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Trang 3 PrefaceThis text contains the solutions to all of the practice problems in the 10th chapter of thelecture notes An Introduction to Complex Analysis [1].. It is a translation of t

Complex analysis

Exercises with solutions

Ji°í Bouchala (and Ond°ej Bouchala)

The author of the painting Imaginární dºungle on the cover page is Ji°í Bouchala(and it is owned by Ond°ej Bouchala).

This text contains the solutions to all of the practice problems in the 10th chapter of thelecture notes An Introduction to Complex Analysis [1] It is a translation of the Czechtext [3]

The typesetting and all of the pictures are the work of my son Ond°ej He also helped

to improve the text in several places with his comments

It is not possible that we caught all of the mistakes during the proof-reading We aregrateful for your leniency and for letting us know about any and all remarks.1

We enjoyed working on this text We wish the same to the reader

In Orlová, 2022

Ji°í Bouchala(and Ond°ej Bouchala)

1 Please send all of the remarks (notes, recommendations, threats and gifts) to my e-mail address jiri.bouchala@vsb.cz.

2 Solution:

3−2i;f) z = 3−i

2+i.Solution:

a)

√3i

)︃

b) z = i = cosπ

2 + i sinπ2.c) z = − 8 = 8(cos π + i sin π)

−√3i

−1α

φ

sin α =

√3

)︃

+ i sin

(︃

−2π3

)︃)︃

φ

|z| = 113

√

16 + 49 =

√65

13 , tan φ =

7 13 4 13

Exercise 3.

Prove the de Moivre's theorem

(∀n ∈ N) (∀φ ∈ R) : (︁ cos φ + i sin φ)︁n = cos(nφ) + i sin(nφ)using mathematical induction

Solution:

1) We start by checking that the formula holds for n = 1:

(cos φ + i sin φ)1 = cos (1 · φ) + i sin (1 · φ)

2) Now we prove the implication (cos φ + i sin φ)n = cos(nφ) + i sin(nφ)⇒?

?

⇒ (cos φ + i sin φ)n+1 = cos((n + 1)φ) + i sin((n + 1)φ):

(cos φ + i sin φ)n+1 i.p.= (cos (nφ) + i sin (nφ)) (cos φ + i sin φ) =

= (cos(nφ) cos φ − sin(nφ) sin φ) + i (sin(nφ) cos φ + cos(nφ) sin φ) ,

and now it suces to apply the known trigonometric identities:

cos(nφ) cos φ − sin(nφ) sin φ = cos(nφ + φ) = cos((n + 1)φ),sin(nφ) cos φ + cos(nφ) sin φ = sin(nφ + φ) = sin((n + 1)φ)

Exercise 4

Let φ ∈ R Express sin(4φ) and cos(4φ) using sin φ and cos φ

Solution:

cos(4φ) + i sin(4φ) = (cos φ + i sin φ)4 =

= (︁cos2φ + 2i sin φ cos φ − sin2φ)︁2 =

= cos4φ − 4 sin2φ cos2φ + sin4φ +

+ 4i sin φ cos3φ − 2 cos2φ sin2φ − 4i sin3φ cos φ =

= cos4φ − 6 sin2φ cos2φ + sin4φ + i(︁4 sin φ cos3

φ − 4 sin3φ cos φ)︁ ,and therefore (it is enough to compare the real and imaginary parts)

cos(4φ) = cos4φ − 6 sin2φ cos2φ + sin4φ,

sin(4φ) = 4 sin φ cos3φ − 4 sin3φ cos φ

1 − i

√3i

(︃

cos(︂−π

4 − π3

)︃

+ i sin

(︃

−24 · 7π12

−5i

−1α

4 ≤arg (z + 2i) ≤ π

2};l) {z ∈ C: |z| + Re z ≤ 1 ∧

{︁z ∈ C: Re(z2) = 2}︁ = {︁x + iy : Re(x2+ 2ixy − y2) = 2}︁ =

={︁x + iy : x2 − y2 = 2}︁ :

√2

−√2

2i

d) {z ∈ C: | Im z| < 1}:

11

−1

−1i

−ii

−i

:

1i

−i

−1 2

−1 2

5 3

1 3

i

j) {z ∈ C: 2 < |z + 2 − 3i| < 4} = {z ∈ C: 2 < |z − (−2 + 3i)| < 4}:

−2

−23i3i

1 2

z2n+1 = (−1)2n+1+ i

2n + 1 = −1 +

i2n + 1 → −1

)︂n

= cos

(︂

nπ4

)︂

+ i sin

(︂

nπ4

)︂

+ i sin

(︂

−π3

)︂)︂6n

=

= cos(−2πn) + i sin(−2πn) → 1

⇕(∀ε > 0) (∃n0 ∈ N) (∀n ∈ N, n > n0) : |zn| < ε;

⇕(∀ε > 0) (∃n0 ∈ N) (∀n ∈ N, n > n0) : |zn| < ε

b) From the assumtions it follows that for all sucienlty large n we have that

zn= |zn| (cos (arg zn) + i sin (arg zn)) ,and the claim follows directly from the continuity of cosine and sine and the theorem ofthe limit of a product

As a counterexample disproving the reverse inequality we can use the sequence

and the choice

)︁2

= 2i;e) z4 = −1;f) z3 = i − 1;

g) z5 = 1;h) z2 = −11 + 60i;i) z2 = 3 + 4i.Solution:

a) z = |z| (cos φ + i sin φ) , 1 = cos 0 + i sin 0

z3 = |z|3(cos (3φ) + i sin (3φ)) = 1 (cos 0 + i sin 0)

⇕(|z|3 = 1) ∧ (∃k ∈ Z : 3φ = 0 + 2kπ)

⇕(|z| = 1) ∧(︁∃k ∈ Z: φ = k2π

3 )︁ ,and therefore

z = zk = cos

(︃

k2π3

)︃

+ i sin

(︃

k2π3

2 , k ∈ {3l + 1 : l ∈ Z} ,

−1

2 − i

√ 3

2 , k ∈ {3l + 2 : l ∈ Z} ,so

2 , −

1

2− i

√32

b) z = |z| (cos φ + i sin φ) , i = cosπ

2 + i sinπ

2,

z2 = |z|2(cos (2φ) + i sin (2φ)) = cosπ2 + i sinπ2

⇕(|z|2 = 1) ∧(︁∃k ∈ Z: 2φ = π

2 + 2kπ)︁ ,and therefore

2 , k ∈ {2l : l ∈ Z} ,

−

√ 2

2 − i

√ 2

2 + i

√2

2 , −

√2

2 − i

√22

which holds if and only if z = x + iy = 3 + 4i or z = −3 − 4i

d) After the change of variables z−1

z+1 =: u = |u| (cos φ + i sin φ) we rstly solve the equation

)︂

+ i sin

(︂π4

)︂)︂

= ±(1 + i),and then easily z−1

z+1 = 1 + i if and only if (z = x + iy)

x + iy − 1 = (1 + i)(x + iy + 1), that is(x − 1) + iy = (x − y + 1) + i(x + y + 1), and therefore(x − 1 = x − y + 1) ∧ (y = x + y + 1), that is

y = 2 ∧ x = −1,and similarily z−1

z+1 = −1 − i if and only if

x + iy − 1 = −(1 + i)(x + iy + 1),(x − 1 = −x + y − 1) ∧ (2y = −x − 1), and therefore

y = −2

5 ∧ x = −1

5.Summary:

)︂

+ i sin(︂π

4 + k

π2

)︃

+ i sin(︃ 3π

4)︃)︃

if and only if

(︃

|z| = 3

√︂√2

(︃

cos(︃ π

4 +

2kπ3

)︃

+ i sin(︃ π

4 +

2kπ3

)︃)︃

: k ∈ {0, 1, 2}

}︃

and therefore

z2 = 3 + 4i ⇔ z = ±(2 + i)

Exercise 12.

Find and draw the set M = {︁1

z: z ∈ Ω}︁

, ifa) Ω = {z ∈ C: arg z = α}, α ∈ (−π, π⟩;

M

−α

α = π ⇒ M = Ω = {z ∈ C : arg z = π}

2}, f (z) := ez;c) Ω = {z ∈ C: 0 < Re z < π ∧ Im z > 0}, f(z) := eiz;

d) Ω = {z ∈ C: Im z = 1

2}, f (z) := z2

π 6

π 6

M

M

π 3

π 3

i

1M

i

M

i

−1 4

−1 4

+ i sin(︃ π

2 + arctan

53

)︃)︃,

)︄)︄,

= (2 ±

√2)i

(2 ±√

2)√2

= 3 ±

√7

2 · (−1 + i),and therefore

iz = Ln

(︄

3 ±√7

√2)︄

, k ∈ Z

;d) i3

4;

e) (−1)√3;f) (−√3i + 1)−3.Solution:

2 − i

√22

(−√3i + 1)−3 = e−3 Ln(−

√ 3i+1) = e−3(ln 2−π3i+2kπi) =

z.Solution:

(Re f )(x, y) = x3− 3xy2+ 5x − 1,(Im f )(x, y) = 3x2y − y3+ 5y.

f)

f (x + iy) = x − iy

x2+ y2,therefore

(Re f )(x, y) = x

x2+ y2,

(Im f )(x, y) = − y

x2+ y2

)︁):

z1, z2 ∈ Ω

z13 = z23 ⇒ |z1| = |z2|

φ1 = φ2 ⇒ z1 = z2,therefore the function f is injective on Ω

Alternatively we can use the continuity of the function f(z) := z+2

z+i at the point i:

a) φ(t) := 1 − it, Dφ = ⟨0, 2⟩.

1i

b) Ω is a line segment with the endpoints a, b ∈ C, a =/ b; φ(t) := a + (b − a)t, t ∈ ⟨0, 1⟩.

Ω = ⟨φ⟩

1 2

1 2

−1 3

2ii2i

21

Ω is open and connected set, and therefore Ω is a domain

f) Ω = {︁z ∈ C: |z| < 1 ∧ arg z ∈ (−π, π⟩∖ {0}}︁

Ω

Ω

11

Ω is open and connected set, and therefore Ω is a domain

b) f(x + iy) = |(x + iy)2| = (|x + iy|)2 = x2+ y2 So f = u + iv, where u(x, y) := x2+ y2

and at the same time the functions u and v are dierentiable in R2, and therefore

f has a derivative (only) in the point 0 ant it is not holomorphic anywhere

∂x = ∂v∂y and ∂u

∂y = −∂v∂x in R2, and therefore f is holomorphic everywhere in C and

It remains to prove or disprove the existence of the derivative at the point 0:

(x2+ y2)2,

∂u

∂y(x, y) =

−x22y(x2 + y2)2,

∂v

∂x(x, y) =

−y(x2+ y2) + xy2x(x2+ y2)2 = x

2y − y3

(x2+ y2)2,and therefore the derivative can exist only in the points x + iy where

(︃

2xy2(x2+ y2)2 = x(−x

2+ y2)(x2+ y2)2

)︃

∧

(︃

2x2y(x2+ y2)2 = y(x

2− y2)(x2+ y2)2

)︃

,

xy2(x2+ y2)2 = −x3

(x2+ y2)2

)︃

It is easy to observe that this system of equations has no solution

Summary: the function f does not have a derivative at any point, and therefore it is notholomorphic at any point

The functions u and v are dierentiable in R2, and for every (x, y) ∈ R2 we have that

f (x + iy) = (x3 − 3xy2 − 2y) + i(3x2y − y3+ 2x + c), c ∈ R.

(︁

You can simplify this to f(z) = z3

+ 2zi + ci, c ∈ R.)︁

−∂u

∂y(x, y) =

2xy(x2+ y2)2 = ∂v

∂x(x, y) ⇒ v(x, y) =

∫︂

2xy(x2+ y2)2 dx

After the change of variables x2+ y2 = t (2x dx = dt) we get

∫︂

2xy(x2+ y2)2 dx = y

∆u(x, y) =/ 0 for every (x, y) v R2

.Function u is not harmonic on Ω, and therefore the required function f does not exist.d)

∂u

∂x(x, y) = 2x + 5 +

2xy(x2+ y2)2 = ∂v

∂y(x, y)

⇓v(x, y) = 2xy + 5y − x

f (z) = z2+ (5 − i)z − i

z + ci, c ∈ R

Similarly to the solution to Exercise 25 a) we can nd out that

f (x + iy) = (x3− 3xy2− 2y + 2) + i(3x2y − y3+ 2x + c), where c ∈ R

a) The requirement

f (0) = f (0 + 0i) = 2 + ic = iobviously cannot be satised by any choice of c ∈ R The function f with the requiredproperties does not exist

b) We want to satisfy the condition

f (1) = f (1 + 0i) = 3 + i(2 + c) = 3 − i,

and therefore 2 + c = −1, which is c = −3 So the required function exists, it is

f (x + iy) = (x3− 3xy2− 2y + 2) + i(3x2y − y3 + 2x − 3)

f (x + iy) = (−3x2y + y3+ c) + i(−3xy2+ x3+ 5), c ∈ R,which is

f (z) = c + iz3+ 5i, c ∈ R

The sought function on the set Ω is

f (x + iy) = ln√︁x2+ y2+ c + i arg(x + iy), c ∈ R,which is

f (z) = c + ln z, c ∈ R

Exercise 28

Let Ω := {z ∈ C : Re z > 0} Let v(x, y) := 1 + arctany

x Find (if it exists) a holomorphicfunction f = u + iv, f : Ω → C, where

2− x2)(x2+ y2)2,

and therefore (the partial derivatives with respects to y are analogous)

∆v(x, y) = 2(y

2 − x2)(x2+ y2)2 + 2(x

2− y2)(x2+ y2)2 = 0

We've proven that the function v is on C∖ {0} harmonic

Let us assume, for contradiction, that there is a function u with the properties statedabove Then for (x, y) ∈ R2, such that x + iy ∈ Ω1 := {z ∈ C : Re z > 0}, we have

At the same time the function u is continuous on R2∖{(0, 0)} (at every point R2∖{(0, 0)}

it must be dierentiable), and therefore

2π = c1− c2 = −(c2− c1) = −2π,which is an contradiction The sought function u does not exist

e < 1 impliesthat it is a contraction).

From this we get

|f′(z0)| = 75 extensibility coecient of the function f at z0

(75 > 1, therefore it is a dilatation),arg f′(z0) = −π + 2 arctan3

4 . rotational angle of the function f at the point z0.c)

1 < |f′(z0)| = 2 extensibility coecient of f at z0 (dilatation),

arg f′(z0) = π

2 . rotational angle f at z0

b) f′(z) exists in C∖ {x + iy : y = 0 ∧ x ≤ −4} =: Ω For every z ∈ Ω we have that

z+3;e) Ω = U(1, 2), f(z) := z−1

2z−6;f) Ω = {z ∈ C: Re z < 1}, f(z) := 1

z;g) Ω = {z ∈ C: Re z < 1}, f(z) := z

z−1+i;h) Ω = {z ∈ C: Re z < 1}, f(z) := z

z−2;i) Ω = {z ∈ C: Re z < 0 ∧ Im z < 0}, f(z) := 1

z;j) Ω = {z ∈ C: Re z > 0 ∧ Im z > 0}, f(z) := z−1

z+1;k) Ω = {z ∈ C: −1 < Re z < 0 ∧ Im z < 0}, f(z) := z−i

z+i;l) Ω = {z ∈ C: |z| < 1 ∧ Re z < 0 ∧ Im z > 0}, f(z) := z

z−i.Solution:

a) Ω = U(1, 2), f(z) := 1 − 2iz, f(Ω) = U(1 − 2i, 4)

ΩΩ11

2 A hint for some of the following exercises Realize (and prove) that:

f is conformal in the set Ω ⊂ C ∞ ,

A, B ⊂ Ω

}︄

⇒ f (A ∩ B) = f (A) ∩ f (B).

b) Ω = {z ∈ C: Re z < 1}, f(z) := (1 + i)z + 1,

f (1) = 1 + i + 1 = 2 + i,

f (1 + i) = 2i + 1,

f (0) = 1,and therefore (think it through!) f(Ω) = {z ∈ C: Re z + Im z < 3}

1

−1

ΩΩ

i2i

i2i

)︃

ΩΩ

1 5

−1 3

ΩΩ

−1

4i

1 4

−1

ΩΩ

1

−1

ΩΩ

i

f (Ω)

f (Ω)

1

i1i

−1 2

1 2

−1 2

11

}︃

ΩΩ

where a, b, c, d ∈ C, ad =/ bc

From the given conditions we get the system of equations

−a + b

−c + d = 0,

ai + b

ci + d = 2i,a(1 + i) + bc(1 + i) + d = 1 − i.

From the rst equation follows that a = b We can choose (think about why!) a = b = 1.The remaining two equations then become

i + 1

ci + d = 2i,

2 + ic(1 + i) + d = 1 − i,

2z + 24iz + 5 − i, z ∈ C,1

From the condition

−1 − 2i = −

2

5 − i

5, z = ∞.

i

1

√2

√2i

22i

11i

ΩΩ

f (Ω)

f (Ω)

11

We start with nding the linear fractional function

√

5 < 1,

2 We can choose for example

f1

that is the function

(which would lead us to the denition f := −f∗)

Because f∗(i) = 1+ii+1 = 1 (the rst possibility is realized), we choose

Let us rstly consider the mapping z ↦→ z4.

Exercise 39.

Find the images of the lines parallel to the real and imaginary axes by the mapping

f (z) := 1z (consider the lines together with the point ∞)

Exercise 40.

Find the images of the sets

Mα = {z ∈ C : arg z = α} and Nr = {z ∈ C : |z| = r},where α ∈ (−π, π⟩ and r ∈ R+, by the mapping f(z) := ln z

2 − t)︁, t ∈ ⟨π

2,π2 + 3⟩,

t − π2 − 3, t ∈ ⟨π2 + 3,π2 + 6⟩.Solution:

ti dt +

∫︂ 3 0

t dt =

= 9i[sin t]

π 2

0 + 9[cos t]

π 2

0 + (1 − i)[︃ t2

2]︃3

2i.

−2 ∈int γ, i ∈ int γ, 1 ∈ ext γ.

But we were supposed to use the Cauchy's integral formulas Which we can do forexample as

z − 2dz =

= 2πi[︃ sin z

z − 5]︃

z − (−π)dz +

∫︂

γ 2

cos z z+π

z − πdz =

= 2πi(︃[︃ cos z

z − π]︃

z=−π

+[︃ cos z

z + π]︃

= 0

32 i.

g)

⟨γ⟩

3 2

γ

1 (z−1) 3

dz(1 − z)(z + 2)(z − i)2 =

γ 1

dz(1 − z)(z + 2)(z − i)2 +

γ 2

dz(1 − z)(z + 2)(z − i)2 =

=

∫︂

γ 1

1 (1−z)(z−i) 2

z + 2 dz +

∫︂

γ 2

1 (1−z)(z+2)

(z − i)2 dz =

= 2πi

(︃[︃

1(1 − z)(z − i)2

[︃

2z + 1(−z2− z + 2)2

)︃

= 2πi(︃ 3 − 4i

(1 + 2i)(8 + 6i)100

ezdz = [ez]1+i0 = e1(cos 1 + i sin 1) − 1 =

= e cos 1 − 1 + i(sin 1)e

b)

∫︂ 1+i 0

z3dz =[︃ z4

4]︃1+i

0

= 1

4(2i)

2 = −1

∫︂ i 0

z2sin z dz =

∫︂ i 0

z sin z dz = [−z cos z]i0+ [sin z]i0 =

= −i cos i + sin i =

= −i cosh 1 + i sinh 1 =

= i(sinh 1 − cosh 1) = −1

ei(we again integrated by parts)

2 →

√2

and therefore the given series converges absolutely for every z ∈ C such that ⃓⃓z+1

z−1

⃓

⃓< 1,and diverges for every z ∈ C for which ⃓⃓z+1

)︁n

convergesabsolutely

Summary: the given series converges (absolutely) for every

1

|z|,the series ∑︁∞

n(z) ∈ C for every z ∈ C, and therefore forevery z ∈ C we have that

lim sn(z) ∈ C ⇔ lim s∗∗n (z) ∈ C

Summary: the given series covnerges (absolutely) on the set {z ∈ C: |z| > 1}.

2(it is enough to realize that for n ≥ 3 we have that 1 ≤ √n

3n − 2 ≤ √n

n ·√n

n → 1), andtherefore

R =

√2

3 .

→ e,and therefore

R = 1

e.f)

R = 1

e.g)

R = 1

h)

1 (n+9)!

1 (n+8)!

n + 9 → 0,and therefore

n → 1, and therefore the radius of convergence of a given series is 1.

For every z ∈ C, |z| < 1 we have that

(1 − z)2

b) n

√︂

1

n → 1, therefore the radius of convergence is 1

Let us dene the function f(z) := ∑︁∞

|z| < 1 ⇒ 1 − z ∈ Ω := {w ∈ C : |w − 1| < 1} ,

ln′w = 1

w v Ω,

11

−1

Ω

there is a c ∈ C such that for each z ∈ C, |z| < 1 we have that

f (z) = − ln(1 − z) + c

Furthermore f(0) = − ln 1 + c = 0, and therefore c = 0

Summary: for each z ∈ C, |z| < 1 we have that

z − 1+

12

1

z + 1.From that it follows that there is c ∈ C, such that for each z ∈ C, |z| < 1 we have that

f (z) = z − ln(1 + z) + c,and because 0 = f(0) = c we have

For every z ∈ C, |z| < 1, we have that

and therefore for every z ∈ C, |z| < 1, we have that

=

= 1z

z 2

e) f(z) := sin(3z2 + 2), z0 = 0;f) f(z) := 1

(z−1) 3, z0 = 3;g) f(z) := sin2z, z0 = 0

a) f(z) = 2

3

1 z+5+ 13 z−11 ,

2 i)︂,

1

and therefore the sought Taylor series has the radius of convergence 1

For every z ∈ C, |z| < 1, we have that

f (z) = z

i · 1

1 + zi2 =

zi

c) Because clearly

11

From this it follows that for every z ∈ C we have that

f (z) = sin(3z2) cos 2 + cos(3z2) sin 2 =

Clearly the radius of convergence is 2 For every z ∈ U(3, 2) we have that

g) The radius of convergence is ∞ and for any z ∈ C we have that

f (z) = sin2z = 1 − cos(2z)

= 1

2− 12

have found out that

2i

i 2

2i

2

1 2

(n+1) 2 +1 1

n 2 +1 converges absolutely

We have (also) found that

z(z−2), 1 < |z − 2| < 2;

f) f(z) := z

(z 2 +1) 2, 0 < |z − i| < 2;g) f(z) := z−sin z

z 4 , 0 < |z| < ∞;h) f(z) := z+2

z 2 −4z+3, 2 < |z − 1| < ∞;i) f(z) := 1

z(z−3) 2, 1 < |z − 1| < 2.Solution:

a) For every z ∈ C, 0 < |z| < 1 we have that

(This holds even for every z ∈ C∖ {0}.)

b) For every z ∈ C, |z| > 1 we have that

1 − 2z5 =

12z

∞

∑︂

n=0

(︃ 52

(This holds even for every z ∈ C such that 0 < |z − 2| < 2.)

f) Because for every z ∈ C, 0 < |z − i| < 2 we have that

(z2+ 1)2 = 1

(z − i)2

z + i − i(z + i)2 = 1

12i· 1

For every z ∈ C, 0 < |z − i| < 2 we have

g) For every z ∈ C we have that

i) Because for every z ∈ C, 1 < |z − 1| < 2, we have that

1 3

ii2i

(−1)n−1(z − 1)n +

(−1)n−1(z − 1)n +

1

z + 2 +

1(z − 1)2,and because f is clearly holomorphic on C∖ {−2, 1}, we have precisely three maximalanuli: