Solutions-II Masamichi Takesaki

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the new contour Γ′ looks like the figure on the right. The singularity z = 3 fallsoutside of Γ′, so it does not contribute to the integral. With

f(z) =cos z

z − 3; f ′(z) =

−(z − 3) sin z − cos z

(z − 3)2

we have ∮

Γ

cos z

z2(z − 3)dz = 2πif ′(0) = −

2πi

9.

Page 161, Problem 8. If f is analytic inside and on the circle C : |z − α| = r,then

f (n)(α) =n!

2πrn

∫ 2π

0

f (α + reiθ)e−inθdθ, n = 0, 1, 2, · · · .

Proof. Parameterize the circle C by C : z(θ) = reiθ, 0 ≤ θ ≤ 2π, and compute theCauch Integral Formula for the n-th derivative:

f (n)(α) =n!

2πi

C

f (z)

(z − α)n+1dz =

n!

2πi

C

f (α + reiθ)

(reiθ)n+1rieiθdθ

=n!

2πrn

∫ 2π

0

f (α + reiθ)

einθdθ.

This proves the assertion. ♥

Page 161, Problem 10. If f is analytic on and inside a simple closed curve Γ,then ∮

Γ

f ′(z)

z − z0dz =

Γ

f (z)

(z − z0)2dz

for all z0 not on the curve Γ.

Proof. Assume that Γ is positively oriented. If z0 lies outside of the curve Γ, thenthe both sides are zero as the integrands are both analytic on and inside the curveΓ. If z0 lies inside the curve Γ, then the Cauchy Integral Formula states

Γ

f ′(z)

z − z0dz = 2πif ′(z0) =

Γ

f(z)

(z − z0)2dz.

If the curve Γ is negatively oriented, then the above integrals both give −2πif ′(z0)for z0 inside Γ, and zero for z0 outside of Γ. This completes the proof. ♥

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Page 161, Problem 13. With g a continuous function on a simple closed curveΓ, the limit value of the analytic function:

G(z) =1

2πi

∮g(ζ)

ζ − zdζ

on the boundary Γ of the domain enclosed by Γ need not be the original function g.

Example. Let Γ : |z| = 1 be the positively oriented unit circle and g(z) = 1z. Then

we get for all z with |z| < 1 that:

G(z) =1

2πi

|ζ|=1

ζ − zdζ =

1

2πi

|ζ|=1

1

ζ(ζ − z)dζ

=1

2πi

|ζ|=1

1

z

(1

ζ − z−

1

ζ

)dζ, z 6= 0,

=1

2πi

1

z

(∮

|ζ|=1

1

ζ − zdζ −

1

2πi

|ζ|=1

1

ζdζ

)

=1

2πiz(2πi − 2πi) = 0.

Thus the function G is constantly zero inside the unit disk Γ. But g on the circle Γis not a zero function. Thus g(z) 6= limr↗1 G(rz) for z ∈ Γ. ♥Page 162, Problem 15. If f is an anlytic function on the unit disk D : |z| ≤ 1and f(0) = 0, then the function F defined by

F (z) =

{ f(z)z z 6= 0;

f ′(0) z = 0,

is analytic on the disk D.

Proof. Clearly, the function F is analytic on the disk D possibly except the originz = 0. But if we define a new function G by the integral:

G(z) =1

2πi

|ζ|=1

f(ζ)ζ

ζ − zdz =

1

2πi

|ζ|=1

f (ζ)

ζ(ζ − z)dz

=1

2πi

|ζ|=1

1

z

(f (ζ)

ζ − z−

f(ζ)

ζ

)dζ, z 6= 0,

=1

2πiz

|ζ|=1

f (ζ)

ζ − zdζ − 1

2πiz

|ζ|=1

f (ζ)

ζdz

=f (z)

z− f (0)

z=

f (z)

z,

then it is analytic throughout the disk D and agrees with F (z) for non-zero z. TheCauchy Integral Formula tells that if z = 0, then G(0) = f ′(0). This completes theproof. ♥

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Page 162, Problem 16. If an analytic function f is never zero on a simplyconnected domain D, then there exists a single-valued branch of log(f (z)) on thedomain D.

Proof. First recall that a simply connecteddomain D means that every closed curve canbe shrunk to a point within the domain D.This means that the interal of any analyticfunction on the domain along every closedcontour is zero because the line integral of

z 0

z

D

Γz0

z

an analytic function does not change along a continuous deformation of the basecurve and the integral over any point curve is zero. This means that every analyticfunction f over D has an anti-derivative F , whose value at z ∈ D can be definedby the contour integral

F (z) =

Γzz0

f(ζ)dζ

along a contour Γzz0

connecting a fixed point z0 ∈ D to the point z ∈ D. Now thefunction:

g(z) =f ′(z)

f(z), z ∈ D,

is analytic throughout the domain D due to the assumption that f does not vanishon D. Thus g has the anti-derivative G on the domain D given by

G(z) =

Γzz0

g(ζ)dζ =

Γzz0

f(ζ)

f ′(ζ)dζ.

The derivative of the function f (z)e−G(z) is zero throughout the Domain D, so thatf (z)e−G(z) is constant, say c. Hence we get f (z) = ceG(z). Thus G + Log(c) is abranch of log(f (z)). ♥

Page 162, Problem 17. There exists a single - valued branch of (z3 − 1)12 in the

disk D : |z| < 1.

Proof. The unit disk D is simply closed and the function: z3 − 1 does not hit zeroon the disk D. Therefore, there exists a single - valued branch G of log(z3 − 1) onthe disk D. Set

g(z) = e12 G(z), z ∈ D,

to obtain a single -valued analytic function g such that g(z)2 = z3 − 1 for all z ∈ D.This completes the proof. ♥

Page 10: Complex Analysis for Applications, Math 132/1, Home Work Solutions …mt/132.1.01w/sol-2.pdf · Complex Analysis for Applications, Math 132/1, Home Work Solutions-II Masamichi Takesaki

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Page 167, Problem 1. If f (z) = 1(1−z)2 and 0 < R < 1, then

i)

max|z|=R

|f (z)| =1

(1 − R)2;

ii)

f (n)(0) = (n + 1)! ;

iii)

(n + 1)! ≤n!

Rn(1 −R)2.

Proof. i) Observe that the maximum of |f(z)| onthe circle CR(0) = {z ∈ C : |z| = R} means thedenominator |(1 − z)2| is minimum, which is thesquare of the distance between the circle CR(0)and the point z = 1. Obviously, it is (1 − R)2.Thus the maximum of |f(z)| on the circle CR(0)is 1

(1−R)2.

CR(0)

0 1R

R

ii) It is easy to see that an anti-derivative of f (z) is

F (z) =1

1 − z,

which has the power series expansion on the disk DR(0) = {z ∈ C : |z| < R}:

F (z) = 1 + z + z2 + · · · =

∞∑

n=0

zn, |z| < 1.

Therefore, we get

f(z) = F ′(z) = 1 + 2z + 3z2 + 4z3 + · · · =

∞∑

n=0

(n + 1)zn =

∞∑

n=0

anzn;

withf (n)(0)

n!= an = (n + 1).

Thus, we conclude the second assertion of the problem.

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iii) The Cauchy Integral Formula applied to f(z) on the disk DR(0) around theorigin gives the following:

f (n)(0)

n!=

1

2πi

|z|=R

f(z)

zn+1dz;

∣∣∣∣f (n)(0)

n!

∣∣∣∣ =

∣∣∣∣∣1

2πi

|z|=R

f(z)

zn+1dz

∣∣∣∣∣ ≤1

2πmax|z|=R

|f (z)| 2πR

Rn+1

=max|z|=R |f(z)|

Rn=

1

(1 − R)2Rn;

∴ (n + 1)! ≤ n!

Rn(1 − R)2.

This completes the proof. ♥Page 167, Problem 2. If f is analytic on the unit disk D = {z ∈ C : |z| < 1}and |f (z)| < 1

1−|z| , then

|f (n)(0)| ≤ n!

Rn(1 − R).

Proof. The Cauchy Integral Formula applied to f , the circle CR(0) = {z ∈ C : |z| =R}, 0 < R < 1, and the point α = 0 gives:

f (n)(0)

n!=

1

2πi

CR(0)

f (z)

zn+1dz;

∣∣∣∣f (n)(0)

n!

∣∣∣∣ =

∣∣∣∣∣1

2πi

CR(0)

f (z)

zn+1dz

∣∣∣∣∣ ≤1

max|z|=R |f(z)|Rn+1

2πR

≤ 1

Rn(1 − R);

∴ |f (n)(0)| ≤n!

Rn(1 −R).

This completes the proof. ♥Page 168, Problem 4. If p(z) = a0 + a1z + · · ·+ anzn and max|z|=1 |p(z)| = M ,then |ak| ≤ M, 0 ≤ k ≤ n.

Proof. The Cauchy Inegral Formula applied to p, the unit circle C = {z ∈ C : |z| =1} and the point α = 0 gives:

ak =1

2πi

C

p(z)

zk+1dz

|ak| =

∣∣∣∣1

2πi

C

p(z)

zk+1dz

∣∣∣∣ ≤1

2πmax|z|=1

|p(z)|2π = M.

This completes the proof. ♥

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Page 168, Problem 5. If f is entire and <f(z) ≤ M,z ∈ C, then f must beconstant.

Proof. Set g(z) = ef(z). Then g is also entire. Now we compute the modulus |g(z)|:

|g(z)| = |ef(z)| = e<f(z) ≤ eM .

Thus g is a bounded entire function. Liouville’s Theorem states that every boundedentire function is a constant. So g must be a constant, equivalently f is constant.This completes the proof. ♥

Page 168, Problem 7. If f is entire and there exists r0 such that |f (z)| ≤ z2 forevery z with |z| ≥ r0, then f is a polynomial of at most degree two.

Proof. Fix z ∈ C and choose R > r0 + |z|. Apply the Cauchy Integral Formula tof , the circle: CR(z) = {w ∈ C : |w − z| = R} and the point z:

f (3)(z)

3!=

1

2πi

CR(z)

f (w)

(w − z)4dz

Observe that if w ∈ CR(z), then

R + |z| ≥ |w| ≥ R − |z| ≥ r0,

so that

|f(w)| ≤ |w2| = |w|2 ≤ (R + |z|)2, w ∈ CR(z).

0

R

z

r0R-|z|

CR(z)

Cr0(0)

Thus we get the estimate:

∣∣∣∣f (3)(z)

3!

∣∣∣∣ =

∣∣∣∣∣1

2πi

CR(z)

f(w)

(w − z)4dz

∣∣∣∣∣ ≤1

(R + |z|)2

R42πR

≤ (R + |z|)2

R3−→ 0 as R → +∞.

Thus f (3)(z) = 0 for every z, which means that f is a polynomial of degree at most two. ♥

Page 168, Problem 9. If f is analytic on the closed disk DR(z0) = {z ∈ C : |z − z0| ≤ R} and|f (z0)| ≥ |f (z)|, z ∈ CR(z0) = {z ∈ C : |z − z0| = R}, then there is no point z1 on the circle CR(z0)such that |f(z0)| > |f(z1)|.

Proof. By the maximum modulus principle, f is a constant on the closed disk. This means thatthere is no point z on the closed disk DR(z0) such that f (z1) 6= f(z0). In particular, there is nopoint z1 on the circle CR(z0) with |f(z1)| 6= |f(z0)|. ♥

Page 13: Complex Analysis for Applications, Math 132/1, Home Work Solutions …mt/132.1.01w/sol-2.pdf · Complex Analysis for Applications, Math 132/1, Home Work Solutions-II Masamichi Takesaki

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Page 195, Problem 1. Prove the following Taylor expansions:

a)

e−z =

∞∑

n=0

(−z)n

n!= 1 − z +

z2

2!−

z3

3!+ · · · , z0 = 0;

c)

sinh z =

∞∑

n=0

z2n+1

(2n + 1)!= z +

z3

3!+

z5

5!+ · · · , z0 = 0;

e)

Log(1 − z) =

∞∑

n=1

−zn

n, z0 = 0.

Proof. a) Let f (z) = ez and g(z) = e−z. Then g(z) = f(−z) and

f ′(z) = f (z) and g′(z) = −f ′(−z) = −f(−z) = −g(z), z ∈ C.

Therefore we get

f (n)(0) = f(0) = 1, n = 1, 2, · · · ;

g(n)(z) = (−1)ng(z) ⇒ g(n)(0) = (−1)ng(0) = (−1)n;

g(z) =

∞∑

n=0

g(n)(0)

n!zn =

∞∑

n=0

(−1)n

n!zn =

∞∑

n=0

(−z)n

n!.

b) By definition,

sinh z =ez − e−z

2.

Hence we get

sinh z =1

2

( ∞∑

n=0

zn

n!−

∞∑

m=0

(−z)m

m!

)=

1

2

∞∑

n=0

(zn − (−z)n

n!

).

In the summation, all even terms hit zero as z2n − (−z)2n = z2n − z2n = 0, so that

sinh z =

∞∑

n=0

z2n+1

(2n + 1)!.

The expansion can also be obtained by observing

sinh′ z = cosh z and cosh′ z = sinh z;

sinh(2n+1) z = cosh z and sinh(2n) z = sinh z;

sin(2n+1)(0) = 1 and sin(2n)(0) = 0.

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14

e) First observe that with f(z) = Log(1 − z)

f ′(z) = −1

1 − z= −

∞∑

n=0

zn, |z| < 1.

Thus

f (n+1)(0) = −n! ⇒ f (n)(0)

n!= − 1

n, |z| < 1.

Therefore we get the expansion:

f (z) =∑

n=0

f (n)(0)

n!=

∞∑

n=1

−zn

n, |z| < 1.

This proves the expansions. ♥Page 195, Problem 4. With α ∈ C, if (1 + z)α = eαLog(1+z) for |z| < 1, then

(1 + z)α =

∞∑

n=0

α(α − 1) · · · (α − (n − 1))

n!zn.

Proof. First we observe that the function: f (z) =1

1+z is analytic on the open unit disk D = {z ∈ C :

|z| < 1}. Thus the integral:

F (z) =

Γ(z)

1

1 + zdz

along any contour Γ(z) connecting the 0 to thepoint z ∈ D does not depend on the curve Γ(z)but only on the end point z. This function F (z)

z

0

Γ(z)

is precisely the principal logarithm function Log(1 + z). Now let

f(z) = eαF (z) = eαLog(1+z) = (1 + z)α

and compute its higher derivatives:

f ′(z) = αF ′(z)eαF (z) =α

1 + zf (z) ⇒ f ′(0) = α;

f ′′(z) = − α

(1 + z)2f (z) +

α

1 + zf ′(z) = − α

(1 + z)2f (z) +

1 + z

)2

f(z)

=α(α − 1)

(1 + z)2f(z) ⇒ f ′′(0) = α(α − 1);

f (3)(z) = −2α(α − 1)

(1 + z)3f(z) +

α(α − 1)

(1 + z)2f ′(z)

= −2α(α − 1)

(1 + z)3f(z) +

α(α − 1)

(1 + z)2α

(1 + z)f (z)

=α(α − 1)(α − 2)

(1 + z)3f (z) ⇒ f (3)(0) = α(α − 1)(α − 2);

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15

If f (n)(z) = α(α−1)···(α−(n−1))(1+z)n f(z), then

f (n+1)(z) = −nα(α − 1) · · · (α − (n − 1))

(1 + z)n+1f (z)

+α(α − 1) · · · (α − (n − 1))

(1 + z)nf ′(z)

= −nα(α − 1) · · · (α − (n − 1))

(1 + z)n+1f (z)

+α(α − 1) · · · (α − (n − 1))

(1 + z)n

α

1 + zf(z)

=α(α − 1) · · · (α − (n − 1))(α − n)

(1 + z)n+1f (z).

By induction, we get

f (n)(z) =α(α − 1) · · · (α − (n − 1))

(1 + z)nf (z)

and

f (n)(0) = α(α − 1) · · · (α − (n − 1))

for all n ∈ N. This gives the required Taylor expansion of f on the open unit disk.This completes the proof. ♥

Page 195, Problem 5. Find and determine the convergence radius of the Taylorseries of the following functions around z0:

a) 11+z

, z0 = 0;

c) z3 sin 3z z0 = 0;g) z

(1−z)2 z0 = 0.

Answer. a) z = −1 is the only singularity of the function f(z) = 11+z . Thus the

largest disk with center 0 which does not contain the singularity is the open unitdisk D = {z ∈ C : |z| < 1}. Thus the convergence radius of the Taylor series of faround 0 is 1 and

1

1 + z=

1

1 − (−z)=

∞∑

n=0

(−z)n =

∞∑

n=0

(−1)nzn for |z| < 1.

c) The function f(z) = z3sin 3z is entire, so that the Taylor series converges

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16

everywhere. Now we compute

z3 sin 3z = z3 e3iz − e−3iz

2i=

z3

2i

∞∑

n=0

((3iz)n − (−3iz)n

n!

)

=z3

2i

∞∑

n=0

((3iz)2n+1 + (3iz)2n+1

(2n + 1)!

)

= 3z4∞∑

n=0

(−9z2)n

(2n + 1)!=

∞∑

n=0

(−1)n3 · 9nz4+2n

(2n + 1)!

= 3z4 − 9z6 +3 · 92z8

5!− 3 · 93z10

7!+ · · · .

g) The only sigularity of f (z) = z(1−z)2 is z = 1. Thus the largest circle with

center z0 = 0 which does not encloses the singularity is the unit circle |z| = 1. Theconvergence radius of the Taylor series of f around the origin is 1. Now we computethe expansion with the fact 1

(1−z)2= d

dz( 11−z

) in mind:

z

(1 − z)2= z

d

dz

(1 + z + z2 + z3 + · · ·

)= z(1 + 2z + 3z2 + · · · + nzn−1 + · · · )

= z + 2z2 + 3z3 + · · · nzn + · · · =

∞∑

n=1

nzn

This completes the answer. ♥Page 195, Problem 6. The Taylor expansion of 1

ζ−zaround z0 6= z is given by

1

ζ − z=

∞∑

n=0

(z − z0)n

(ζ − z0)n+1

for z ∈ D|ζ−z0|(z0) = {z ∈ C : |z − z0| < |ζ − z0|}.Proof.

z0

| − z0 |

The singularity of the function f(z) = 1ζ−z

is only z = ζ. Thus the largest disk

centered at z0 which excludes the singularity ζ is the open disk D|ζ−z0|(z0) of radius|ζ − z0|. So the Taylor expansion of f around z0 converges on the disk D|ζ−z0|(z0).Now we compute the expansion

1

ζ − z=

1

(ζ − z0) − (z − z0)=

1

ζ − z0· 1

1 −(

z−z0

ζ−z0

)

=1

ζ − z0

∞∑

n=0

(z − z0

ζ − z0

)n

=

∞∑

n=0

(z − z0)n

(ζ − z0)n+1

to complete the proof. ♥

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17

Page 196, Problem 20. If f is analytic on the closed disk DR(0) of radius R withcenter 0, then

f (z) −n∑

k=0

f (k)(0)

k!zk =

1

2πi

|ζ|=R

zn+1f (ζ)

ζn+1(ζ − z)dζ.

Proof. First observe for ζ on the unit circle |ζ| = 1, we have

1

ζ − z=

1

ζ· 1

1 − zζ

=

n∑

k=0

zk

ζk+1+

∞∑

k=n+1

zk

ζk+1

=n∑

k=0

zk

ζk+1+

zn+1

ζn+2

∞∑

k=0

zk

ζk=

n∑

k=0

zk

ζk+1+

zn+1

ζn+2

1

1 − zζ

=

n∑

k=0

zk

ζk+1+

zn+1

ζn+1

1

ζ − z.

We then substitute the above to the Cauchy Integral Formula:

f(z) =1

2πi

|ζ|=R

f (ζ)

ζ − zdζ =

1

2πi

|ζ|=R

f (ζ)1

ζ − zdζ

=1

2πi

|ζ|=R

f(ζ)

(n∑

k=0

zk

ζk+1+

zn+1

ζn+1

1

ζ − z

)dζ

=

n∑

k=0

zk

2πi

|ζ|=R

f (ζ)

ζk+1dζ +

1

2πi

|ζ|=R

zn+1

ζn+1

f(ζ)

(ζ − z)dζ

=

n∑

k=0

f (k)(0)

k!zk +

1

2πi

|ζ|=R

zn+1f (ζ)

ζn+1(ζ − z)dζ, |z| < R.

This completes the proof. ♥

Page 217, Problem 1-a, 1-d. Find the Laurent series for the function 1z+z2 in

each of the following domains:

a) 0 < |z| < 1;d) 1 < |1 + z|.

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18

Answer. a) and d) We simply compute:

1

z + z2=

1

z

1

1 + z=

1

z−

1

1 + z=

1

z−

( ∞∑

n=0

(−z)n

)

=1

z− 1 + z − z2 + z3 − z4 + · · · + (−1)n−1zn + · · · , |z| < 1;

=1

(1 + z) − 1−

1

1 + z=

1

1 + z

1

1 −(

11+z

) −1

1 + z

=1

1 + z

∞∑

n=0

1

(1 + z)n−

1

1 + z=

∞∑

n=2

1

(1 + z)n, |1 + z| > 1.

This completes the answer. ♥Page 217, Problem 3. Find the Laurent series of the function:

z

(z + 1)(z − 2)

around the origin.

Answer. First the singularities of the functions are z = 1 and z = 2. Thus theLaurent series has three regeions to consider:

a) |z| < 1, b) 1 < |z| < 2, c) 2 < |z|.Second decompose the given function:

z

(z − 1)(z − 2)= z

(1

1 − z− 1

2 − z

)

a) We compute:

z

(z − 1)(z − 2)= z

( ∞∑

n=0

zn −1

2

1

1 − z2

)= z

( ∞∑

n=0

zn −1

2

∞∑

n=0

(z

2

)n)

= z∞∑

n=0

(2n+1 − 1

2n+1

)zn =

∞∑

n=1

(2n − 1

2n

)zn, |z| < 1.

b) We continue:

z

(z − 1)(z − 2)= z

(1

1 − z− 1

2 − z

)= z

(1

z

(1

1z− 1

)− 1

2 − z

)

= z

(1

z

(−

∞∑

n=0

1

zn

)− 1

2

∞∑

n=0

(z

2

)n)

= −∞∑

n=0

1

zn−

∞∑

n=1

zn

2n

= − · · · − 1

z2− 1

z− 1 − z

2− z2

4− · · · , 1 < |z| < 2.

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19

c) For the region |z| > 2, we compute

z

(z − 1)(z − 2)= z

(1

z − 2−

1

z − 1

)= z

(1

1

1 − 2z

−1

1

1 − 1z

)

=∞∑

n=0

((2

z

)n

−(

1

z

)n)=

∞∑

n=0

2n − 1

zn, |z| > 2.

This completes the answer. ♥

Page 218, Problem 5. Find the Laurent series for

z + 1

z(z − 4)3in 0 < |z − 4| < 4.

Answer. The singularities of the function are z =0 and z = 4. The punctured open disk D∗

4(4) ={z ∈ C : 0 < |z − 4| < 4} does no contain any of

0 4

the singularities. We now compute:

z + 1

z(z − 4)3=

(z − 4) + 5

((z − 4) + 4)(z − 4)3=

1

(z − 4)3·(z − 4) + 5

4 − (4 − z)

=1

(z − 4)3· 1

4· 1

1 − 4−z4

((z − 4) + 5)

=1

(z − 4)3· 1

4·( ∞∑

n=0

(4 − z

4

)n)

((z − 4) + 5)

=1

(z − 4)3

( ∞∑

n=0

(−

z − 4

4

)n+1

+5

4

(−

z − 4

4

)n)

=1

(z − 4)3

(5

4+

∞∑

n=1

1

4· (−1)n

(z − 4

4

)n)

=5

4

1

(z − 4)3+

∞∑

n=−2

(−1)n+1 (z − 4)n

4n+4

This finishes the expansion. ♥

Page 218, Problem 10. The Laurent expansion of the function:

f (z) = exp

2

(z −

1

z

)], z 6= 0

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20

is given by∞∑

k=−∞

Jk(λ)zk, (∗)

where

Jk(λ) = (−1)kJ−k(λ) =1

∫ 2π

0

cos(kθ − λ sin θ)dθ.

Proof. Obviously the singularity of f is z = 0 only. So it has the Laurent expansionof the form (∗) which converges everywhere except at the origin. Each coefficientJk(λ) is computed by the Cauchy integral:

Jk(λ) =1

2πi

|z|=1

f(z)

zk+1dz =

1

2πi

∫ π

−π

exp(

λ2(eiθ − e−iθ)

)

ei(k+1)θieiθdθ

=1

∫ π

−π

exp(λi sin θ)e−ikθdθ =1

∫ π

−π

ei(λ sin θ−kθ)dθ

=1

∫ π

−π

(cos(λ sin θ − kθ) + i sin(λ sin θ − kθ))dθ

=1

∫ π

−π

cos(λ sin θ − kθ)dθ,

where the last step follows from the fact that the function sin(λ sin θ − kθ) is anodd function of θ and hence its integral over the interval [−π, π] vanishes.

The Laurent series applied to −z and 1z

gives:

f (−z) =

∞∑

k=−∞

Jk(λ)(−z)k =

∞∑

k=−∞

(−1)kJk(λ)zk;

f

(1

z

)=

∞∑

k=−∞Jk(λ)z−k =

∞∑

k=−∞J−k(λ)zk.

But f (−z) = f (1/z), which implies

J−k(λ) = (−1)kJk(λ).

This completes the proof. ♥

Page 226, Problem 2. Determine the order of the pole of

f(z) =1

(2 cos z − 2 + z2)2

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21

at z = 0.

Answer. Setting g(z) = 1f(z) , we will determine the order of the zero of g at z = 0.

To determine the degree of the zero of g at 0, we consider instead the order of thezero of h(z) = 2 cos z − 2 + z2 at 0, which is a half of the order of the zero of g at0. So we compute the derivatives:

h′(z) = −2 sin z + 2z ⇒ h′(0) = 0; h′′(z) = −2 cos z + 2 ⇒ h′′(0) = 0;

h(3)(z) = 2 sin z ⇒ h(3)(0) = 0; h(4)(z) = −2 cos z ⇒ h(4)(0) = −2 6= 0.

Thus the order of zero of h at 0 is 4, so that that of g at 0 is 8. Hence the order ofthe pole of f at 0 is 8. This answers the question. ♥

Page 226, Problem 9. Does there exist a function f (z)with an essential singu-larity at z0 which is bounded along some line segment emanating from z0?

Answer. Yes, there exists such a function f . For example, the origin 0 is an essentialsingularity of the function

f(z) = e1z .

It is bounded by 1 along the imaginary axis. ♥

Page 227, Problem 17 (Schwarz’s Lemma). If f is an analytic function onthe unit disk D = {z ∈ C : |z| < 1} and satisfies the condition:

f (0) = 0 and |f (z)| ≤ 1 for all z ∈ D,

then |f(z)| ≤ |z| for all z ∈ D.

Proof. Set

G(z) =f (z)

z, z 6= 0.

First the function G is analytic on the punctured disk D∗ = {z ∈ C : 0 < |z| < 1},so that the origin 0 is an isolated singularity of G. But we have

limz→0

G(z) = limz→0

f (z) − f(0)

z= f ′(0).

Thus the singularity 0 is removable and the function

F (z) =

{G(z), z 6= 0

f ′(0), z = 0,

is analytic on the entire unit disk D.

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22

For any ζ ∈ C = {z ∈ C : |z| = 1}, we have

|F (rζ)| =|f (rζ)|

r|ζ| =|f (rζ)|

r≤

1

r, 0 < r < 1.

Hence for any 0 < r < 1, we have

supζ∈C

|F (rζ)| ≤ 1

r,

i.e., the maximum value of |F (z)| on the circle Cr of radius r, 0 < r < 1, with center0 is at most 1/r. The maximum modulus principle for analytic functions shows that

sup|z|≤r

|F (z)| ≤1

r, 0 < r < 1,

which is equivalent to the assertion of Schwarz’s lemma. ♥Page 230, Problem 1-a, 1-g, 1-i. Classify the behavior of the following functionsat the point at infinity:

a) ez; g)sin z

z2; i) etan 1

z .

Answer. For all problems, we consider the behavior of the functions at 0 afterreplacing z by 1/z.

a) First let f(z) = e1z . Then 0 is an isolated singularity of f . When z goes to 0

along the imaginary axis, f stay bounded. But z goes to 0 along the real axis fromthe positive side, then it goes to +∞. So f(z) does not converge to anywhere norto ∞. This means that 0 is an essential singularity of f . Thus the point at infinityis an essential singularity of ez.

b) Let

f(z) =sin 1

z(1z

)2 = z2 sin1

z= z2 e

iz − e−

iz

2i.

As z → 0 along the real axis, the limit of f(z) is 0. But z → 0 along the imaginary

axis from the above, z2e−iz → 0, but z2e

iz → ∞, so that f (z) → ∞. Thus f(z)

does not converge to any point on the extended complex plane C∪ {∞}. Thus thepoint at infinity is an essential singularity of the function.

i) Letf (z) = etan z.

As z → 0, tan z = sin zcos z converges to 0, so that

limz→0

f(z) = 1,

and indeed f(0) = 1. Thus the point at infinity is a removable singularity and the

assigned value of etan 1z at the point at infinity is 0. ♥

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23

Page 231, Problem 2. If f is analytic at ∞, then it has a series expansion ofthe form:

f(z) =

∞∑

n=0

an

zn

which converges uniformly outside some disk.

Proof. With g(z) = f(1/z), g is analytic at the origin 0, so it has a Taylor series:

g(z) =

∞∑

n=0

anzn

which converges unitormly on a disk: Dr = {z ∈ C : |z| < r}, r > 0. Hence theoriginal function f is of the form:

f (z) = g

(1

z

)=

∞∑

n=0

an1

zn

which converges uniformly outside the closed disk D 1r

= {z ∈ C : |z| ≤ 1r}, i.e., on

the area: {z ∈ C : |z| > 1r }. ♥

Page 231, Problem 3. Find the power series expansion of the following functionsaround ∞:

a)z − 1

z + 1; b)

z2

z2 + 1; c)

1

z3 − i.

Answer. As we are concerned with the behavior of the functions near ∞, we lookat z with large |z| for all problems.

a) We simply compute:

z − 1

z + 1= 1 − 2

z + 1= 1 − 2

1

z

∞∑

n=0

(−1

z

)n

= 1 + 2

∞∑

n=1

(−1)n

zn, |z| > 1.

b) We complute:

z2

z2 + 1=

1

1 + 1z2

=

∞∑

n=0

(−1)n

z2n, |z| > 1.

c) This is also easily computed:

1

z3 − i=

1

z3

1

1 − iz3

=1

z3

∞∑

n=0

in

(z3)n=

∞∑

n=0

in

z3(n+1), |z| > 1.

This answers the questions. ♥

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24

Page 231, Problem 1. The unit sphere:

S2 : x21 + x2

2 + x23 = 1

is projected to the extended complex plane C∪{∞}identified with the (x1, x2)-plane:

P : x3 = 0

by connecting each point (x1, x2, x3) ∈ S2 to thepoint on P by the straitline line starting from thenorth pole (0, 0, 1), which is called the stereogaphic

N = (0,0,1)

P = (x1, x2 , x3 )

z = x + iy

x

y

0

projection. Then

x1 =2x

|z|2 + 1, x2 =

2y

|z|2 + 1, x3 =

|z|2 − 1

|z|2 + 1;

z =x1 + ix2

1 − x3.

Proof. In general the equation of the line connecting (x1, x2, x3) and (y1, y2, y3) is given by

(p1(t), p2(t), p3(t)) = t(x1, x2, x3) + (1 − t)(y1, y2, y3), t ∈ R.

With (y1, y2, y3) = (0, 0, 1) and (p1(t), p2(t), p3(t)) = (x, y, 0) we solve the equation:

x = tx1, y = tx2, 0 = tx3 + (1 − t),

equivalentlyz = t(x1 + ix2), tx3 + (1 − t) = 0.

Hence|z|2 = t2(x2

1 + x22) = t2(1 − x2

3) = t2 − t2x23 = t2 − (t − 1)2

= 2t− 1.

∴ t =|z|2 + 1

2; t − 1 =

|z|2 − 1

2

∴ x1 =2x

|z|2 + 1, x2 =

2y

|z|2 + 1,

x3 =t − 1

t=

|z|2 − 1

|z|2 + 1.

The inverse map: (x1, x2, x3) ∈ S2 7→ z = x + iy ∈ C is given by:

tx3 + (1 − t) = 0 ⇒ t(x3 − 1) + 1 = 0 ∴ t =1

1 − x3;

x = tx1 =x1

1 − x3; y = txx =

x2

1 − x3;

∴ z =x1 + ix2

1 − x3.

This completes the proof. ♥

Page 25: Complex Analysis for Applications, Math 132/1, Home Work Solutions …mt/132.1.01w/sol-2.pdf · Complex Analysis for Applications, Math 132/1, Home Work Solutions-II Masamichi Takesaki

25

Page 251, Problem 1-a, 1-c, 1-h. Determine all the isolated singularities of each of the followingfunctions and compute the residue of each singularity.

a)e3z

z − 2; c)

cos z

z2; h)

z − 1

sin z.

Answer. a) The singularities are z = 2 and z = ∞ by inspection. The singularity z = 2 is a simplepole. So we get

Res

(e3z

z − 2, 2

)= lim

z→2(z − 2)

e3z

z − 2= e3·2 = e6.

The singularity z = ∞ is an essential singularity. So we have to compute the power series expansionaround ∞. To this end, let

f(z) =e

3z

1z− 2

=ze

3z

1 − 2z=

(1

1 − 2z

)ze

3z .

We want to find the residue of the essential singularity 0 of the function f , which is determined bylooking at the Laurent series of f near 0:

f (z) =

( ∞∑

n=0

(2z)n

)z

∞∑

n=0

(3n

n!zn

)

=

(1 + 2z + (2z)2 + · · ·

)z

(1 +

3

z+

9

2!z2+

33

3!z3· · · + 3m

m!zm+ · · ·

);

=

(z + 2z2 + 4z3 + · · · + 2n−1zn + · · ·

)(1 +

3

z+

32

2!z2+ · · · + 3n

n!zn+ · · ·

)

=

( ∞∑

n=1

bnzn

) ( ∞∑

m=0

c−mz−m

)=

∞∑

n=0

anzn

with bn = 2n−1, n ≥ 1 and c−m = 3m

m!, m ≥ 0. The coefficients ak are determined by the following:

a0 =

∞∑

n=1

bnc−n =

∞∑

n=1

2n−13n

n!=

1

2

∞∑

n=1

6n

n!=

e6 − 1

2;

ak =

∞∑

n=0

bn+kc−n =

∞∑

n=0

2n+k−13n

n!= 2k−1

∞∑

n=0

6n

n!= 2k−1e6, k ≥ 1;

a−k =∞∑

n=1

bnc−k−n =∞∑

n=1

2n−13n+k

(n + k)!=

3k

2

∞∑

n=1

6n

(n + k)!, k ≥ 1.

Page 26: Complex Analysis for Applications, Math 132/1, Home Work Solutions …mt/132.1.01w/sol-2.pdf · Complex Analysis for Applications, Math 132/1, Home Work Solutions-II Masamichi Takesaki

26

In particular, we get

Res(f ; 0) = a−1 =3

2

∞∑

n=1

6n

(n + 1)!=

3

2 · 6

∞∑

n=1

6n+1

(n + 1)!

=1

4(e6 − 1 − 6) =

e6 − 7

4;

Res

(e3z

z − 2;∞

)=

e6 − 7

4.

c) The singularities of f (z) = cos zz2 are z = 0 and z = ∞. As limz→0 f(z) = ∞, the singularity

z = 0 is a pole. The numerator cos z does not vanish at z = 0, the order of the pole 0 is 2 andindeed we have

limz→0

zf(z) = limz→0

cos z

z= ∞; lim

z→0z2f (z) = lim

z→0cos z = 1.

So we have the Laurent series expansion of f near 0:

f(z) =cos z

z2=

1

z2+

a−1

z+ a0 + a1z + · · · + anzn + · · · ;

z2f(z) = cos z = 1 + a−1z + z0z2 + · · · + anzn+2 + · · · ;

a−1 =d

dz

(z2f(z)

)∣∣∣∣z=0

= − sin z|z=0 = −1;

Res(cos z

z2; 0

)= −1.

To investigate the singularity ∞ of f , set g(z) = f (1/z), i.e.,

g(z) = z2 cos1

z=

z2(eiz + e−

iz )

2.

Since g is bounded as z → 0 along the real axis and unbounded as z → 0 along the imaginary axis,0 is an essential singularity of g, i.e., ∞ is an essential singularity of the original function f .

To find the residue of f at ∞, we expand g near the origin:

g(z) =z2

2

( ∞∑

n=0

in

n!zn+

∞∑

n=0

(−i)n

n!zn

)=

z2

2

( ∞∑

n=0

(in

n!zn+

(−i)n

n!zn

))

= z2

( ∞∑

n=0

(−1)n

(2n)!z2n

)= z2 − 1

2+

1

4!z2− · · · .

Thus a simple inspection yields that Res(g; 0) = 0. Hence the residue of f at ∞ is 0.h) The denominator sin z of the function f(z) = z−1

sin zis zero at z = nπ, n ∈ Z but the numerator

z − 1 is not zero at these points. Hence these are singularities of f and indeed simple poles as

Page 27: Complex Analysis for Applications, Math 132/1, Home Work Solutions …mt/132.1.01w/sol-2.pdf · Complex Analysis for Applications, Math 132/1, Home Work Solutions-II Masamichi Takesaki

27

limz→nπ f(z) = ∞. Also z = ∞ is a singularity. But as limn→∞ nπ = ∞, the singularity ∞ is notisolated. Thus it is an essential singularity. The residue of f at a non-isolated singularity is notdefined. So we want just to calculate the residue of these poles.

Res

(z − 1

sin z; nπ

)= lim

z→nπ(z − nπ)

(z − 1)

sin z= (nπ − 1) lim

z→nπ

z − nπ

sin z

= (nπ − 1)1

cos z

∣∣∣∣z=nπ

(as

d

dz(sin z) = cos z

)

= (−1)n(nπ − 1).

This completes the answer. ♥

Page 251, 3-a, 3-d. Evalueate the integrals:

a)

|z|=5

sin z

z2 − 4dz; d)

|z|=3

eiz

z2(z − 2)(z + 5i)dz.

Answer. a) The function f (z) = sin zz2−4

has simple poles at z = ±2, which are both inside the contour:

|z| = 5. We first determine the residues of these poles:

Res(f ; 2) = limz→2

(z − 2)sin z

z2 − 4= lim

z→2

sin z

z + 2=

sin 2

4;

Res(f ;−2) = limz→−2

(z + 2)sin z

z2 − 4= lim

z→−2

sin z

z − 2=

sin(−2)

−4=

sin 2

4.

By the Residue Theorem, we get

|z|=5

sin z

z2 − 4dz = 2πi (Res(f ; 2) + Res(f ;−2)) = πi sin 2.

d) The denominator’s zero of the function f (z) = eiz

z2(z−2)(z+5i) are at z = 0, 2 and z = 5i, of

which only z = 0 and z = 2 are inside the contour |z| = 3. The singularity z = 0 is a double pole off and the singularity z = 2 is a simple pole. So we compute:

Res(f ; 0) =d

dzz2f(z)

∣∣∣∣z=0

=d

dz

(eiz

(z − 2)(z − 5i)

)∣∣∣∣z=0

=ieiz(z − 2)(z − 5i) + eiz(z − 5i) + eiz(z − 2)

(z − 2)2(z − 5i)2

∣∣∣∣z=0

=i(−2)(−5i) + (−5i) + (−2)

(−2)2(−5i)2=

12 + 5i

100;

Res(f ; 2) = limz→2

(z − 2)f(z) = limz→2

eiz

z2(z − 5i)=

e2i

4(2 − 5i)=

e2i(2 + 5i)

4 · 29.

Page 28: Complex Analysis for Applications, Math 132/1, Home Work Solutions …mt/132.1.01w/sol-2.pdf · Complex Analysis for Applications, Math 132/1, Home Work Solutions-II Masamichi Takesaki

28

Finally we evaluate the integral

|z|=3

eiz

z2(z − 2)(z − 5i)dz = 2πi (Res(f ; 0) + Res(f ; 2))

= 2πi

(12 + 5i

100+

e2i(2 + 5i)

4 · 29

)= πi

(12 + 5i

50+

e2i(2 + 5i)

58

).

This completes the answer. ♥