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Complex Analysis

Problems with solutions

Juan Carlos Ponce Campuzano

Copyright c

2016 Juan Carlos Ponce Campuzano

PUBLISHED BY JUAN CARLOS PON CE CAMPUZANO

ISBN 978-0-6485736-1-6

This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike

4.0 International License.

License at: https://creativecommons.org/licenses/by-nc-sa/4.0/

First e-book version, August 2015

Contents

Foreword .................................................... 5

1Complex Numbers ......................................... 7

1.1 Basic algebraic and geometric properties 7

1.2 Modulus 10

1.3 Exponential and Polar Form, Complex roots 13

2Functions ................................................... 19

2.1 Basic notions 19

2.2 Limits, Continuity and Differentiation 27

2.3 Analytic functions 31

2.3.1 Harmonicfunctions ............................................ 36

3Complex Integrals ........................................ 39

3.1 Contour integrals 39

3.2 Cauchy Integral Theorem and Cauchy Integral Formula 43

3.3 Improper integrals 56

4Series ....................................................... 59

4.1 Taylor and Laurent series 59

4.2 Classiﬁcation of singularities 68

Foreword

This text constitutes a collection of problems for using as an additional learning resource

for those who are taking an introductory course in complex analysis. The problems are

numbered and allocated in four chapters corresponding to different subject areas: Complex

Numbers,Functions,Complex Integrals and Series. The majority of problems are provided

with answers, detailed procedures and hints (sometimes incomplete solutions).

Of course, no project such as this can be free from errors and incompleteness. I will

be grateful to everyone who points out any typos, incorrect solutions, or sends any other

suggestion for improving this manuscript.

Contact: j.ponce@uq.edu.au

2016

1. Complex Numbers

1.1 Basic algebraic and geometric properties

1. Verify that

(a) √2−i−i1−√2i=−2i

(b) (2−3i)(−2+i) = −1+8i

Solution. We have

√2−i−i1−√2i=√2−i−i+√2=−2i,

and

(2−3i)(−2+i) = −4+2i+6i−3i2=−4+3+8i=−1+8i.

2. Reduce the quantity

5i

(1−i)(2−i)(3−i)

to a real number.

Solution. We have

5i

(1−i)(2−i)(3−i)=5i

(1−i)(5−5i)=i

(1−i)2=i

−2i=1

2

8Chapter 1. Complex Numbers

3. Show that

(a) Re(iz) = −Im(z);

(b) Im(iz) = Re(z).

Proof. Let z=x+yi with x=Re(z)and y=Im(z). Then

Re(iz) = Re(−y+xi) = −y=−Im(z)

and

Im(iz) = Im(−y+xi) = x=Re(z).

4.

Verify the associative law for multiplication of complex numbers. That is, show that

(z1z2)z3=z1(z2z3)

for all z1,z2,z3∈C.

Proof. Let zk=xk+iykfor k=1,2,3. Then

(z1z2)z3= ((x1+y1i)(x2+y2i))(x3+y3i)

= ((x1x2−y1y2) + i(x2y1+x1y2))(x3+y3i)

= (x1x2x3−x3y1y2−x2y1y3−x1y2y3)

+i(x2x3y1+x1x3y2+x1x2y3−y1y2y3)

and

z1(z2z3) = (x1+y1i)((x2+y2i))(x3+y3i))

= (x1+y1i)((x2x3−y2y3) + i(x2y3+x3y2))

= (x1x2x3−x3y1y2−x2y1y3−x1y2y3)

+i(x2x3y1+x1x3y2+x1x2y3−y1y2y3)

Therefore,

(z1z2)z3=z1(z2z3)

5. Compute

(a) 2+i

2−i;

(b) (1−2i)4.

Answer: (a) (3+4i)/5, (b) −7+24i.

1.1 Basic algebraic and geometric properties 9

6. Let fbe the map sending each complex number

z=x+yi −→ x y

−y x

Show that f(z1z2) = f(z1)f(z2)for all z1,z2∈C.

Proof. Let zk=xk+ykifor k=1,2. Then

z1z2= (x1+y1i)(x2+y2i) = (x1x2−y1y2) + i(x2y1+x1y2)

and hence

f(z1z2) = x1x2−y1y2x2y1+x1y2

−x2y1−x1y2x1x2−y1y2.

On the other hand,

f(z1)f(z2) = x1y1

−y1x1 x2y2

−y2x2=x1x2−y1y2x2y1+x1y2

−x2y1−x1y2x1x2−y1y2.

Therefore, f(z1z2) = f(z1)f(z2).

7. Use binomial theorem

(a+b)n=n

0an+n

1an−1b+... +n

n−1abn−1+n

nbn

=

n

∑

k=0n

kan−kbk

to expand

(a) (1+√3i)2011;

(b) (1+√3i)−2011.

Solution. By binomial theorem,

(1+√3i)2011 =

2011

∑

k=02011

k(√3i)k=

2011

∑

k=02011

k3k/2ik.

Since ik= (−1)mfor k=2meven and ik= (−1)mifor k=2m+1 odd,

(1+√3i)2011 =∑

0≤2m≤2011 2011

2m3m(−1)m

+i∑

0≤2m+1≤2011 2011

2m+13m√3(−1)m

=

1005

∑

m=02011

2m(−3)m+i

1005

∑

m=02011

2m+1(−3)m√3.

10 Chapter 1. Complex Numbers

Similarly,

(1+√3i)−2011 =1

1+√3i2011

= 1−√3i

4!2011

=1

42011

2011

∑

k=02011

k(−√3i)k

=1

42011

1005

∑

m=02011

2m(−3)m

−i

42011

1005

∑

m=02011

2m+1(−3)m√3.

8.

Suppose that

z1

and

z2

are complex numbers, with

z1z2

real and non-zero. Show that

there exists a real number rsuch that z1=rz2.

Proof. Let z1=x1+iy1and z2=x2+iy2with x1,x2,y1,y2∈R. Thus

z1z2=x1x2−y1y2+ (x1y2+y1x2)i

Since z1z2is real and non-zero, z16=0, z26=0, and

x1x2−y1y26=0 and x1y2+y1x2=0.

Thus, since z26=0, then

z1

z2

=x1+iy1

x2−iy2·x2+iy2

x2+iy2

=x1x2−y1y2+ (x1y2+y1x2)i

x2

2+y2

2

=x1x2−y1y2

x2

2+y2

2

.

By setting r=x1x2−y1y2

x2

2+y2

2

, we have the result.

1.2 Modulus

1. Show that

|z1−z2|2+|z1+z2|2=2(|z1|2+|z2|2)

for all z1,z2∈C.

1.2 Modulus 11

Proof. We have

|z1−z2|2+|z1+z2|2

= (z1−z2)(z1−z2)+(z1+z2)(z1+z2)

= (z1−z2)(z1−z2)+(z1+z2)(z1+z2)

= ((z1z1+z2z2)−(z1z2+z2z1)) + ((z1z1+z2z2)+(z1z2+z2z1))

=2(z1z1+z2z2) = 2(|z1|2+|z2|2).

2. Verify that √2|z| ≥ |Rez|+|Im z|.

Hint: Reduce this inequality to (|x|−|y|)2≥0.

Solution. Note that

0≤(|Rez|+|Im z|)2=|Rez|2−2|Re z||Im z|+|Im z|2.

Thus

2|Rez||Imz| ≤ |Re z|2+|Im z|2,

and then

|Rez|2+2|Re z||Im z|+|Im z|2≤2(|Re z|2+|Im z|2).

That is

(|Rez|+|Im z|)2≤2(|Rez|2+|Im z|2) = 2|z|2,

and therefore,

|Rez|+|Im z| ≤ √2|z|.

3. Sketch the curves in the complex plane given by

(a) Im(z) = −1;

(b) |z−1|=|z+i|;

(c) 2|z|=|z−2|.

Solution. Let z=x+yi.

(a) {Im(z) = −1}={y=−1}

is the horizontal line passing through the point

−i

.

(b) Since

|z−1|=|z+i| ⇔ |(x−1) + yi|=|x+ (y+1)i|

⇔ |(x−1) + yi|2=|x+ (y+1)i|2

⇔(x−1)2+y2=x2+ (y+1)2

⇔x+y=0,

the curve is the line x+y=0.

12 Chapter 1. Complex Numbers

(c) Since

2|z|=|z−2| ⇔ 2|x+yi|=|(x−2) + yi|

⇔4|x+yi|2=|(x−2) + yi|2

⇔4(x2+y2) = (x−2)2+y2

⇔3x2+4x+3y2=4

⇔x+2

32

+y2=16

9

⇔

z+2

3

=4

3

the curve is the circle with centre at −2/3 and radius 4/3.

4. Show that

R4−R

R2+R+1≤

z4+iz

z2+z+1≤R4+R

(R−1)2

for all zsatisfying |z|=R>1.

Proof. When |z|=R>1,

|z4+iz| ≥ |z4|−|iz|=|z|4−|i||z|=R4−R

and

|z2+z+1| ≤ |z2|+|z|+|1|=|z|2+|z|+1=R2+R+1

by triangle inequality. Hence

z4+iz

z2+z+1≥R4−R

R2+R+1.

On the other hand,

|z4+iz| ≤ |z4|+|iz|=|z|4+|i||z|=R4+R

and

|z2+z+1|= z−−1+√3i

2! z−−1−√3i

2!

=

z−−1+√3i

2

z−−1−√3i

2

≥ |z|−

−1+√3i

2! |z|−

−1−√3i

2!

= (R−1)(R−1) = (R−1)2

1.3 Exponential and Polar Form, Complex roots 13

Therefore,

z4+iz

z2+z+1≤R4+R

(R−1)2.

5. Show that

|Log(z)| ≤ |ln|z||+π(1.1)

for all z6=0.

Proof. Since Log(z) = ln|z|+iArg(z)for −π<Arg(z)≤π,

|Log(z)|=|ln|z|+iArg(z)| ≤ |ln |z||+|iArg(z)| ≤ |ln|z||+π.

1.3 Exponential and Polar Form, Complex roots

1. Express the following in the form x+iy, with x,y∈R:

(a) i

1−i+1−i

i;

(b) all the 3rd roots of −8i;

(c) i+1

√21337

Solution. (a)

i

1−i+1−i

i=i2+ (1−i)2

(1−i)i

=−1−2i

1−i·1−i

1−i

=−1+i−2i−2

2

=−3−i

2=−3

2−i

2

(b) We have that

−8i=23exp−iπ

2

Thus the cube roots are

2exp −iπ

6,2exp iπ

2and 2exp 7iπ

6.

That is √3−i,2,−√3−i

14 Chapter 1. Complex Numbers

(c)

i+1

√21337

=exp iπ

41337

=exp 1337πi

4

=exp167 ·2πi+π

4i

=exp π

4i=1+i

√2.

2. Find the principal argument and exponential form of

(a) z=i

1+i;

(b) z=√3+i;

(c) z=2−i.

Answer:

(a) Arg(z) = π/4 and z= (√2/2)exp(πi/4).

(b) Arg(z) = π/6 and z=2exp(πi/6).

(c) Arg(z) = −tan−1(1/2)and z=√5exp(−tan−1(1/2)i).

3. Find all the complex roots of the equations:

(a) z6=−9;

(b) z2+2z+ (1−i) = 0.

Solution. (a) The roots are

z=6

√−9=6

√9eπi=3

√3eπi/6e2mπi/6(m=0,1,2,3,4,5)

=35/6

2+

3

√3

2i,3

√3i,−35/6

2+

3

√3

2i,−35/6

2−

3

√3

2i,−3

√3i,35/6

2−

3

√3

2i.

(b) The roots are

z=−2+p4−4(1−i)

2=−1+√i

=−1+peπi/2=−1+eπi/4e2mπi/2(m=0,1)

= −1+√2

2!+√2

2i, −1−√2

2!−√2

2i.

4.

Find the four roots of the polynomial

z4+16

and use these to factor

z4+16

into two

quadratic polynomials with real coefﬁcients.

1.3 Exponential and Polar Form, Complex roots 15

Solution. The four roots of z4+16 are given by

4

√−16 =4

√16eπi=4

√16eπi/4e2mπi/4

=2eπi/4,2e3πi/4,2e5πi/4,2e7πi/4

for m=0,1,2,3. We see that these roots appear in conjugate pairs:

2eπi/4=2e7πi/4and 2e3πi/4=2e5πi/4.

This gives the way to factor

z4+16

into two quadratic polynomials of real coefﬁ-

cients:

z4+16 = (z−2eπi/4)(z−2e3πi/4)(z−2e5πi/4)(z−2e7πi/4)

=(z−2eπi/4)(z−2e7πi/4)(z−2e3πi/4)(z−2e5πi/4)

= (z2−2Re(2eπi/4)z+4)(z2−2 Re(2e3πi/4)z+4)

= (z2−2√2z+4)(z2+2√2z+4)

5. Do the following:

(a) Use exponential form to compute

i. (1+√3i)2011;

ii. (1+√3i)−2011.

(b) Prove that

1005

∑

m=02011

2m(−3)m=22010

and

1005

∑

m=02011

2m+1(−3)m=22010.

Solution. Since

1+√3i=2 1

2+√3

2i!=2exp πi

3,

we have

(1+√3i)2011 =22011 exp2011πi

3=22011 exp2011πi

3

=22011 exp670πi+πi

3

=22011 expπi

3=22011 1

2+√3

2i!

=22010(1+√3i).

16 Chapter 1. Complex Numbers

Similarly,

(1+√3i)−2011 =2−2013(1−√3i).

By Problem 7 in section 1.1, we have

22010(1+√3i) = (1+√3i)2011

=

1005

∑

m=02011

2m(−3)m+i

1005

∑

m=02011

2m+1(−3)m√3.

It follows that

1005

∑

m=02011

2m(−3)m=

1005

∑

m=02011

2m+1(−3)m=22010.

6. Establish the identity

1+z+z2+···+zn=1−zn+1

1−z(z6=1)

and then use it to derive Lagrange’s trigonometric identity:

1+cosθ+cos 2θ···+cosnθ=1

2+sin (2n+1)θ

2

2sin θ

2

(0<θ<2π).

Hint:

As for the ﬁrst identity, write

S=1+z+z2+···+zn

and consider the

difference S−zS. To derive the second identity, write z=eiθin the ﬁrst one.

Proof. If z6=1, then

(1−z)(1+z+···+zn) = 1+z+···+zn−(z+z2+···+zn+1)

=1−zn+1

Thus

1+z+z2+···+zn=

1−zn+1

1−z,if z6=1

n+1,if z=1.

Taking z=eiθ, where 0 <θ<2π, then z6=1. Thus

1+eiθ+e2iθ+···+eniθ=1−e(n+1)θ

1−eiθ=1−e(n+1)θ

−eiθ/2eiθ/2−e−iθ/2

=−e−iθ/2(1−e(n+1)θ)

2isin(θ/2)

=

ie−iθ/2−e(n+1

2)iθ

2sin(θ/2)

=1

2+sin[(n+1

2)θ]

2sin(θ/2)+icos(θ/2)−cos[(n+1

2)θ]

2sin(θ/2)

1.3 Exponential and Polar Form, Complex roots 17

Equating real and imaginary parts, we obtain

1+cosθ+cos 2θ···+cosnθ=1

2+sin[(n+1

2)θ]

2sin(θ/2)

and

sinθ+sin 2θ···+sinnθ=cos(θ/2)−cos[(n+1

2)θ]

2sin(θ/2).

7.

Use complex numbers to prove the

Law of Cosine

: Let

∆ABC

be a triangle with

|BC|=a,|CA|=b,|AB|=cand ∠BCA =θ. Then

a2+b2−2abcos θ=c2.

Hint: Place Cat the origin, Bat z1and Aat z2. Prove that

z1z2+z2z1=2|z1z2|cosθ.

Proof.

Following the hint, we let

C=0

,

B=z1

and

A=z2

. Then

a=|z1|

,

b=|z2|

and c=|z2−z1|. So

a2+b2−c2=|z1|2+|z2|2−|z2−z1|2

= (z1z1+z2z2)−(z2−z1)(z2−z1)

= (z1z1+z2z2)−(z2−z1)(z2−z1)

= (z1z1+z2z2)−(z1z1+z2z2−z1z2−z2z1)

=z1z2+z2z1.

Let z1=r1eiθ1and z2=r2eiθ2. Then

z1z2+z2z1=r1eiθ1r2eiθ2+r2eiθ2r1eiθ1

= (r1eiθ1)(r2e−iθ2)+(r2eiθ2)(r1e−iθ1)

=r1r2ei(θ1−θ2)+r1r2ei(θ2−θ1)

=2r1r2cos(θ1−θ2) = 2|z1||z2|cosθ=2ab cosθ.

Therefore, we have

a2+b2−c2=z1z2+z2z1=2abcos θ

and hence

a2+b2−2abcos θ=c2.

2. Functions

2.1 Basic notions

1.

Write the following functions

f(z)

in the forms

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

under Carte-

sian coordinates with u(x,y) = Re(f(z)) and v(x,y) = Im(f(z)):

(a) f(z) = z3+z+1

(b) f(z) = z3−z;

(c) f(z) = 1

i−z;

(d) f(z) = exp(z2).

Solution. (a)

f(z) = (x+iy)3+ (x+iy) + 1

= (x+iy)(x2−y2+2ixy) + x+iy +1

=x3−xy2+2ix2y+ix2y−iy3−2xy2+x+iy +1

=x3−3xy2+x+1+i(3x2y−y3+y).

(b)

f(z) = z3−z= (x+yi)3−(x+yi)

= (x3+3x2yi −3xy2−y3i)−(x+yi)

= (x3−3xy2−x) + i(3x2y−y3−y),

20 Chapter 2. Functions

(c)

f(z) = 1

i−z=1

−x+ (1−y)i

=−x−(1−y)i

x2+ (1−y)2

=−x

x2+ (1−y)2−i1−y

x2+ (1−y)2

(d)

f(z) = exp(z2) = exp((x+yi)2)

=exp((x2−y2) + 2xyi)

=ex2−y2(cos(2xy) + isin(2xy))

=ex2−y2cos(2xy)−iex2−y2sin(2xy)

2.

Suppose that

f(z) = x2−y2−2y+i(2x−2xy)

, where

z=x+iy

. Use the expressions

x=z+z

2and y=z−z

2i

to write f(z)in terms of zand simplify the result.

Solution. We have

f(z) = x2−y2−2y+i(2x−2xy)

=x2−y2+i2x−i2xy −2y

= (x−iy)2+i(2x+2iy)

=z2+2iz.

3. Suppose p(z)is a polynomial with real coefﬁcients. Prove that

(a) p(z) = p(z);

(b) p(z) = 0 if and only if p(z) = 0;

(c)

the roots of

p(z) = 0

appear in conjugate pairs, i.e., if

z0

is a root of

p(z) = 0

,

so is z0.

Proof. Let p(z) = a0+a1z+... +anznfor a0,a1,..., an∈R. Then

p(z) = a0+a1z+···+anzn

=a0+a1z+···+anzn

=a0+ (a1)z+···+ (an)zn

=a0+a1z+···+anzn=p(z).

If

p(z) = 0

, then

p(z) = 0

and hence

p(z) = p(z) = 0

; on the other hand, if

p(z) = 0

,

then p(z) = p(z) = 0 and hence p(z) = 0.

By the above,

p(z0) = 0

if and only if

p(z0) = 0

. Therefore,

z0

is a root of

p(z) = 0

if and only if z0is.

2.1 Basic notions 21

4. Let

T(z) = z

z+1.

Find the inverse image of the disk |z|<1/2 under Tand sketch it.

Solution. Let D={|z|<1/2}. The inverse image of Dunder Tis

T−1(D) = {z∈C:T(z)∈D}={|T(z)|<1

2}

=z:

z

z+1

<1

2={2|z|<|z+1|}.

Let z=x+yi. Then

2|z|<|z+1| ⇔ 4(x2+y2)<(x+1)2+y2

⇔3x2−2x+3y2<1

⇔x−1

32

+y2<4

9

⇔

z−1

3

<2

3

So

T−1(D) = z:

z−1

3

<2

3

is the disk with centre at 1/3 and radius 2/3.

5.

Sketch the following sets in the complex plane

C

and determine whether they are

open, closed, or neither; bounded; connected. Brieﬂy state your reason.

(a) |z+3|<1;

(b) |Im(z)| ≥ 1;

(c) 1 ≤ |z+3|<2.

Solution.

(a) Since

{|z+3|<1}={(x+3)2+y2−1<0}

and

f(x,y) = (x+3)2+

y2−1

is a continuous function on

R2

, the set is open. It is not closed since the only

sets that are both open and closed in Care /0 and C.

Since

|z|=|z+3−3| ≤ |z+3|+|−3|=|z+3|+3<4

for all |z+3|<1, {|z+3|<1} ⊂ {|z|<4}and hence it is bounded.

It is connected since it is a convex set.

Solution. (b) We have

{|Im(z)| ≥ 1}={|y| ≥ 1}={y≥1}∪{y≤ −1}.

22 Chapter 2. Functions

Since

f(x,y) = y

is continuous on

R2

, both

{y≥1}

and

{y≤ −1}

are closed and

hence

{|Im(z)| ≥ 1}

is closed. It is not open since the only sets that are both open

and closed in Care /0 and C.

Since zn=n+2i∈ {|Im(z)| ≥ 1}for all n∈Zand

lim

n→∞|zn|=lim

n→∞pn2+4=∞,

the set is unbounded.

The set is not connected. Otherwise, let

p=2i

and

q=−2i

. There is a polygonal

path

p0p1∪p1p2∪... ∪pn−1pn

with p0=p,pn=qand pk∈ {|Im(z)| ≥ 1}for all 0 ≤k≤n.

Let

0≤m≤n

be the largest integer such that

pm∈{y≥1}

. Then

pm+1∈{y≤ −1}

.

So

Im(pm)≥1>0

and

Im(pm+1)≤ −1<0

. It follows that there is a point

p∈

pmpm+1

such that

Im(p) = 0

. This is a contradiction since

pmpm+1⊂{|Im(z)|≥ 1}

but p6∈ {|Im(z)| ≥ 1}. Therefore the set is not connected.

Solution.

(c) Since

−2∈ {1≤ |z+3|<2}

and

{|z+2|<r} 6⊂ {1≤ |z+3|<2}

for all

r>0

,

{1≤ |z+3|<2}

is not open. Similarly,

−1

is a point lying on its

complement

{1≤ |z+3|<2}c={|z+3| ≥ 2}∪{|z+3|<1}

and

{|z+1|<r}6⊂ {1≤ |z+3|<2}c

for all

r>0

. Hence

{1≤|z+3|<2}c

is not

open and

{1≤ |z+3|<2}

is not closed. In summary,

{1≤ |z+3|<2}

is neither

open nor closed.

Since

|z|=|z+3−3| ≤ |z+3|+|−3|<5

for all |z+3|<2, {1≤ |z+3|<2} ⊂ {|z|<5}and hence it is bounded.

The set is connected. To see this, we let

p1=−3/2,p2=−3+3i/2,p3=−9/2

and

p4=−3−3i/2

. All these points lie on the circle

{|z+3|=3/2}

and hence lie

in {1≤ |z+3|<2}.

It is easy to check that for every point

p∈{1≤ |z+3|<2}

,

ppk⊂{1≤ |z+3|<2}

for at least one pk∈ {p1,p2,p3,p4}. So the set is connected.

6. Show that

|sinz|2= (sin x)2+ (sinhy)2

for all complex numbers z=x+yi.

Proof.

|sin(z)|2=|sin(x+yi)|2=|sin(x)cos(yi) + cos(x)sin(yi)|2

=|sin(x)cosh(y)−icos(x)sinh(y)|2

=sin2xcosh2y+cos2xsinh2y

=sin2x(1+sinh2y) + cos2xsinh2y

=sin2x+ (cos2x+sin2x)sinh2y= (sinx)2+ (sinh y)2.

2.1 Basic notions 23

7. Show that

|cos(z)|2= (cosx)2+ (sinh y)2

for all z∈C, where x=Re(z)and y=Im(z).

Proof.

|cos(z)|2=|cos(x+yi)|2=|cos(x)cos(yi)−sin(x)sin(yi)|2

=|cos(x)cosh(y)−isin(x)sinh(y)|2

=cos2xcosh2y+sin2xsinh2y

=cos2x(1+sinh2y) + sin2xsinh2y

=cos2x+ (cos2x+sin2x)sinh2y= (cosx)2+ (sinh y)2

8. Show that

tan(z1+z2) = tanz1+tan z2

1−(tanz1)(tan z2)

for all complex numbers

z1

and

z2

satisfying

z1,z2,z1+z26=nπ+π/2

for any integer

n.

Proof. Since

tanz1+tan z2=i(e−iz1−eiz1)

eiz1+e−iz1+i(e−iz2−eiz2)

eiz2+e−iz2

=i(e−iz1−eiz1)(eiz2+e−iz2)+(e−iz2−eiz2)(eiz1+e−iz1)

(eiz1+e−iz1)(eiz2+e−iz2)

=−2iei(z1+z2)−e−i(z1+z2)

(eiz1+e−iz1)(eiz2+e−iz2)

and

1−(tanz1)(tan z2) = 1−i(e−iz1−eiz1)

eiz1+e−iz1i(e−iz2−eiz2)

eiz2+e−iz2

=(e−iz1+eiz1)(e−iz2+eiz2)+(e−iz1−e−iz1)(e−iz2−eiz2)

(e−iz1+eiz1)(e−iz2+eiz2)

=2ei(z1+z2)+e−i(z1+z2)

(eiz1+e−iz1)(eiz2+e−iz2),

we have

tanz1+tan z2

1−(tanz1)(tan z2)=−iei(z1+z2)−e−i(z1+z2)

ei(z1+z2)+e−i(z1+z2)=tan(z1+z2).

24 Chapter 2. Functions

Alternatively, we can argue as follows if we assume that the identity holds for

z1

and

z2real. Let

F(z1,z2) = tan(z1+z2)−tanz1+tan z2

1−(tanz1)(tan z2).

We assume that F(z1,z2) = 0 for all z1,z2∈Rwith z1,z2,z1+z26=nπ+π/2.

Fixing z1∈R, we let f(z) = F(z1,z). Then f(z)is analytic in its domain

C\({nπ+π/2}∪{nπ+π/2−z1}).

And we know that

f(z) = 0

for

z

real. Therefore, by the uniqueness of analytic

functions,

f(z)≡0

in its domain. So

F(z1,z2) = 0

for all

z1∈R

and

z2∈C

in its

domain.

Fixing z2∈C, we let g(z) = F(z,z2). Then g(z)is analytic in its domain

C\({nπ+π/2}∪{nπ+π/2−z2}).

And we have proved that

g(z) = 0

for

z

real. Therefore, by the uniqueness of analytic

functions,

g(z)≡0

in its domain. Hence

F(z1,z2) = 0

for all

z1∈C

and

z2∈C

in

its domain.

9. Find all the complex roots of the equation cosz=3.

Solution.

Since

cosz= (eiz +e−iz )/2

, it comes down to solve the equation

eiz +

e−iz =6, i.e.,

w+w−1=6⇔w2−6w+1=0

if we let

w=eiz

. The roots of

w2−6w+1=0

are

w=3±2√2

. Therefore, the

solutions for cosz=3 are

iz =log(3±2√2)⇔z=−i(ln(3±2√2) + 2nπi) = 2nπ−iln(3±2√2)

for nintegers.

10. Calculate sinπ

4+i.

Solution.

sinπ

4+i=1

2i(ei(π/4+i)−e−i(π/4+i))

=1

2i(e−1eπi/4−ee−πi/4)

=1

2ie−1(cos π

4+isin π

4)−e(cos π

4−isin π

4)

=√2

4e+1

e+√2

4e−1

ei

2.1 Basic notions 25

11. Compute cosπ

3+i.

Solution.

cosπ

3+i=1

2(ei(π/3+i)+e−i(π/3+i))

=1

2(e−1eπi/3+ee−πi/3)

=1

2e−1(cos π

3+isin π

3) + e(cos π

3−isin π

3)

=1

4e+1

e−√3i

4e−1

e

12. Find iiand its principal value.

Solution. We have

ii=eilogi=ei(2nπi+πi/2)=e−2nπ−π/2

for nintegers and its principal value given by

ii=eiLogi=ei(πi/2)=e−π/2.

13. Let f(z)be the principal branch of 3

√z.

(a) Find f(−i).

Solution.

f(−i) = exp(1

3Log(−i)) = exp(1

3(−πi

2)) = exp(−πi

6) = √3

2−i

2

(b) Show that

f(z1)f(z2) = λf(z1z2)

for all z1,z26=0, where λ=1,−1+√3i

2or −1−√3i

2.

26 Chapter 2. Functions

Proof. Since

f(z1)f(z2)

f(z1z2)=exp(1

3Logz1+1

3Logz2−1

3Log(z1z2))

=exp(1

3(Logz1+Log z2−Log(z1z2)))

=exp(i

3(Argz1+Arg z2−Arg(z1z2))) = exp(2nπi

3)

for some integer n,λ=exp(2nπi/3). Therefore,

λ=

1 if n=3k

−1+√3i

2if n=3k+1

−1−√3i

2if n=3k+2

where k∈Z.

14. Let f(z)be the principal branch of z−i.

(a) Find f(i).

Solution.

f(i) = i−i=exp(−iLog(i)) = exp(−i(πi/2)) = eπ/2.

(b) Show that

f(z1)f(z2) = λf(z1z2)

for all z1,z26=0, where λ=1,e2πor e−2π.

Proof. Since

f(z1)f(z2)

f(z1z2)=exp(−iLogz1−iLog z2+iLog(z1z2))

=exp(−i(Logz1+Log z2−Log(z1z2)))

=exp(−i(iArgz1+iArgz2−iArg(z1z2)))

=exp(Argz1+Argz2−Arg(z1z2))

=exp(2nπ)

for some integer n,λ=exp(2nπ). And since

−π<Arg(z1)≤π,−π<Arg(z2)≤π

and

−π<Arg(z1z2)≤π,

we conclude that

−3π<Argz1+Arg z2−Arg(z1z2)<3π

and hence −3<2n<3. So n=−1,0 or 1 and λ=e−2π,1 or e2π.

2.2 Limits, Continuity and Differentiation 27

2.2 Limits, Continuity and Differentiation

1. Compute the following limits if they exist:

(a) lim

z→−i

iz3+1

z2+1;

(b) lim

z→∞

4+z2

(z−1)2.

(c) lim

z→0

Im(z)

z.

Solution. (a)

lim

z→−i

iz3+1

z2+1=lim

z→−i

i(z3+i3)

z2+1

=lim

z→−i

i(z+i)(z2−iz +i2)

(z+i)(z−i)

=lim

z→−i

i(z2−iz +i2)

z−i

=ilimz→−i(z2−iz +i2)

limz→−i(z−i)

=ilimz→−iz2−ilimz→−iz+limz→−ii2

limz→−iz−limz→−ii

=i((−i)2−i(−i) + i2)

−i−i=3

2

(b)

lim

z→∞

4+z2

(z−1)2=lim

z→0

4+z−2

(z−1−1)2

=lim

z→0

4z2+1

(1−z)2=limz→0(4z2+1)

limz→0(1−z)2

=4(limz→0z)2+limz→01

(limz→01−limz→0z)2=1

(c) Since

lim

Re(z)=0,z→0

Im(z)

z=lim

y→0

y

yi =−i

and

lim

Im(z)=0,z→0

Im(z)

z=lim

x→0

0

x=0

the limit does not exist.

28 Chapter 2. Functions

2. Show the following limits:

(a) lim

z→∞

4z5

z5−42z=4;

(b) lim

z→∞

z4

z2+42z=∞;

(c) lim

z→∞

(az +b)3

(cz +d)3=a3

c3, if c6=0.

Solution. (a)

lim

z→∞

4z5

z5−42z=lim

z→0

41

z5

1

z5−421

z=lim

z→0

4

1−42z4=4.

(b) Notice that

lim

z→∞

z4

z2+42z⇐⇒ lim

z→01/z4

1/z2+42/z−1

⇐⇒ lim

z→01

z2+42z3−1

⇐⇒ lim

z→0z2+42z3=0.

Therefore, lim

z→∞

z4

z2+42z=∞.

(c)

lim

z→∞

(az +b)3

(cz +d)3=lim

z→0

(a/z+b)3

(c/z+d)3=lim

z→0

(a+bz)3

(c+dz)3=a3

c3.

3. Show that lim

z→0z/zdoes not exist.

Hint:

Consider what happens to the function at points of the form

x+0i

for

x→0

,

x6=0, and then at points of the form 0 +yi for y→0,y6=0.

Proof. For z=x+0i,x6=0,

z

z=x

x=1→1 as x→0.

On the other hand, for z=0+yi,y6=0,

z

z=yi

−yi =−1→ −1 as y→0.

However,

lim

z→0z/z

must be independent of direction of approach. Hence limit does

not exist.

2.2 Limits, Continuity and Differentiation 29

4. Show that if f(z)is continuous at z0, so is |f(z)|.

Proof.

Let

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

. Since

f(z)

is continuous at

z0=x0+y0i

,

u(x,y)

and v(x,y)are continuous at (x0,y0). Therefore,

(u(x,y))2+ (v(x,y))2

is continuous at

(x0,y0)

since the sums and products of continuous functions are

continuous. It follows that

|f(z)|=q(u(x,y))2+ (v(x,y))2

is continuous at

z0

since the compositions of continuous functions are continuous.

5. Let

f(z) = (z3/z2if z6=0

0 if z=0

Show that

(a) f(z)is continuous everywhere on C;

(b) the complex derivative f0(0)does not exist.

Proof.

Since both

z3

and

z2

are continuous on

C∗=C\{0}

and

z26=0

,

f(z) = z3/z2

is continuous on C∗.

At z=0, we have

lim

z→0|f(z)|=lim

z→0

z3

z2

=lim

z→0|z|=0

and hence

limz→0f(z) = 0=f(0)

. So

f

is also continuous at

0

and hence continuous

everywhere on C.

The complex derivative f0(0), if exists, is given by

lim

z→0

f(z)−f(0)

z−0=lim

z→0

z3

z3.

Let z=x+yi. If y=0 and x→0, then

lim

z=x→0

z3

z3=lim

x→0

x3

x3=1.

On the other hand, if x=0 and y→0, then

lim

z=yi→0

z3

z3=lim

x→0

(−yi)3

(yi)3=−1.

So the limit limz→0z3/z3and hence f0(0)do not exist.

30 Chapter 2. Functions

6.

Show that

f(z)

in problem 5 is actually nowhere differentiable, i.e., the complex

derivative f0(z)does not exist for any z∈C.

Proof. It sufﬁces to show that C-R equations fail at every z6=0:

∂

∂x+i∂

∂yf(z) = ∂

∂x+i∂

∂yz3

z2

=∂

∂xz3

z2+i∂

∂yz3

z2

=3z2

z2−2z3

z3+i−3iz2

z2−2iz3

z3

=6z2

z26=0

for z6=0.

7. Find f0(z)when

(a) f(z) = z2−4z+2;

(b) f(z) = (1−z2)4;

(c) f(z) = z+1

2z+1(z6=−1

2);

(d) f(z) = e1/z(z6=0).

Answer. (a) 2z−4; (b) −8(1−z2)3z; (c) −1/(2z+1)2; (d) −e1/z/z2.

8.

Prove the following version of complex L’Hospital: Let

f(z)

and

g(z)

be two

complex functions deﬁned on

|z−z0|<r

for some

r>0

. Suppose that

f(z0) =

g(z0) = 0, f(z)and g(z)are differentiable at z0and g0(z0)6=0. Then

lim

z→z0

f(z)

g(z)=f0(z0)

g0(z0)

[Refer to: problems 1c and 6 in section 3.1; and problem 9 in section 3.2]

Proof. Since f(z)and g(z)are differentiable at z0, we have

lim

z→z0

f(z)−f(z0)

z−z0

=f0(z0)

and

lim

z→z0

g(z)−g(z0)

z−z0

=g0(z0).

And since g0(z0)6=0,

lim

z→z0

f(z)−f(z0)

g(z)−g(z0)=limz→z0(f(z)−f(z0))/(z−z0)

limz→z0(g(z)−g(z0))/(z−z0)=f0(z0)

g0(z0).

Finally, since f(z0) = g(z0) = 0,

lim

z→z0

f(z)

g(z)=f0(z0)

g0(z0).

2.3 Analytic functions 31

9.

Show that if

f(z)

satisﬁes the Cauchy-Riemann equations at

z0

, so does

(f(z))n

for

every positive integer n.

Proof. Since f(z)satisﬁes the Cauchy-Riemann equations at z0,

∂

∂x+i∂

∂yf(z) = 0

at z0. Therefore,

∂

∂x+i∂

∂y(f(z))n=∂

∂x(f(z))n+i∂

∂y(f(z))n

= ( f(z))n−1∂

∂xf(z) + i(f(z))n−1∂

∂yf(z)

= ( f(z))n−1∂

∂x+i∂

∂yf(z) = 0

at z0.

2.3 Analytic functions

1. Explain why the function f(z) = 2z2−3−zez+e−zis entire.

Proof.

Since every polynomial is entire,

2z2−3

is entire; since both

−z

and

ez

are

entire, their product

−zez

is entire; since

ez

and

−z

are entire, their composition

e−z

is entire. Finally, f(z)is the sum of 2z3−3, −zezand e−zand hence entire.

2.

Let

f(z)

be an analytic function on a connected open set

D

. If there are two constants

c1

and

c2∈C

, not all zero, such that

c1f(z) + c2f(z) = 0

for all

z∈D

, then

f(z)

is

a constant on D.

Proof.

If

c2=0

,

c16=0

since

c1

and

c2

cannot be both zero. Then we have

c1f(z) =

0 and hence f(z) = 0 for all z∈D.

If

c26=0

,

f(z) = −(c1/c2)f(z)

. And since

f(z)

is analytic in

D

,

f(z)

is anlaytic in

D

. So both

f(z)

and

f(z)

are analytic in

D

. Therefore, both

f(z)

and

f(z)

satisfy

Cauchy-Riemann equations in D. Hence

∂

∂x+i∂

∂y(u+vi) = 0

and

∂

∂x+i∂

∂y(u−vi) = 0

in D, where f(z) = u(x,y) + iv(x,y)with u=Re(f)and v=Im(f). It follows that

∂

∂x+i∂

∂yu=∂

∂x+i∂

∂yv=0

and hence

ux=uy=vx=vy=0

in

D

. Therefore,

u

and

v

are constants on

D

and

hence f(z)≡const.

32 Chapter 2. Functions

3. Show that the function sin(z)is nowhere analytic on C.

Proof. Since

∂

∂x+i∂

∂ysin(z) = ∂

∂xsin(z) + i∂

∂ysin(z)

=cos(z)∂z

∂x+icos(z)∂z

∂y

=cos(z) + icos(z)(−i) = 2cos(z)

sin(z)

is not differentiable and hence not analytic at every point

z

satisfying

cos(z)6=

0

. At every point

z0

satisfying

cos(z0) = 0

, i.e.,

z0=nπ+π/2

,

sin(z)

is not differ-

entiable in

|z−z0|<r

for all

r>0

. Hence

sin(z)

is not analytic at

z0=nπ+π/2

either. In conclusion, sin(z)is nowhere analytic.

4.

Let

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

be an entire function satisfying

u(x,y)≤x

for all

z=

x+yi. Show that f(z)is a polynomial of degree at most one.

Proof.

Let

g(z) = exp(f(z)−z)

. Then

|g(z)|=exp(u(x,y)−x)

. Since

u(x,y)≤x

,

|g(z)| ≤ 1

for all

z

. And since

g(z)

is entire,

g(z)

must be constant by Louville’s

theorem. Therefore,

g0(z)≡0

. That is,

(f0(z)−1)exp(f(z)−z)≡0

and hence

f0(z) = 1 for all z. So f(z)≡z+cfor some constant c.

5. Show that

|exp(z3+i) + exp(−iz2)| ≤ ex3−3xy2+e2xy

where x=Re(z)and y=Im(z).

Proof. Note that |ez|=eRe(z). Therefore,

|exp(z3+i) + exp(−iz2)| ≤ |exp(z3+i)|+|exp(−iz2)|

=exp(Re(z3+i)) + exp(Re(−iz2))

=exp(Re((x3−3xy2)+(3x2y−y3+1)i))

+exp(Re(2xy −(x2−y2)i))

=ex3−3xy2+e2xy.

6. Let f(z) = u(x,y) + iv(x,y)be an entire function satisfying

v(x,y)≥x

for all

z=x+yi

, where

u(x,y) = Re(f(z))

and

v(x,y) = Im(f(z))

. Show that

f(z)

is a polynomial of degree 1.

2.3 Analytic functions 33

Proof. Let g(z) = exp(i f (z) + z). Then

|g(z)|=|exp((−v(x,y) + iu(x,y)) + (x+iy))|=exp(x−v(x,y)).

Since

v(x,y)≤x

,

x−v(x,y)≤0

and

|g(z)| ≤ 1

for all

z

. And since

g(z)

is entire,

g(z)

must be constant by Louville’s theorem. Therefore,

g0(z)≡0

. That is,

(i f 0(z) +

1)exp(i f (z)+ z)≡0

and hence

f0(z) = i

for all

z

. So

f(z)≡iz+c

for some constant

c.

7. Show that the entire function cosh(z)takes every value in Cinﬁnitely many times.

Proof.

For every

w0∈C

, the quadratic equation

y2−2w0y+1=0

has a complex

root

y0

. We cannot have

y0=0

since

02−2w0·0+16=0

. Therefore,

y06=0

and

there is z0∈Csuch that ez0=y0. Then

cosh(z0) = ez0+e−z0

2=y2

0+1

2y0

=2w0y0

2y0

=w0.

And since

cosh(z+2πi) = cosh(z)

,

cosh(z0+2nπi) = w0

for all integers

n

. There-

fore, cosh(z)takes every value w0inﬁnitely many times.

8.

Determine which of the following functions

f(z)

are entire and which are not? You

must justify your answer. Also ﬁnd the complex derivative

f0(z)

of

f(z)

if

f(z)

is

entire. Here z=x+yi with x=Re(z)and y=Im(z).

(a) f(z) = 1

1+|z|2

Solution. Since

u(x,y) = Re(f(z)) = (1+x2+y2)−1

and

v(x,y) = 0

,

ux=

2x(1+x2+y2)−26=0=vy

. Hence the Cauchy-Riemann equations fail for

f(z)

and f(z)is not entire.

(b) f(z) = 2(3z)

(here

2z

and

3z

are taken to be the principle values of

2z

and

3z

,

respectively, by convention)

Solution. Let

g(z) = 2z

and

h(z) = 23z

. Since both

g(z)

and

h(z)

are entire,

f(z) = g(h(z)) is entire and

f0(z) = g0(h(z))h0(z) = 23z3z(ln2)(ln 3)

by chain rule.

(c) f(z) = (x2−y2)−2xyi

Solution. Since

∂

∂x+i∂

∂yf= (2x−2yi) + i(−2y−2xi) = 4x−4yi 6=0,

the Cauchy-Riemann equations fail for f(z)and hence f(z)is not entire.

34 Chapter 2. Functions

(d) f(z) = (x2−y2) + 2xyi

Solution. Since

∂

∂x+i∂

∂yf= (2x−2yi) + i(2y+2xi) = 0,

the Cauchy-Riemann equations hold for

f(z)

everywhere. And since

fx

and

fy

are continous, f(z)is analytic on C. And f0(z) = fx=2x+2yi =2z.

9. Let CRdenote the upper half of the circle |z|=Rfor some R>1. Show that

eiz

z2+z+1≤1

(R−1)2

for all zlying on CR.

Proof. For z∈CR,|z|=Rand Im(z)≥0. Let z=x+yi. Since y=Im(z)≥0,

|eiz|=|ei(x+yi)|=|e−y+xi|=e−y≤1

for z∈CR. And since

|z2+z+1|= z−−1+√3i

2! z−−1−√3i

2!

=

z−−1+√3i

2

z−−1−√3i

2

≥ |z|−

−1+√3i

2! |z|−

−1−√3i

2!

= (R−1)(R−1) = (R−1)2,

we obtain

eiz

z2+z+1≤1

(R−1)2

for z∈CR.

10. Let

f(z) = (z2/|z|if z6=0

0 if z=0

Show that f(z)is continuous everywhere but nowhere analytic on C.

2.3 Analytic functions 35

Proof.

Since both

z

and

|z|

are continuous on

C

,

z2/|z|

is continuous on

C∗

. There-

fore,

f(z)

is continuous on

C∗

. To see that it is continuous at

0

, we just have to show

that

lim

z→0f(z) = lim

z→0

z2

|z|=f(0) = 0.

This follows from

lim

z→0

z2

|z|

=lim

z→0|z|2

|z|=lim

z→0|z|=0.

Therefore, f(z)is continuous everywhere on C.

To show that

f(z)

is nowhere analytic, it sufﬁces to show that the Cauchy-Riemann

equations fail for f(z)on C∗. This follows from

∂

∂x+i∂

∂yz2

|z|

=2z

|z|−xz2

|z|3+i−2iz

|z|−yz2

|z|3

=(4z−x−iy)z2

|z|3=3z

|z|6=0

for z6=0. Consequently, f(z)is nowhere analytic.

11. Find where

tan−1(z) = i

2Log i+z

i−z

is analytic?

Solution. The branch locus of tan−1(z)is

z:i+z

i−z=w∈(−∞,0]=z:z=iw−1

w+1,w∈(−∞,0].

For w∈(−∞,0],

w−1

w+1=1−2

w+1∈(−∞,−1]∪(1,∞)

so tan−1(z)is analytic in

C\{z: Re(z) = 0,Im(z)∈(−∞,−1]∪[1,∞)}.

36 Chapter 2. Functions

2.3.1 Harmonic functions

1.

Verify that the following functions

u

are harmonic, and in each case give a conjugate

harmonic function v(i.e. v such that u+iv is analytic).

(a) u(x,y) = 3x2y+2x2−y3−2y2,

(b) y(x,y) = ln(x2+y2).

Solution. (a) If u(x,y) = 3x2y+2x2−y3−2y2, then

ux=6xy +4x,uy=3x2−3x2−4y

uxx =6y+4, uyy =−6y−4

Thus

∆u=uxx +uyy =6y+4+ (−6y−4) = 0.

Hence, uis harmonic.

The harmonic conjugate of

u

will satisfy the Cauchy-Riemann equations and have

continuos partials of all orders. By Cauchy-Riemann equations

ux=vy,vx=−uy,

we have that vy=6xy +4x. Thus

v=Z(6xy +4x)dy =3xy2+4xy +g(x).

Thus

vx=3y2+4y+g0(x)

Since vx=−uy,

3y2+4y+g0(x) = −3x2+3y2+4y

g0(x) = −3x2

g(x) = −x3

Therefore, the harmonic conjugate is

v(x,y) = 3xy2+4xy −x3.

(b) If u(x,y) = ln(x2−y2), then

ux=2x

x2+y2,uy=2y

x2+y2

uxx =2(y2−x2)

(x2+y2)2,uyy =2(x2−y2)

(x2+y2)2

Thus

∆u=uxx +uyy =2(y2−x2)

x2+y2+2(x2−y2)

x2+y2x2+y2=0.

Hence, uis harmonic.

2.3 Analytic functions 37

Similarly to the previous part, by Cauchy-Riemann equations

ux=vy,vx=−uy,

we have that vy=2x

x2+y2. Thus

v=Z2x

x2+y2dy =2 arctan y

x+g(x)

So we have

vx=−2y

x2+y2+g0(x)

Since vx=−uy,

−2y

x2+y2+g0(x) = −2y

x2+y2

g0(x) = 0

g(x) = c c ∈R

Hence the harmonic conjugate is

v(x,y) = 2arctan y

x

Notice that uis deﬁned on C\{0}and vis not deﬁned if x=0.

3. Complex Integrals

3.1 Contour integrals

1. Evaluate the following integrals:

(a) Z2

1t2+i2dt;

(b) Zπ/4

0

e−2it dt;

(c) Z∞

0

tezt dt when Re(z)<0.

Solution.

(a)

Z2

1t2+i2dt =Z2

1

(t4+2it2+i2)dt

=t5

5+2it3

3−t

2

1

=26

5+14

3i

(b)

Zπ/4

0

e−2it dt =−e−2it

2i

π/4

0

=1+i

2i=1

2−i

2

40 Chapter 3. Complex Integrals

(c)

Z∞

0

tezt dt =1

zZ∞

0

td(ezt )

=1

ztezt

∞

0−Z∞

0

ezt dt

=1

zlim

t→∞tezt −1

zlim

t→∞ezt −1

=1

z2

where

lim

t→∞tezt =lim

t→∞ezt =0

because

lim

t→∞|tezt |=lim

t→∞tetRe(z)=lim

t→∞

t

e−xt =−lim

t→∞

1

xe−xt =0

by L’Hospital (see Problem 8 in section 2.2), and

lim

t→∞|ezt |=lim

t→∞etRe(z)=lim

t→∞

1

e−xt =0

as x=Re(z)<0.

2. Find the contour integral Rγzdz for

(a) γ

is the triangle

ABC

oriented counterclockwise, where

A=0

,

B=1+i

and

C=−2;

(b) γis the circle |z−i|=2 oriented counterclockwise.

Solution. (a)

Zγ

zdz =ZAB

zdz +ZBC

zdz +ZCA

zdz

=Z1

0

t(1+i)d(t(1+i))

+Z1

0

(1−t)(1+i)−2td((1−t)(1+i)−2t)

+Z1

0−2(1−t)d(−2(1−t))

=Z1

0

2tdt +Z1

0

((2i−4) + 10t)dt +Z1

0

4(t−1)dt

=1+ (2i−4) + 5−2=2i

(b)

Z2π

0

i+2eit d(i+2eit) = Z2π

0

2i(−i+2e−it )eitdt =8πi.

3.1 Contour integrals 41

3. Compute the following contour integral

ZL

zdz,

where

L

is the boundary of the triangle

ABC

with

A=0

,

B=1

and

C=i

, oriented

counter-clockwise.

Solution.

ZL

zdz =ZAB

zdz +ZBC

zdz +ZCA

zdz

=Z1

0

tdt +Z1

0

(1−t) + tid((1−t) + ti)

+Z1

0

(1−t)id((1−t)i)

=Z1

0

tdt + (−1+i)Z1

0

((1−t)−ti)dt −Z1

0

(1−t)dt =i

4. Evaluate the contour integral

ZC

f(z)dz

using the parametric representations for C, where

f(z) = z2−1

z

and the curve Cis

(a) the semicircle z=2eiθ(0 ≤θ≤π);

(b) the semicircle z=2eiθ(π≤θ≤2π);

(c) the circle z=2eiθ(0 ≤θ≤2π).

Solution.

(a)

ZC

f(z)dz =Zπ

0

4e2iθ−1

2eiθd(2eiθ) = (2e2iθ−i)

π

0=−πi

(b)

ZC

f(z)dz =Z2π

π

4e2iθ−1

2eiθd(2eiθ) = (2e2iθ−i)

2π

π=−πi

(c) Adding (a) and (b), we have −2πi.

42 Chapter 3. Complex Integrals

5. Redo previous Problem 4 using an antiderivative of f(z).

Solution. For (a),

ZC

f(z)dz =z2

2

−2

2− lim

z→−2

Im(z)>0

Log(z)−Log(2)!

=−(ln 2 +πi−ln2) = −πi.

For (b),

ZC

f(z)dz =z2

2

2

−2− Log(2)−lim

z→−2

Im(z)<0

Log(z)!

=−(ln 2 −(ln 2 −πi)) = −πi.

6. Let CRbe the circle |z|=R(R>1) oriented counterclockwise. Show that

ZCR

Log(z2)

z2dz

<4ππ+lnR

R

and then

lim

R→∞ZCR

Log(z2)

z2dz =0.

Proof. Using expression (1.1) in Problem 5, we have

|Log(z2)| ≤ ln |z2|+π=2 lnR+π

for |z|=R>1. Therefore,

ZCR

Log(z2)

z2dz≤2πRπ+2 ln R

R2

=4ππ/2+lnR

R<4ππ+lnR

R.

And since

lim

R→∞4ππ+lnR

R=4πlim

R→∞

1

R=0

by L’Hospital (see Problem 8 in section 2.2),

lim

R→∞ZCR

Log(z2)

z2dz =0.

3.2 Cauchy Integral Theorem and Cauchy Integral Formula 43

7. Without evaluating the integral, show that

ZC

dz

z2+z+1≤9π

16

where

C

is the arc of the circle

|z|=3

from

z=3

to

z=3i

lying in the ﬁrst quadrant.

Proof. Since

|z2+z+1| ≥ |z2|−|z|−1=|z|2−|z|−1=5

for |z|=3,

1

z2+z+1≤1

5.

Therefore,

ZC

dz

z2+z+1≤6π

41

5=3π

10 <9π

16 .

3.2 Cauchy Integral Theorem and Cauchy Integral Formula

1.

Let

C

be the boundary of the triangle with vertices at the points

0

,

3i

and

−4

oriented

counterclockwise. Compute the contour integral