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Complex Analysis
Problems with solutions
Juan Carlos Ponce Campuzano

Copyright c 2016 Juan Carlos Ponce Campuzano
P UBLISHED BY J UAN C ARLOS P ONCE C AMPUZANO
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike
4.0 International License.
License at: />First e-book version, August 2015


Contents

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

1

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

1.1

Basic algebraic and geometric properties

1.2

Modulus

13

1.3

Exponential and Polar Form, Complex roots


16

2

Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.1

Basic notions

21

2.2

Limits, Continuity and Differentiation

35

2.3

Analytic functions

39

2.3.1

Harmonic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3


Complex Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

3.1

Contour integrals

49

3.2

Cauchy Integral Theorem and Cauchy Integral Formula

55

3.3

Improper integrals

71

4

Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

4.1

Taylor and Laurent series

75


4.2

Classification of singularities

85

7


4.3

Applications of residues

91

4.3.1

Improper integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100


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:
2016



1. Complex Numbers

1.1

Basic algebraic and geometric properties
1. Verify that
√
√
2 − i − i 1 − 2i = −2i
(a)
(b) (2 − 3i) (−2 + i) = −1 + 8i
Solution. We have
√
√
√
√
2 − i − i 1 − 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
5i
i
i
1
=
=
=
=
2
(1 − i)(2 − i)(3 − i) (1 − i)(5 − 5i) (1 − i)
−2i 2


Chapter 1. Complex Numbers

8
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

(z1 z2 )z3 = z1 (z2 z3 )
for all z1 , z2 , z3 ∈ C.
Proof. Let zk = xk + iyk for k = 1, 2, 3. Then
(z1 z2 )z3 = ((x1 + y1 i)(x2 + y2 i))(x3 + y3 i)
= ((x1 x2 − y1 y2 ) + i(x2 y1 + x1 y2 ))(x3 + y3 i)
= (x1 x2 x3 − x3 y1 y2 − x2 y1 y3 − x1 y2 y3 )
+ i(x2 x3 y1 + x1 x3 y2 + x1 x2 y3 − y1 y2 y3 )
and
z1 (z2 z3 ) = (x1 + y1 i)((x2 + y2 i))(x3 + y3 i))
= (x1 + y1 i)((x2 x3 − y2 y3 ) + i(x2 y3 + x3 y2 ))
= (x1 x2 x3 − x3 y1 y2 − x2 y1 y3 − x1 y2 y3 )
+ i(x2 x3 y1 + x1 x3 y2 + x1 x2 y3 − y1 y2 y3 )
Therefore,
(z1 z2 )z3 = z1 (z2 z3 )

5. Compute
2+i
;
(a)
2−i
(b) (1 − 2i)4 .
Answer: (a) (3 + 4i)/5, (b) −7 + 24i.


1.1 Basic algebraic and geometric properties
6. Let f be the map sending each complex number
z = x + yi →
−

x y

−y x

Show that f (z1 z2 ) = f (z1 ) f (z2 ) for all z1 , z2 ∈ C.
Proof. Let zk = xk + yk i for k = 1, 2. Then
z1 z2 = (x1 + y1 i)(x2 + y2 i) = (x1 x2 − y1 y2 ) + i(x2 y1 + x1 y2 )
and hence
f (z1 z2 ) =

x1 x2 − y1 y2 x2 y1 + x1 y2
.
−x2 y1 − x1 y2 x1 x2 − y1 y2

On the other hand,
x2 y2
x1 x2 − y1 y2 x2 y1 + x1 y2
=
.
−y2 x2
−x2 y1 − x1 y2 x1 x2 − y1 y2

x1 y1
−y1 x1

f (z1 ) f (z2 ) =

Therefore, f (z1 z2 ) = f (z1 ) f (z2 ).
7. Use binomial theorem
n n
n
n n−1

n n
b
abn−1 +
a b + ... +
a +
n
n−1
1
0
n
n n−k k
=∑
a b
k=0 k

(a + b)n =

to expand √
(a) (1 + √3i)2011 ;
(b) (1 + 3i)−2011 .
Solution. By binomial theorem,
2011
√
2011 √ k 2011 2011 k/2 k
(1 + 3i)2011 = ∑
( 3i) = ∑
3 i.
k
k
k=0

k=0

Since ik = (−1)m for k = 2m even and ik = (−1)m i for k = 2m + 1 odd,
√
(1 + 3i)2011 =

∑

0≤2m≤2011

+i

∑

2011 m
3 (−1)m
2m

0≤2m+1≤2011
1005

=

∑

m=0

√
2011
3m 3(−1)m

2m + 1

1005
√
2011
2011
(−3)m + i ∑
(−3)m 3.
2m
m=0 2m + 1

9


Chapter 1. Complex Numbers

10
Similarly,
√
(1 + 3i)−2011 =
=
=

1
√
1 + 3i
2011

1
42011

1

∑
k=0
1005

∑

42011 m=0
−

i

1005

∑

42011 m=0

2011

=

√
1 − 3i
4

2011

√

2011
(− 3i)k
k
2011
(−3)m
2m
√
2011
(−3)m 3.
2m + 1

8. Graph the following regions in the complex plane:
(a) {z : Re z ≥ 2 Im z};
(b) {z : π/2 < Arg z ≤ 3π/4};
(c) {z : |z − 4i + 2| > 2}.
Solution. (a)

Figure 1.1:


1.1 Basic algebraic and geometric properties
(b)

Figure 1.2:
(c)

Figure 1.3:

11



Chapter 1. Complex Numbers

12

9. Find all complex solutions of the following equations:
(a) z = z;
(b) z + z = 0;
9
(c) z = .
z
Solution. (a) Let z = z + iy. Thus
z
x + iy
x − iy
−iy
y

=
=
=
=
=

z
x + iy
x + iy
iy
0


Hence, z = z if and only if Im z = 0.
(b) Let z = z + iy. Thus
z+z
x + iy + z + iy
x − iy + x + iy
2x
x

=
=
=
=
=

0
0
0
0
0

Hence, z + z if and only if Re z = 0.
(c) In this part we have
9
z=
⇐⇒ zz = 9 ⇐⇒
z
9
Hence, z = if and only if |z| = 3.
z


|z|2 = 9

⇐⇒

|z| = 3.

10. Suppose that z1 and z2 are complex numbers, with z1 z2 real and non-zero. Show that
there exists a real number r such that z1 = rz2 .
Proof. Let z1 = x1 + iy1 and z2 = x2 + iy2 with x1 , x2 , y1 , y2 ∈ R. Thus
z1 z2 = x1 x2 − y1 y2 + (x1 y2 + y1 x2 )i
Since z1 z2 is real and non-zero, z1 = 0, z2 = 0, and
x1 x2 − y1 y2 = 0

and x1 y2 + y1 x2 = 0.

Thus, since z2 = 0, then
x1 + iy1 x2 + iy2
z1
=
·
z2
x2 − iy2 x2 + iy2
x1 x2 − y1 y2 + (x1 y2 + y1 x2 )i
=
x22 + y22
x1 x2 − y1 y2
=
.
x22 + y22
x1 x2 − y1 y2

By setting r =
, we have the result.
x22 + y22


1.2 Modulus

13

√
√
√
2 is defined by Q
2 = p + q 2 : p, q ∈ Q .
√
2 is a field.
(a) Show that Q
√
√
(b) Is 3 ∈ Q
2 ?

11. The set Q adjoin

√
√
√
Proof. (a) Let p + q 2, r + s 2 ∈ Q
2 . Since Q ⊂ R and R is a field, we have
the following:

Closure under (+):
√
√
√
√
p + q 2 + r + s 2 = (p + r) + (q + s) 2 ∈ Q
2
Closure under (·):
√
√
p+q 2 · r+s 2
(b) Suppose that

√
√
= (pr + 2sq) + (rq + ps) 2 ∈ Q
2

√
√
√
3 = a+b 2 ∈ Q
2 . Note that b = 0. Thus we have

√
√
3−b 2 = a
√ √
3 − 2 2 3b + 2b2 = a2
√

2 6b = 3 − a2 .
Since b = 0,

That is,

1.2

√
3 − a2
6=
.
2b

√
√
√
6 ∈ Q, which is a contradiction. Therefore, 3 ∈
/Q
2

Modulus
1. Show that
|z1 − z2 |2 + |z1 + z2 |2 = 2(|z1 |2 + |z2 |2 )
for all z1 , z2 ∈ C.
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 )
= ((z1 z1 + z2 z2 ) − (z1 z2 + z2 z1 )) + ((z1 z1 + z2 z2 ) + (z1 z2 + z2 z1 ))
= 2(z1 z1 + z2 z2 ) = 2(|z1 |2 + |z2 |2 ).



Chapter 1. Complex Numbers

14

√
2. Verify that 2|z| ≥ | Re z| + | Im z|.
Hint: Reduce this inequality to (|x| − |y|)2 ≥ 0.
Solution. Note that
0 ≤ (| Re z| + | Im z|)2 = | Re z|2 − 2| Re z|| Im z| + | Im z|2 .
Thus
2| Re z|| Im z| ≤ | Re z|2 + | Im z|2 ,
and then
| Re z|2 + 2| Re z|| Im z| + | Im z|2 ≤ 2(| Re z|2 + | Im z|2 ).
That is
(| Re z| + | Im z|)2 ≤ 2(| Re z|2 + | Im z|2 ) = 2|z|2 ,
and therefore,
| Re z| + | 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.
(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
2 2
16
⇔ x+
+ y2 =
3
9
2
4
⇔ z+ =
3
3
the curve is the circle with centre at −2/3 and radius 4/3.


1.2 Modulus

15

4. Show that
z4 + iz

R4 − R
R4 + R
≤
≤
R2 + R + 1
z2 + z + 1
(R − 1)2
for all z satisfying |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
R4 − R
z4 + iz
≥
.
z2 + z + 1
R2 + R + 1
On the other hand,
|z4 + iz| ≤ |z4 | + |iz| = |z|4 + |i||z| = R4 + R
and
√
√
−1 + 3i
−1 − 3i
|z + z + 1| = z −
z−
2
2

√
√
−1 + 3i
−1 − 3i
= z−
z−
2
2
√
√
−1 + 3i
−1 − 3i
≥ |z| −
|z| −
2
2
2

= (R − 1)(R − 1) = (R − 1)2
Therefore,
z4 + iz
R4 + R
≤
.
z2 + z + 1
(R − 1)2

5. Show that
| Log(z)| ≤ |ln |z|| + π
for all z = 0.

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

(1.1)


Chapter 1. Complex Numbers

16

1.3

Exponential and Polar Form, Complex roots
1. Express the following in the form x + iy, with x, y ∈ R:
i
1−i
(a)
+
;
1−i
i
(b) all the 3rd roots of −8i;
i + 1 1337
√
(c)
2
Solution. (a)
1−i
i2 + (1 − i)2
i

+
=
1−i
i
(1 − i)i
−1 − 2i 1 − i
=
·
1−i 1−i
−1 + i − 2i − 2
=
2
3 i
−3 − i
=− −
=
2
2 2
(b) We have that
−iπ
2

−8i = 23 exp
Thus the cube roots are
2 exp

−iπ
6

That is


,

2 exp

√
3 − i,

iπ
2

and

2,

√
− 3−i

(c)
i+1
√
2

1337

iπ 1337
4
1337πi
= exp
4

=

exp

π
= exp 167 · 2πi + i
4
π
1+i
= exp i = √ .
4
2

2. Find the principal argument and exponential form of
i
(a) z =
;
1+i
√
(b) z = 3 + i;
(c) z = 2 − i.

2 exp

7iπ
6

.



1.3 Exponential and Polar Form, Complex roots

17

Answer:
√
(a) Arg(z) = π/4 and z = ( 2/2) exp(πi/4).
(b) Arg(z) = π/6 and z = 2 exp(πi/6).
√
(c) Arg(z) = − tan−1 (1/2) and z = 5 exp(− 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
√
√
√
6
3
z = 6 −9 = 9eπi = 3eπi/6 e2mπi/6 (m = 0, 1, 2, 3, 4, 5)
√
√
√
√
3
3
3
3

√
35/6
3 √
35/6
3
35/6
3
35/6
3
3
3
=
+
i, 3i, −
+
i, −
−
i, − 3i,
−
i.
2
2
2
2
2
2
2
2

(b) The roots are

√
4 − 4(1 − i)
= −1 + i
2
πi/2
= −1 + e
= −1 + eπi/4 e2mπi/2 (m = 0, 1)
√
√
√
√
2
2
2
2
= −1 +
i, −1 −
i.
+
−
2
2
2
2

z=

−2 +

4. Find the four roots of the polynomial z4 + 16 and use these to factor z4 + 16 into two

quadratic polynomials with real coefficients.
Solution. The four roots of z4 + 16 are given by
√
√
√
4
4
4
−16 = 16eπi = 16eπi/4 e2mπi/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/4 and 2e3πi/4 = 2e5πi/4 .
This gives the way to factor z4 + 16 into two quadratic polynomials of real coefficients:
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 − 2 Re(2eπi/4 )z + 4)(z2 − 2 Re(2e3πi/4 )z + 4)
√
√
= (z2 − 2 2z + 4)(z2 + 2 2z + 4)


Chapter 1. Complex Numbers

18
5. Do the following:
(a) Use exponential
form to compute

√ 2011
;
i. (1 + √3i)
ii. (1 + 3i)−2011 .
(b) Prove that
1005

∑

m=0

2011
(−3)m = 22010
2m

and
1005

∑

m=0

2011
(−3)m = 22010 .
2m + 1

Solution. Since
√
1 + 3i = 2


√
1
3
+
i
2
2

= 2 exp

πi
,
3

we have
√
2011πi
(1 + 3i)2011 = 22011 exp
3
= 22011 exp 670πi +

= 22011 exp
πi
3

πi
=2
exp
= 22011
3

√
= 22010 (1 + 3i).
2011

2011πi
3

√
1
3
+
i
2
2

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=0


1005
√
2011
2011
(−3)m + i ∑
(−3)m 3.
2m
m=0 2m + 1

It follows that
1005

∑

m=0

1005
2011
2011
m
(−3) = ∑
(−3)m = 22010 .
2m
m=0 2m + 1


1.3 Exponential and Polar Form, Complex roots

19


6. Establish the identity
1 − zn+1
(z = 1)
1−z
and then use it to derive Lagrange’s trigonometric identity:
1 + z + z2 + · · · + zn =

(2n+1)θ

1 sin 2
1 + cos θ + cos 2θ · · · + cos nθ = +
2
2 sin θ2

(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 = eiθ in the first one.
Proof. If z = 1, then
(1 − z)(1 + z + · · · + zn ) = 1 + z + · · · + zn − (z + z2 + · · · + zn+1 )
= 1 − zn+1
Thus


 1 − zn+1
, if z = 1
1 + z + z2 + · · · + zn =
1
−

z
 n + 1,
if z = 1.

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

2iθ

1+e +e

niθ

+···+e

1 − e(n+1)θ
1 − e(n+1)θ
=
=
1 − eiθ
−eiθ /2 eiθ /2 − e−iθ /2
−e−iθ /2 (1 − e(n+1)θ )
=
2i sin(θ /2)
1

i e−iθ /2 − e(n+ 2 )iθ
=

2 sin(θ /2)

cos(θ /2) − cos[(n + 12 )θ ]
1 sin[(n + 12 )θ ]
=
+
+i
2
2 sin(θ /2)
2 sin(θ /2)
Equating real and imaginary parts, we obtain
1 sin[(n + 12 )θ ]
1 + cos θ + cos 2θ · · · + cos nθ = +
2
2 sin(θ /2)
and

cos(θ /2) − cos[(n + 12 )θ ]
sin θ + sin 2θ · · · + sin nθ =
.
2 sin(θ /2)

7. Use complex numbers to prove the Law of Cosine: Let ∆ABC be a triangle with
|BC| = a, |CA| = b, |AB| = c and ∠BCA = θ . Then
a2 + b2 − 2ab cos θ = c2 .
Hint: Place C at the origin, B at z1 and A at z2 . Prove that
z1 z2 + z2 z1 = 2|z1 z2 | cos θ .


Chapter 1. Complex Numbers

20


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
= (z1 z1 + z2 z2 ) − (z2 − z1 )(z2 − z1 )
= (z1 z1 + z2 z2 ) − (z2 − z1 )(z2 − z1 )
= (z1 z1 + z2 z2 ) − (z1 z1 + z2 z2 − z1 z2 − z2 z1 )
= z1 z2 + z2 z1 .
Let z1 = r1 eiθ1 and z2 = r2 eiθ2 . Then
z1 z2 + z2 z1 = r1 eiθ1 r2 eiθ2 + r2 eiθ2 r1 eiθ1
= (r1 eiθ1 )(r2 e−iθ2 ) + (r2 eiθ2 )(r1 e−iθ1 )
= r1 r2 ei(θ1 −θ2 ) + r1 r2 ei(θ2 −θ1 )
= 2r1 r2 cos(θ1 − θ2 ) = 2|z1 ||z2 | cos θ = 2ab cos θ .
Therefore, we have
a2 + b2 − c2 = z1 z2 + z2 z1 = 2ab cos θ
and hence
a2 + b2 − 2ab cos 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 Cartesian 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;
1
(c) f (z) =
;

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 + 2ix2 y + ix2 y − iy3 − 2xy2 + x + iy + 1
x3 − 3xy2 + x + 1 + i(3x2 y − y3 + y).

(b)
f (z) = z3 − z = (x + yi)3 − (x + yi)
= (x3 + 3x2 yi − 3xy2 − y3 i) − (x + yi)
= (x3 − 3xy2 − x) + i(3x2 y − y3 − y),


Chapter 2. Functions

22
(c)
1
1
=
i − z −x + (1 − y)i
−x − (1 − y)i
= 2

x + (1 − y)2
x
1−y
=− 2
−i 2
2
x + (1 − y)
x + (1 − y)2

f (z) =

(d)
f (z) = exp(z2 ) = exp((x + yi)2 )
= exp((x2 − y2 ) + 2xyi)
= ex

2 −y2

x2 −y2

=e

(cos(2xy) + i sin(2xy))
cos(2xy) − iex

2 −y2

sin(2xy)

2. Suppose that f (z) = x2 − y2 − 2y + i(2x − 2xy), where z = x + iy. Use the expressions

z+z
z−z
and y =
2
2i
to write f (z) in terms of z and simplify the result.
x=

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 + a1 z + ... + an zn for a0 , a1 , ..., an ∈ R. Then
p(z) = a0 + a1 z + · · · + an zn
= a0 + a1 z + · · · + an zn
= a0 + (a1 )z + · · · + (an )zn
= a0 + a1 z + · · · + an zn = 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 z0 is.


2.1 Basic notions

23

4. Let

z
.
z+1
Find the inverse image of the disk |z| < 1/2 under T and sketch it.
T (z) =

Solution. Let D = {|z| < 1/2}. The inverse image of D under T is
1
T −1 (D) = {z ∈ C : T (z) ∈ D} = {|T (z)| < }
2
z
1
= z:
<
= {2|z| < |z + 1|}.
z+1
2
Let z = x + yi. Then
2|z| < |z + 1| ⇔ 4(x2 + y2 ) < (x + 1)2 + y2

⇔ 3x2 − 2x + 3y2 < 1
1 2
4
+ y2 <
3
9
1
2
⇔ z− <
3
3

⇔ x−

So
T −1 (D) = z : z −

2
1
<
3
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 C are 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}.


Chapter 2. Functions

24

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 C are 0/ and C.
Since zn = n + 2i ∈ {| Im(z)| ≥ 1} for all n ∈ Z and
lim |zn | = lim

n→∞

n→∞

n2 + 4 = ∞,

the set is unbounded.
The set is not connected. Otherwise, let p = 2i and q = −2i. There is a polygonal
path
p0 p1 ∪ p1 p2 ∪ ... ∪ pn−1 pn

with p0 = p, pn = q and 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 ∈
pm pm+1 such that Im(p) = 0. This is a contradiction since pm pm+1 ⊂ {| Im(z)| ≥ 1}
but p ∈ {| Im(z)| ≥ 1}. Therefore the set is not connected.
Solution. (c) Since −2 ∈ {1 ≤ |z + 3| < 2} and {|z + 2| < r} ⊂ {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} ⊂ {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
| sin z|2 = (sin x)2 + (sinh y)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) − i cos(x) sinh(y)|2
= sin2 x cosh2 y + cos2 x sinh2 y
= sin2 x(1 + sinh2 y) + cos2 x sinh2 y
= sin2 x + (cos2 x + sin2 x) sinh2 y = (sin x)2 + (sinh y)2 .