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Complex Analysis: Problems with solutions

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Abstract

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).
!
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
4.3 Applications of residues 74
4.3.1 Improperintegrals .............................................. 79
Bibliography ............................................... 83
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) 2ii12i=2i
(b) (23i)(2+i) = 1+8i
Solution. We have
2ii12i=2ii+2=2i,
and
(23i)(2+i) = 4+2i+6i3i2=4+3+8i=1+8i.
2. Reduce the quantity
5i
(1i)(2i)(3i)
to a real number.
Solution. We have
5i
(1i)(2i)(3i)=5i
(1i)(55i)=i
(1i)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,z3C.
Proof. Let zk=xk+iykfor k=1,2,3. Then
(z1z2)z3= ((x1+y1i)(x2+y2i))(x3+y3i)
= ((x1x2y1y2) + i(x2y1+x1y2))(x3+y3i)
= (x1x2x3x3y1y2x2y1y3x1y2y3)
+i(x2x3y1+x1x3y2+x1x2y3y1y2y3)
and
z1(z2z3) = (x1+y1i)((x2+y2i))(x3+y3i))
= (x1+y1i)((x2x3y2y3) + i(x2y3+x3y2))
= (x1x2x3x3y1y2x2y1y3x1y2y3)
+i(x2x3y1+x1x3y2+x1x2y3y1y2y3)
Therefore,
(z1z2)z3=z1(z2z3)
5. Compute
(a) 2+i
2i;
(b) (12i)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,z2C.
Proof. Let zk=xk+ykifor k=1,2. Then
z1z2= (x1+y1i)(x2+y2i) = (x1x2y1y2) + i(x2y1+x1y2)
and hence
f(z1z2) = x1x2y1y2x2y1+x1y2
x2y1x1y2x1x2y1y2.
On the other hand,
f(z1)f(z2) = x1y1
y1x1x2y2
y2x2=x1x2y1y2x2y1+x1y2
x2y1x1y2x1x2y1y2.
Therefore, f(z1z2) = f(z1)f(z2).
7. Use binomial theorem
(a+b)n=n
0an+n
1an1b+... +n
n1abn1+n
nbn
=
n
k=0n
kankbk
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 =
02m2011 2011
2m3m(1)m
+i
02m+12011 2011
2m+13m3(1)m
=
1005
m=02011
2m(3)m+i
1005
m=02011
2m+1(3)m3.
10 Chapter 1. Complex Numbers
Similarly,
(1+3i)2011 =1
1+3i2011
= 13i
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)m3.
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,y2R. Thus
z1z2=x1x2y1y2+ (x1y2+y1x2)i
Since z1z2is real and non-zero, z16=0, z26=0, and
x1x2y1y26=0 and x1y2+y1x2=0.
Thus, since z26=0, then
z1
z2
=x1+iy1
x2iy2·x2+iy2
x2+iy2
=x1x2y1y2+ (x1y2+y1x2)i
x2
2+y2
2
=x1x2y1y2
x2
2+y2
2
.
By setting r=x1x2y1y2
x2
2+y2
2
, we have the result.
1.2 Modulus
1. Show that
|z1z2|2+|z1+z2|2=2(|z1|2+|z2|2)
for all z1,z2C.
1.2 Modulus 11
Proof. We have
|z1z2|2+|z1+z2|2
= (z1z2)(z1z2)+(z1+z2)(z1+z2)
= (z1z2)(z1z2)+(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|)20.
Solution. Note that
0(|Rez|+|Im z|)2=|Rez|22|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|22(|Re z|2+|Im z|2).
That is
(|Rez|+|Im z|)22(|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) |z1|=|z+i|;
(c) 2|z|=|z2|.
Solution. Let z=x+yi.
(a) {Im(z) = 1}={y=1}
is the horizontal line passing through the point
i
.
(b) Since
|z1|=|z+i| ⇔ |(x1) + yi|=|x+ (y+1)i|
⇔ |(x1) + yi|2=|x+ (y+1)i|2
(x1)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|=|z2| ⇔ 2|x+yi|=|(x2) + yi|
4|x+yi|2=|(x2) + yi|2
4(x2+y2) = (x2)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
R4R
R2+R+1
z4+iz
z2+z+1R4+R
(R1)2
for all zsatisfying |z|=R>1.
Proof. When |z|=R>1,
|z4+iz| ≥ |z4|−|iz|=|z|4|i||z|=R4R
and
|z2+z+1| ≤ |z2|+|z|+|1|=|z|2+|z|+1=R2+R+1
by triangle inequality. Hence
z4+iz
z2+z+1R4R
R2+R+1.
On the other hand,
|z4+iz| ≤ |z4|+|iz|=|z|4+|i||z|=R4+R
and
|z2+z+1|= z1+3i
2! z13i
2!
=
z1+3i
2
z13i
2
|z|
1+3i
2! |z|
13i
2!
= (R1)(R1) = (R1)2
1.3 Exponential and Polar Form, Complex roots 13
Therefore,
z4+iz
z2+z+1R4+R
(R1)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,yR:
(a) i
1i+1i
i;
(b) all the 3rd roots of 8i;
(c) i+1
21337
Solution. (a)
i
1i+1i
i=i2+ (1i)2
(1i)i
=12i
1i·1i
1i
=1+i2i2
2
=3i
2=3
2i
2
(b) We have that
8i=23expiπ
2
Thus the cube roots are
2exp iπ
6,2exp iπ
2and 2exp 7iπ
6.
That is 3i,2,3i
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=2i.
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) = tan1(1/2)and z=5exp(tan1(1/2)i).
3. Find all the complex roots of the equations:
(a) z6=9;
(b) z2+2z+ (1i) = 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+p44(1i)
2=1+i
=1+peπi/2=1+eπi/4e2mπi/2(m=0,1)
= 1+2
2!+2
2i, 12
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 = (z2eπi/4)(z2e3πi/4)(z2e5πi/4)(z2e7πi/4)
=(z2eπi/4)(z2e7πi/4)(z2e3πi/4)(z2e5πi/4)
= (z22Re(2eπi/4)z+4)(z22 Re(2e3πi/4)z+4)
= (z222z+4)(z2+22z+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 =22013(13i).
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)m3.
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=1zn+1
1z(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 SzS. To derive the second identity, write z=eiθin the first one.
Proof. If z6=1, then
(1z)(1+z+···+zn) = 1+z+···+zn(z+z2+···+zn+1)
=1zn+1
Thus
1+z+z2+···+zn=
1zn+1
1z,if z6=1
n+1,if z=1.
Taking z=eiθ, where 0 <θ<2π, then z6=1. Thus
1+eiθ+e2iθ+···+eniθ=1e(n+1)θ
1eiθ=1e(n+1)θ
eiθ/2eiθ/2eiθ/2
=eiθ/2(1e(n+1)θ)
2isin(θ/2)
=
ieiθ/2e(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+b22abcos 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=|z2z1|. So
a2+b2c2=|z1|2+|z2|2|z2z1|2
= (z1z1+z2z2)(z2z1)(z2z1)
= (z1z1+z2z2)(z2z1)(z2z1)
= (z1z1+z2z2)(z1z1+z2z2z1z2z2z1)
=z1z2+z2z1.
Let z1=r1eiθ1and z2=r2eiθ2. Then
z1z2+z2z1=r1eiθ1r2eiθ2+r2eiθ2r1eiθ1
= (r1eiθ1)(r2eiθ2)+(r2eiθ2)(r1eiθ1)
=r1r2ei(θ1θ2)+r1r2ei(θ2θ1)
=2r1r2cos(θ1θ2) = 2|z1||z2|cosθ=2ab cosθ.
Therefore, we have
a2+b2c2=z1z2+z2z1=2abcos θ
and hence
a2+b22abcos 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) = z3z;
(c) f(z) = 1
iz;
(d) f(z) = exp(z2).
Solution. (a)
f(z) = (x+iy)3+ (x+iy) + 1
= (x+iy)(x2y2+2ixy) + x+iy +1
=x3xy2+2ix2y+ix2yiy32xy2+x+iy +1
=x33xy2+x+1+i(3x2yy3+y).
(b)
f(z) = z3z= (x+yi)3(x+yi)
= (x3+3x2yi 3xy2y3i)(x+yi)
= (x33xy2x) + i(3x2yy3y),
20 Chapter 2. Functions
(c)
f(z) = 1
iz=1
x+ (1y)i
=x(1y)i
x2+ (1y)2
=x
x2+ (1y)2i1y
x2+ (1y)2
(d)
f(z) = exp(z2) = exp((x+yi)2)
=exp((x2y2) + 2xyi)
=ex2y2(cos(2xy) + isin(2xy))
=ex2y2cos(2xy)iex2y2sin(2xy)
2.
Suppose that
f(z) = x2y22y+i(2x2xy)
, where
z=x+iy
. Use the expressions
x=z+z
2and y=zz
2i
to write f(z)in terms of zand simplify the result.
Solution. We have
f(z) = x2y22y+i(2x2xy)
=x2y2+i2xi2xy 2y
= (xiy)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,..., anR. 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
T1(D) = {zC: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
3x22x+3y2<1
x1
32
+y2<4
9
z1
3
<2
3
So
T1(D) = z:
z1
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+y21<0}
and
f(x,y) = (x+3)2+
y21
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+33| ≤ |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}={y1}{y≤ −1}.
22 Chapter 2. Functions
Since
f(x,y) = y
is continuous on
R2
, both
{y1}
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 nZand
lim
n|zn|=lim
npn2+4=,
the set is unbounded.
The set is not connected. Otherwise, let
p=2i
and
q=2i
. There is a polygonal
path
p0p1p1p2... pn1pn
with p0=p,pn=qand pk∈ {|Im(z)| ≥ 1}for all 0 kn.
Let
0mn
be the largest integer such that
pm{y1}
. 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+33| ≤ |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=33i/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 zC, 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(eiz1eiz1)
eiz1+eiz1+i(eiz2eiz2)
eiz2+eiz2
=i(eiz1eiz1)(eiz2+eiz2)+(eiz2eiz2)(eiz1+eiz1)
(eiz1+eiz1)(eiz2+eiz2)
=2iei(z1+z2)ei(z1+z2)
(eiz1+eiz1)(eiz2+eiz2)
and
1(tanz1)(tan z2) = 1i(eiz1eiz1)
eiz1+eiz1i(eiz2eiz2)
eiz2+eiz2
=(eiz1+eiz1)(eiz2+eiz2)+(eiz1eiz1)(eiz2eiz2)
(eiz1+eiz1)(eiz2+eiz2)
=2ei(z1+z2)+ei(z1+z2)
(eiz1+eiz1)(eiz2+eiz2),
we have
tanz1+tan z2
1(tanz1)(tan z2)=iei(z1+z2)ei(z1+z2)
ei(z1+z2)+ei(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,z2Rwith z1,z2,z1+z26=nπ+π/2.
Fixing z1R, we let f(z) = F(z1,z). Then f(z)is analytic in its domain
C\({nπ+π/2}{nπ+π/2z1}).
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
z1R
and
z2C
in its
domain.
Fixing z2C, we let g(z) = F(z,z2). Then g(z)is analytic in its domain
C\({nπ+π/2}{nπ+π/2z2}).
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
z1C
and
z2C
in
its domain.
9. Find all the complex roots of the equation cosz=3.
Solution.
Since
cosz= (eiz +eiz )/2
, it comes down to solve the equation
eiz +
eiz =6, i.e.,
w+w1=6w26w+1=0
if we let
w=eiz
. The roots of
w26w+1=0
are
w=3±22
. Therefore, the
solutions for cosz=3 are
iz =log(3±22)z=i(ln(3±22) + 2nπi) = 2nπiln(3±22)
for nintegers.
10. Calculate sinπ
4+i.
Solution.
sinπ
4+i=1
2i(ei(π/4+i)ei(π/4+i))
=1
2i(e1eπi/4eeπi/4)
=1
2ie1(cos π
4+isin π
4)e(cos π
4isin π
4)
=2
4e+1
e+2
4e1
ei
2.1 Basic notions 25
11. Compute cosπ
3+i.
Solution.
cosπ
3+i=1
2(ei(π/3+i)+ei(π/3+i))
=1
2(e1eπi/3+eeπi/3)
=1
2e1(cos π
3+isin π
3) + e(cos π
3isin π
3)
=1
4e+1
e3i
4e1
e
12. Find iiand its principal value.
Solution. We have
ii=eilogi=ei(2nπi+πi/2)=e2nππ/2
for nintegers and its principal value given by
ii=eiLogi=ei(πi/2)=eπ/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
2i
2
(b) Show that
f(z1)f(z2) = λf(z1z2)
for all z1,z26=0, where λ=1,1+3i
2or 13i
2.
26 Chapter 2. Functions
Proof. Since
f(z1)f(z2)
f(z1z2)=exp(1
3Logz1+1
3Logz21
3Log(z1z2))
=exp(1
3(Logz1+Log z2Log(z1z2)))
=exp(i
3(Argz1+Arg z2Arg(z1z2))) = exp(2nπi
3)
for some integer n,λ=exp(2nπi/3). Therefore,
λ=
1 if n=3k
1+3i
2if n=3k+1
13i
2if n=3k+2
where kZ.
14. Let f(z)be the principal branch of zi.
(a) Find f(i).
Solution.
f(i) = ii=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 e2π.
Proof. Since
f(z1)f(z2)
f(z1z2)=exp(iLogz1iLog z2+iLog(z1z2))
=exp(i(Logz1+Log z2Log(z1z2)))
=exp(i(iArgz1+iArgz2iArg(z1z2)))
=exp(Argz1+Argz2Arg(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 z2Arg(z1z2)<3π
and hence 3<2n<3. So n=1,0 or 1 and λ=e2π,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
(z1)2.
(c) lim
z0
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)(z2iz +i2)
(z+i)(zi)
=lim
z→−i
i(z2iz +i2)
zi
=ilimz→−i(z2iz +i2)
limz→−i(zi)
=ilimz→−iz2ilimz→−iz+limz→−ii2
limz→−izlimz→−ii
=i((i)2i(i) + i2)
ii=3
2
(b)
lim
z
4+z2
(z1)2=lim
z0
4+z2
(z11)2
=lim
z0
4z2+1
(1z)2=limz0(4z2+1)
limz0(1z)2
=4(limz0z)2+limz01
(limz01limz0z)2=1
(c) Since
lim
Re(z)=0,z0
Im(z)
z=lim
y0
y
yi =i
and
lim
Im(z)=0,z0
Im(z)
z=lim
x0
0
x=0
the limit does not exist.
28 Chapter 2. Functions
2. Show the following limits:
(a) lim
z
4z5
z542z=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
z542z=lim
z0
41
z5
1
z5421
z=lim
z0
4
142z4=4.
(b) Notice that
lim
z
z4
z2+42zlim
z01/z4
1/z2+42/z1
lim
z01
z2+42z31
lim
z0z2+42z3=0.
Therefore, lim
z
z4
z2+42z=.
(c)
lim
z
(az +b)3
(cz +d)3=lim
z0
(a/z+b)3
(c/z+d)3=lim
z0
(a+bz)3
(c+dz)3=a3
c3.
3. Show that lim
z0z/zdoes not exist.
Hint:
Consider what happens to the function at points of the form
x+0i
for
x0
,
x6=0, and then at points of the form 0 +yi for y0,y6=0.
Proof. For z=x+0i,x6=0,
z
z=x
x=11 as x0.
On the other hand, for z=0+yi,y6=0,
z
z=yi
yi =1→ −1 as y0.
However,
lim
z0z/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
z0|f(z)|=lim
z0
z3
z2
=lim
z0|z|=0
and hence
limz0f(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
z0
f(z)f(0)
z0=lim
z0
z3
z3.
Let z=x+yi. If y=0 and x0, then
lim
z=x0
z3
z3=lim
x0
x3
x3=1.
On the other hand, if x=0 and y0, then
lim
z=yi0
z3
z3=lim
x0
(yi)3
(yi)3=1.
So the limit limz0z3/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 zC.
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
z22z3
z3+i3iz2
z22iz3
z3
=6z2
z26=0
for z6=0.
7. Find f0(z)when
(a) f(z) = z24z+2;
(b) f(z) = (1z2)4;
(c) f(z) = z+1
2z+1(z6=1
2);
(d) f(z) = e1/z(z6=0).
Answer. (a) 2z4; (b) 8(1z2)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
|zz0|<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
zz0
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
zz0
f(z)f(z0)
zz0
=f0(z0)
and
lim
zz0
g(z)g(z0)
zz0
=g0(z0).
And since g0(z0)6=0,
lim
zz0
f(z)f(z0)
g(z)g(z0)=limzz0(f(z)f(z0))/(zz0)
limzz0(g(z)g(z0))/(zz0)=f0(z0)
g0(z0).
Finally, since f(z0) = g(z0) = 0,
lim
zz0
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))n1
xf(z) + i(f(z))n1
yf(z)
= ( f(z))n1
x+i
yf(z) = 0
at z0.
2.3 Analytic functions
1. Explain why the function f(z) = 2z23zez+ezis entire.
Proof.
Since every polynomial is entire,
2z23
is entire; since both
z
and
ez
are
entire, their product
zez
is entire; since
ez
and
z
are entire, their composition
ez
is entire. Finally, f(z)is the sum of 2z33, zezand ezand hence entire.
2.
Let
f(z)
be an analytic function on a connected open set
D
. If there are two constants
c1
and
c2C
, not all zero, such that
c1f(z) + c2f(z) = 0
for all
zD
, 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 zD.
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(uvi) = 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
|zz0|<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)| ≤ ex33xy2+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((x33xy2)+(3x2yy3+1)i))
+exp(Re(2xy (x2y2)i))
=ex33xy2+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(xv(x,y)).
Since
v(x,y)x
,
xv(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
w0C
, the quadratic equation
y22w0y+1=0
has a complex
root
y0
. We cannot have
y0=0
since
022w0·0+16=0
. Therefore,
y06=0
and
there is z0Csuch that ez0=y0. Then
cosh(z0) = ez0+ez0
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) = (x2y2)2xyi
Solution. Since
x+i
yf= (2x2yi) + i(2y2xi) = 4x4yi 6=0,
the Cauchy-Riemann equations fail for f(z)and hence f(z)is not entire.
34 Chapter 2. Functions
(d) f(z) = (x2y2) + 2xyi
Solution. Since
x+i
yf= (2x2yi) + 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+11
(R1)2
for all zlying on CR.
Proof. For zCR,|z|=Rand Im(z)0. Let z=x+yi. Since y=Im(z)0,
|eiz|=|ei(x+yi)|=|ey+xi|=ey1
for zCR. And since
|z2+z+1|= z1+3i
2! z13i
2!
=
z1+3i
2
z13i
2
|z|
1+3i
2! |z|
13i
2!
= (R1)(R1) = (R1)2,
we obtain
eiz
z2+z+11
(R1)2
for zCR.
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
z0f(z) = lim
z0
z2
|z|=f(0) = 0.
This follows from
lim
z0
z2
|z|
=lim
z0|z|2
|z|=lim
z0|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+i2iz
|z|yz2
|z|3
=(4zxiy)z2
|z|3=3z
|z|6=0
for z6=0. Consequently, f(z)is nowhere analytic.
11. Find where
tan1(z) = i
2Log i+z
iz
is analytic?
Solution. The branch locus of tan1(z)is
z:i+z
iz=w(,0]=z:z=iw1
w+1,w(,0].
For w(,0],
w1
w+1=12
w+1(,1](1,)
so tan1(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+2x2y32y2,
(b) y(x,y) = ln(x2+y2).
Solution. (a) If u(x,y) = 3x2y+2x2y32y2, then
ux=6xy +4x,uy=3x23x24y
uxx =6y+4, uyy =6y4
Thus
u=uxx +uyy =6y+4+ (6y4) = 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(x2y2), then
ux=2x
x2+y2,uy=2y
x2+y2
uxx =2(y2x2)
(x2+y2)2,uyy =2(x2y2)
(x2+y2)2
Thus
u=uxx +uyy =2(y2x2)
x2+y2+2(x2y2)
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
e2it dt;
(c) Z
0
tezt dt when Re(z)<0.
Solution.
(a)
Z2
1t2+i2dt =Z2
1
(t4+2it2+i2)dt
=t5
5+2it3
3t
2
1
=26
5+14
3i
(b)
Zπ/4
0
e2it dt =e2it
2i
π/4
0
=1+i
2i=1
2i
2
40 Chapter 3. Complex Integrals
(c)
Z
0
tezt dt =1
zZ
0
td(ezt )
=1
ztezt
0Z
0
ezt dt
=1
zlim
ttezt 1
zlim
tezt 1
=1
z2
where
lim
ttezt =lim
tezt =0
because
lim
t|tezt |=lim
ttetRe(z)=lim
t
t
ext =lim
t
1
xext =0
by L’Hospital (see Problem 8 in section 2.2), and
lim
t|ezt |=lim
tetRe(z)=lim
t
1
ext =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 |zi|=2 oriented counterclockwise.
Solution. (a)
Zγ
zdz =ZAB
zdz +ZBC
zdz +ZCA
zdz
=Z1
0
t(1+i)d(t(1+i))
+Z1
0
(1t)(1+i)2td((1t)(1+i)2t)
+Z1
02(1t)d(2(1t))
=Z1
0
2tdt +Z1
0
((2i4) + 10t)dt +Z1
0
4(t1)dt
=1+ (2i4) + 52=2i
(b)
Z2π
0
i+2eit d(i+2eit) = Z2π
0
2i(i+2eit )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
(1t) + tid((1t) + ti)
+Z1
0
(1t)id((1t)i)
=Z1
0
tdt + (1+i)Z1
0
((1t)ti)dt Z1
0
(1t)dt =i
4. Evaluate the contour integral
ZC
f(z)dz
using the parametric representations for C, where
f(z) = z21
z
and the curve Cis
(a) the semicircle z=2eiθ(0 θπ);
(b) the semicircle z=2eiθ(πθ2π);
(c) the circle z=2eiθ(0 θ2π).
Solution.
(a)
ZC
f(z)dz =Zπ
0
4e2iθ1
2eiθd(2eiθ) = (2e2iθi)
π
0=πi
(b)
ZC
f(z)dz =Z2π
π
4e2iθ1
2eiθd(2eiθ) = (2e2iθi)
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 +πiln2) = 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
RZCR
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)
z2dz2πRπ+2 ln R
R2
=4ππ/2+lnR
R<4ππ+lnR
R.
And since
lim
R4ππ+lnR
R=4πlim
R
1
R=0
by L’Hospital (see Problem 8 in section 2.2),
lim
RZCR
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+19π
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+11
5.
Therefore,
ZC
dz
z2+z+16π
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