Content uploaded by Juan Carlos Ponce Campuzano
Author content
All content in this area was uploaded by Juan Carlos Ponce Campuzano on Jun 01, 2019
Content may be subject to copyright.
!
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