Complex analysis
Exe cises wi h solu ions
Jiří Bouchala
(and Ondřej Bouchala)
17. lis opadu 2172/15
708 00 Os a a-Po uba
Czech Republic
uni [email p o ec ed]
www. sb.cz T ans o ma ion o he s uc u e
and con en o highe educa ion
a VŠB-TUO
NPO_VŠB-TUO_MSMT-16605/2022
This wo k is licenced unde CC BY-SA 4.0 cba.
The au ho o he pain ing Imaginá ní džungle on he co e page is Jiří Bouchala
(and i is owned by Ondřej Bouchala).
P e ace
This ex con ains he solu ions o all o he p ac ice p oblems in he 10 h chap e o he
lec u e no es “An In oduc ion o Complex Analysis” [1]. I is a ansla ion o he Czech
ex [3].
The ypese ing and all o he pic u es a e he wo k o my son Ondřej. He also helped
o imp o e he ex in se e al places wi h his commen s.
I is no possible ha we caugh all o he mis akes du ing he p oo eading. We a e
g a e ul o you leniency and o le ing us know abou any and all ema ks.1
We enjoyed wo king on his ex . We wish he same o he eade .
In O lo á, 2024
Jiří Bouchala
(and Ondřej Bouchala)
1Please send all o he ema ks (no es, ecommenda ions, h ea s and gi s) o my e-mail add ess
[email p o ec ed].
Exe cise 1.
Find he eal and imagina y pa o he complex numbe
a) z= (1 + i)(3 −2i) ;
b) z=2−3i
3+4i;
c) z=1+i
1−i;
d) z= 2i−2−4i
2.
Solu ion:
a) z= (3 + 2) + i; Re z= 5,Im z= 1.
b) z=2−3i
3+4i=(2−3i)(3−4i)
9+16 =6−12−9i−8i
25 ; Re z=−6
25,Im z=−17
25.
c) z=1+i
1−i=(1+i)2
2=1+2i−1
2; Re z= 0,Im z= 1.
d) z= 2i−2−4i
2= 2i−2+4i
2=−1; Re z=−1,Im z= 0.
Exe cise 2.
W i e he gi en complex numbe in he igonome ic o m
a) z=−1 + √3i;
b) z=i;
c) z=−8 ;
d) z=−1−√3i;
e) z=2+i
3−2i;
) z=3−i
2+i.
Solu ion:
a)
√3i
−1
αφ
cos α=√3
2, α =π
6, φ =π
2+α=π
2+π
6=2
3π;
z=−1 + √3i=√1+3(︃cos 2π
3+isin 2π
3)︃= 2 (︃cos 2π
3+isin 2π
3)︃.
b) z=i= cos π
2+isin π
2.
c) z=−8 = 8(cos π+isin π).
1

d)
−√3i
−1
α
φ
sin α=√3
2, α =π
3, φ =π+α=4
3π;
z=−1−√3i= 2 (︃cos 4
3π+isin 4
3π)︃= 2 (︃cos (︃−2π
3)︃+isin (︃−2π
3)︃)︃.
e) z=2+i
3−2i=(2+i)(3+2i)
9+4 =4
13 +7
13i,
7
13i
4
13
φ
|z|=1
13√16 + 49 = √65
13 , an φ=
7
13
4
13
=7
4, φ = a c an 7
4;
z=√65
13 (︃cos (︃a c an 7
4)︃+isin (︃a c an 7
4)︃)︃.
) z=3−i
2+i=(3−i)(2−i)
5=5−5i
5= 1 −i,
−i
1
z=√2(︂cos (︂−π
4)︂+isin (︂−π
4)︂)︂.
2
Exe cise 3.
P o e he de Moi e’s heo em
(∀n∈N) (∀φ∈R) : (︁cos φ+isin φ)︁n= cos(nφ) + isin(nφ)
using ma hema ical induc ion.
Solu ion:
1) We s a by checking ha he o mula holds o n= 1:
(cos φ+isin φ)1= cos (1 ·φ) + isin (1 ·φ).
2) Now we p o e he implica ion (cos φ+isin φ)n= cos(nφ) + isin(nφ)?
⇒
?
⇒(cos φ+isin φ)n+1 = cos((n+ 1)φ) + isin((n+ 1)φ):
(cos φ+isin φ)n+1 i.p.
= (cos (nφ) + isin (nφ)) (cos φ+isin φ) =
= (cos(nφ) cos φ−sin(nφ) sin φ) + i(sin(nφ) cos φ+ cos(nφ) sin φ),
and now i su ices o apply he known “ igonome ic iden i ies”:
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).
Exe cise 4.
Le φ∈R. Exp ess sin(4φ)and cos(4φ)using sin φand cos φ.
Solu ion:
cos(4φ) + isin(4φ) = (cos φ+isin φ)4=
=(︁cos2φ+ 2isin φcos φ−sin2φ)︁2=
= cos4φ−4 sin2φcos2φ+ sin4φ+
+ 4isin φcos3φ−2 cos2φsin2φ−4isin3φcos φ=
= cos4φ−6 sin2φcos2φ+ sin4φ+i(︁4 sin φcos3φ−4 sin3φcos φ)︁,
and he e o e (i is enough o compa e he eal and imagina y pa s)
cos(4φ) = cos4φ−6 sin2φcos2φ+ sin4φ,
sin(4φ) = 4 sin φcos3φ−4 sin3φcos φ.
3
Exe cise 5.
Find Re zand Im z o z=(︂1−i
1+√3i)︂24
.
Solu ion:
1 + √3i
1−i
√3i
−i
1
1−i
1 + √3i=√2(︁cos (︁−π
4)︁+isin (︁−π
4)︁)︁
2(︁cos π
3+isin π
3)︁=1
√2(︃cos (︂−π
4−π
3)︂+isin (︃−7
12π)︃)︃,
z=1
212 (︃cos (︃−24 ·7π
12 )︃+isin (︃−24 ·7π
12 )︃)︃=1
212 ;
Re z=1
212 ,Im z= 0.
Exe cise 6.
Find A g zand a g z o
a) z=(︁√3 + i)︁126 ;
b) z= (1 + i)137 ;
c) z=−1−5i.
Solu ion:
a) z= (√3 + i)126 =(︁2(cos π
6+isin π
6))︁126 = 2126 (cos(21π) + isin(21π)) = −2126;
A g z={π+ 2kπ :k∈Z},a g z=π.
b) z= 2137
2(︁cos (︁137π
4)︁+isin (︁137π
4)︁)︁= 2137
2(︁cos π
4+isin π
4)︁;
A g z={︂π
4+ 2kπ :k∈Z}︂,a g z=π
4.
4
c)
−5i
−1
α
an α=5
1, α = a c an 5;
A g z={−π+ a c an 5 + 2kπ :k∈Z},a g z=−π+ a c an 5.
Exe cise 7.
D aw in he complex plane he se
a) {z∈C: Re z≤1};
b) {z∈C: Re(z2) = 2};
c) {z∈C: Im 1
z=1
4};
d) {z∈C:|Im z|<1};
e) {z∈C:|z|= Re z+ 1};
) {z∈C:|z−2|=|1−2z|};
g) {z∈C:z−2
z−3= 1};
h) {z∈C:|1 + z|<|1−z|};
i) {z∈C:|z+ 1|= 2|z−1|};
j) {z∈C: 2 <|z+ 2 −3i|<4};
k) {z∈C:π
4≤a g(z+ 2i)≤π
2};
l) {z∈C:|z|+ Re z≤1∧
∧−π
2≤a gz≤π
4}.
Solu ion:
a) {z∈C: Re z≤1}:
1
1 2
5
b) F om he assump ions i ollows ha o all su icien ly la ge nwe ha e ha
zn=|zn|(cos (a g zn) + isin (a g zn)) ,
and he claim ollows di ec ly om he con inui y o cosine and sine and he heo em o
he limi o a p oduc .
As a coun e example disp o ing he e e se inequali y, we can use he sequence
zn:= cos (︃π+(−1)n
n)︃+isin (︃π+(−1)n
n)︃
and he choice
= 1, φ =π.
Exe cise 11.
Find all z∈Csuch ha
a) z3= 1;
b) z2=i;
c) z2= 24i−7;
d) (︁z−1
z+1)︁2= 2i;
e) z4=−1;
) z3=i−1;
g) z5= 1;
h) z2=−11 + 60i;
i) z2= 3 + 4i.
Solu ion:
a) z=|z|(cos φ+isin φ),1 = cos 0 + isin 0.
z3=|z|3(cos (3φ) + isin (3φ)) = 1 (cos 0 + isin 0)
⇕
(|z|3= 1) ∧(∃k∈Z: 3φ= 0 + 2kπ)
⇕
(|z|= 1) ∧(︁∃k∈Z:φ=k2π
3)︁,
and he e o e
z=zk= cos (︃k2π
3)︃+isin (︃k2π
3)︃=⎧
⎨
⎩
1, k ∈ {3l:l∈Z},
−1
2+i√3
2, k ∈ {3l+ 1: l∈Z},
−1
2−i√3
2, k ∈ {3l+ 2: l∈Z},
so
z3= 1 ⇔z∈{︄1,−1
2+i√3
2,−1
2−i√3
2}︄.
1 = z0
1 = z0
z1
z2=z−1
12

b) z=|z|(cos φ+isin φ), i = cos π
2+isin π
2,
z2=|z|2(cos (2φ) + isin (2φ)) = cos π
2+isin π
2
⇕
(|z|2= 1) ∧(︁∃k∈Z: 2φ=π
2+ 2kπ)︁,
and he e o e
z=zk= cos (︂π
4+kπ)︂+isin (︂π
4+kπ)︂=
={︄√2
2+i√2
2, k ∈ {2l:l∈Z},
−√2
2−i√2
2, k ∈ {2l+ 1: l∈Z}.
z1
z0
z2=i⇔z∈{︄√2
2+i√2
2,−√2
2−i√2
2}︄.
c) Le z=x+iy. Then
z2=x2+ 2ixy −y2= 24i−7⇔(︃x2−y2=−7
2xy = 24 )︃⇔
⇔(︃x2−y2=−7
y=12
x)︃⇔
⇔(︃x2−144
x2=−7
y=12
x)︃⇔
⇔(︃x4+ 7x2−144 = 0
y=12
x)︃,
which holds i and only i z=x+iy = 3 + 4io z=−3−4i.
13
d) A e he change o a iables z−1
z+1 =: u=|u|(cos φ+isin φ)we i s ly sol e he equa ion
u2= 2i, ha is
|u|2(cos (2φ) + isin (2φ)) = 2 (︂cos (︂π
2)︂+isin (︂π
2)︂)︂.
The solu ion is
u=±√2(︂cos (︂π
4)︂+isin (︂π
4)︂)︂=±(1 + i),
and hen easily z−1
z+1 = 1 + ii and only i (z=x+iy)
x+iy −1 = (1 + i)(x+iy + 1), ha is
(x−1) + iy = (x−y+ 1) + i(x+y+ 1),and he e o e
(x−1 = x−y+ 1) ∧(y=x+y+ 1), ha is
y= 2 ∧x=−1,
and simila ly z−1
z+1 =−1−ii and only i
x+iy −1 = −(1 + i)(x+iy + 1),
(x−1 = −x+y−1) ∧(2y=−x−1),and he e o e
y=−2
5∧x=−1
5.
Summa y: (︃z−1
z+ 1)︃2
= 2i⇔(︃z=−1+2i∨z=−1
5−2
5i)︃.
e) |z|4(cos (4φ) + isin (4φ)) = cos π+isin πi and only i
z=zk= cos (︂π
4+kπ
2)︂+isin (︂π
4+kπ
2)︂, k ∈Z, ha is
z4=−1⇔z∈{︃1 + i
√2,−1 + i
√2,−1−i
√2,1−i
√2}︃.
11
) |z|3(cos (3φ) + isin (3φ)) = √2(︃cos (︃3π
4)︃+isin (︃3π
4)︃)︃
i and only i (︃|z|=3
√︂√2)︃∧(︃3φ=3π
4+ 2kπ, k ∈Z)︃.
F om his, i easily ollows ha z3=i−1i and only i
z∈{︃6
√2(︃cos (︃π
4+2kπ
3)︃+isin (︃π
4+2kπ
3)︃)︃:k∈ {0,1,2}}︃.
14
g) z= cos (︃2π
5k)︃+isin (︃2π
5k)︃, k ∈ {0,1,2,3,4}.
1
1
h)
z2= (x+iy)2=−11 + 60i
⇕
x2+ 2ixy −y2= 11 + 60i
⇕
x2−y2=−11 ∧2xy = 60
⇕
x2−900
x2=−11 ∧y=30
x
⇕
y=30
x∧x2=−11 ±√121 + 3600
2=⎧
⎪
⎨
⎪
⎩
−11 −√3721
2. . . no possible,
−11 + √3721
2=−11 + 61
2= 25,
and he e o e
z2=−11 + 60i⇔z=±(5 + 6i).
i) Le z=x+iy. Then
z2= (x+iy)2= 3 + 4i
⇕
x2−y2= 3 ∧2xy = 4
⇕
x2−4
x2= 3 ∧y=2
x
⇕
y=2
x∧x2=3±√9 + 16
2={︃3−5
2. . . no possible,
4,
and he e o e
z2= 3 + 4i⇔z=±(2 + i).
15
Exe cise 12.
Find and d aw he se M={︁1
z:z∈Ω}︁, i
a) Ω = {z∈C:a gz=α}, α ∈(−π, π⟩;
b) Ω = {z∈C:|z−1|= 1};
c) Ω = {z∈C: Re z= Im z};
d) Ω = {x+iy ∈C:x= 1};
e) Ω = {x+iy ∈C:y= 0}.
Solu ion:
a) α∈(−π, π)⇒M={z∈C: a g z=−α};
Ω
α
M
−α
α=π⇒M=Ω={z∈C: a g z=π}.
ΩM
16
b)
M={︃u+i :1
u+i ∈Ω}︃∪{∞} =
={︃u+i :
1
u+i −1= 1}︃∪{∞} =
={u+i :|1−u−i |=|u+i |}∪{∞} =
={︁u+i : (1 −u)2+ 2=u2+ 2}︁∪{∞} =
={u+i : 1 −2u= 0}∪{∞} =
={︃u+i :u=1
2}︃∪{∞}.
1
i
2
2
Ω
1
i
1
2
1
2
M
c)
M={︃u+i :1
u+i ∈Ω}︃∪{∞} =
={︃u+i :u−i
u2+ 2∈Ω}︃∪{∞},
and because u
u2+ 2=−
u2+ 2⇔(u=
/0∧u=− ), we ha e ha
M={u+i :u=
/0∧u=− }∪{∞}.
Ω
M
17

d)
M={︃u+i :1
u+i ∈Ω}︃=
={︃u+i :u
u2+ 2= 1}︃=
={︄u+i :(︃u−1
2)︃2
+ 2=1
4}︄
∖
{0}.
1
i
1
1
Ω
i
1
2
1
2
M
e)
M={︃u+i :1
u+i ∈Ω}︃∪{∞} =
={︃u+i :−
u2+ 2= 0}︃∪{∞} =
={u+i : = 0 =
/u}∪{∞}.
ΩM
Exe cise 13.
Find and d aw he se M={ (z): z∈Ω}, i
a) Ω = {z∈C:|a gz| ≤ π
6}, (z) := z2;
b) Ω = {z∈C:|Im z|<π
2}, (z) := ez;
c) Ω = {z∈C: 0 <Re z < π ∧Im z > 0}, (z) := eiz;
d) Ω = {z∈C: Im z=1
2}, (z) := z2.
18
Solu ion:
a)
M={︂z∈C:|a g z| ≤ π
6·2 = π
3}︂.
Ω
Ω
π
6
π
6
M
M
π
3
π
3
b)
M={︂ex+iy :|y|<π
2}︂=
={︂ex(cos (y) + isin (y)): |y|<π
2}︂=
={z∈C: Re z > 0}.
1
1
π
2i
−π
2i
π
2i
−π
2i
Ω
Ω
M
M
c) M={︁ei(x+iy)=e−y(cos x+isin x): 0 < x < π ∧y > 0}︁=
={z∈C:|z|<1∧Im z > 0}.
Ω
Ω
1π
i
1
M
i
M
i
19
d)
M={︄(︃x+1
2i)︃2
:x∈R}︄=
={︃x2−1
4+xi:x∈R}︃=
={︃y2−1
4+yi:y∈R}︃.
Ω
Ω
1
2i
1
2i
1
i
2
i
2
−1
4
−1
4
M
M
Exe cise 14.
Compu e
a) sin(2 −3i);
b) cos i;
c) cosh i;
d) Ln(−5+3i)aln(−5+3i);
e) Ln(−4−√3i)aln(−4−√3i);
) Ln(ie2).
Solu ion:
a)
sin(2 −3i) = ei(2−3i)−e−i(2−3i)
2i=
=e3(cos (2) + isin (2)) −e−3(cos (−2) + isin (−2))
2i=
=e3−e−3
2i·cos 2 + i(e3+e−3)·sin 2
2i=
= cosh 3 ·sin 2 −(sinh 3 ·cos 2)i.
=
.
= 9.15 + 4.17i.
b)
cos i=ei·i+e−i·i
2= cosh 1 .
= 1.54.
20
c)
cosh i=ei+e−i
2=cos 1 + isin 1 + cos (−1) + isin (−1)
2= cos 1 .
= 0.54.
d)
−5+3i=√34 (︃cos (︃π
2+ a c an 5
3)︃+isin (︃π
2+ a c an 5
3)︃)︃,
and he e o e
Ln(−5+3i) = ln √34 + i(︃π
2+ a c an 5
3)︃+ 2kπi, k ∈Z;
ln(−5+3i) = ln √34 + i(︃π
2+ a c an 5
3)︃.
e)
−4−√3i=√19 (︄cos (︄−π+ a c an √3
4)︄+isin (︄−π+ a c an √3
4)︄)︄,
and he e o e
Ln(−4−√3i) = ln √19 + i(︄−π+ a c an √3
4)︄+ 2kπi, k ∈Z;
ln(−4−√3i) = ln √19 + i(︄−π+ a c an √3
4)︄.
)
Ln(ie2) = ln(e2) + iπ
2+ 2kπi =
= 2 + π
2i+ 2kπi, k ∈Z.
Exe cise 15.
Find all z∈C, o which we ha e ha
a) sin z= 3;
b) cos z=√3
2;
c) sin z+ cos z= 2;
d) sin z−cos z= 3;
e) z2+ 2z+ 9 + 6i= 0.
21
whe e φ1, φ2∈⟨︁0,π
4)︁. Then
z3
1=z3
2
⇕
|z1|3(cos (3φ1) + isin (3φ1)) = |z2|(cos (3φ2) + isin (3φ2))
⇕
(|z1|=|z2|)∧(∃k∈Z: 3φ1= 3φ2+ 2kπ).
F om ha , i ollows ha (we a e using he assump ion φ1, φ2∈⟨︁0,π
4)︁):
z1, z2∈Ω
z3
1=z3
2⇒|z1|=|z2|
φ1=φ2⇒z1=z2,
he e o e he unc ion is injec i e on Ω.
Exe cise 19.
Decide i he gi en limi exis s, and i i does compu e i
a) lim
z→0
Re z
z;
b) lim
z→0
Im(z2)
zz ;
c) lim
z→0
zIm z
|z|;
d) lim
z→0
z2
|z|2;
e) lim
z→0
z3
|z|2;
) lim
z→i
z2+z(2−i)−2i
z2+1 ;
g) lim
z→0
Re z
1+|z|.
Solu ion:
a) lim
z→0
Re z
zdoes no exis , because
0 =
/1
n→0∧Re (︁1
n+ 0i)︁
1
n
= 1 →1
and a he same ime
0 =
/i1
n→0∧Re (︁i1
n)︁
1
n
= 0 →0.
b) lim
z→0
Im z2
z·zdoes no exis , because o 0 =
/z=x+iy we ha e ha
Im z2
z·z=2xy
x2+y2={︃1, x =y=
/0,
0, x ·y= 0, x2+y2=
/0.
1
1
0
0
28

c) lim
z→0
zIm z
|z|= 0, because
0 =
/zn→0⇒
znIm zn
|zn|=|Im zn| → 0⇒znIm zn
|zn|→0.
d) lim
z→0
z2
|z|2does no exis , because o 0 =
/z=x+iy we ha e ha
z2
|z|2=x2−y2+ 2ixy
x2+y2={︃i, x =y=
/0,
1, y = 0 =
/x.
i
i
1
1
e) lim
z→0
z3
|z|2= 0, because
lim
z→0
z3
|z|2= lim
z→0|z|= 0.
)
lim
z→i
z2+z(2 −i)−2i
z2+ 1 = lim
z→i
(z−i)(z+ 2)
(z−i)(z+i)= lim
z→i
z+ 2
z+i=
= lim
x+iy→i
x+2+iy
x+i(y+ 1) =
= lim
(x,y)→(0,1)
x(x+ 2) + y(y+ 1)
x2+ (y+ 1)2+
+ilim
(x,y)→(0,1)
xy −(x+ 2)(y+ 1)
x2+ (y+ 1)2=
=1·2
22+i−2·2
4=1
2−i.
Al e na i ely we can use he con inui y o he unc ion (z) := z+2
z+ia he poin i:
lim
z→i
z+ 2
z+i=2 + i
2i=1
2−i.
g)
lim
z→0
Re z
1 + |z|= lim
(x,y)→(0,0)
x
1 + √︁x2+y2=0
1= 0.
29
Exe cise 20.
D aw he se ⟨φ⟩:= {φ( ): ∈Dφ}, i
a) φ( ) := 1 −i , Dφ =⟨0,2⟩;
b) φ( ) := −i 2, Dφ =⟨−1,2⟩;
c) φ( ) := 1 + e−i , Dφ =⟨0,2π⟩;
d) φ( ) := e2i −1, Dφ =⟨0,2π⟩;
e) φ( ) := {︄eiπ , ∈ ⟨0,1),
−2, ∈ ⟨1,3⟩;
) φ( ) := {︄ei , ∈ ⟨−π
2, π),
3
π−4, ∈ ⟨π, 2π⟩.
Solu ion:
a) φ( ) := 1 −i , Dφ =⟨0,2⟩.
1
i
−2i⟨φ⟩
b) φ( ) := −i 2, Dφ =⟨−1,2⟩.
−4i
−12
⟨φ⟩
−i
−i
c) φ( ) := 1 + e−i , Dφ =⟨0,2π⟩.
1⟨φ⟩
d) φ( ) := e2i −1, Dφ =⟨0,2π⟩.
−1
⟨φ⟩
“2×oběhnu ᔓ2×a ound”
30
e) φ( ) := {︄eiπ , ∈ ⟨0,1),
−2, ∈ ⟨1,3⟩.
−1 1
i
i
⟨φ⟩
) φ( ) := {︄ei , ∈ ⟨−π
2, π),
3
π−4, ∈ ⟨π, 2π⟩.
−i
−1 2
i
i
1
1
⟨φ⟩
Exe cise 21.
Find a pa ame iza ion o he se Ω(i.e. ind a cu e φsuch ha ⟨φ⟩= Ω), i
a) Ω = {z∈C:|z−2+3i|= 2};
b) Ωis a line segmen wi h he endpoin s a, b ∈C,a=
/b;
c) Ω = {z∈C: Re z= 2 Im z};
d) Ω = {z∈C: Re (︁1
z)︁= 2}.
Solu ion:
a) Ω = {z∈C:|z−2+3i|= 2};φ( ) := 2 −3i+ 2ei , ∈ ⟨0,2π⟩.
−3i
2
Ω = ⟨φ⟩
31
b) Ωis a line segmen wi h he endpoin s a, b ∈C,a=
/b;φ( ) := a+ (b−a) , ∈ ⟨0,1⟩.
Ω = ⟨φ⟩
a
b
c) Ω = {z∈C: Re z= 2 Im z};φ( ) := +
2i, ∈R.
i
2
Ω = ⟨φ⟩
d)
Ω = {︃z∈C: Re (︃1
z)︃= 2}︃=
={︃x+iy : Re (︃1
x+iy)︃=x
x2+y2= 2}︃=
={︂x+iy ∈C
∖
{0}: 2 (︂x2−x
2+y2)︂= 0}︂=
={︃x+iy ∈C
∖
{0}: 2 (︃(x−1
4)2+y2−1
16)︃= 0}︃=
={︄x+iy ∈C
∖
{0}:(︃x−1
4)︃2
+y2=1
16}︄;
φ( ) := 1
4+1
4ei , ∈(−π, π).
1
4
Ω = ⟨φ⟩
1
2
1
2
32
Exe cise 22.
D aw he se Ω, and decide i Ωis a domain and i i is an open se , whe e
a) Ω = {z∈C:|z−i|<1∨ |z+i|<1};
b) Ω = {z∈C:|z−1|<1∧ |z−2|<2};
c) Ω = {z∈C:|z−1|<|z+ 1|};
d) Ω = {z∈C:|z+ 1|>2|z|};
e) Ω = {z∈C: 1 <|z|<2};
) Ω = {︁z∈C:|z|<1∧a g z∈(−π, π⟩
∖
{0}}︁;
g) Ω = {z∈C:|2z|<|1 + z2|}.
Solu ion:
a) Ω = {z∈C:|z−i|<1∨ |z+i|<1}.
Ω
Ω
i
i
−i
−i
Ωis open, bu no connec ed, and he e o e Ωis no a domain.
b) Ω = {z∈C:|z−1|<1∧ |z−2|<2}.
Ω
Ω
1
14
4
2
2
Ωis open and connec ed se , and he e o e Ωis a domain.
c) Ω = {z∈C:|z−1|<|z+ 1|}.
Ω
Ω
−11
1
Ωis open and connec ed se , and he e o e Ωis a domain.
33

d)
Ω = {z∈C:|z+ 1|>2|z|} =
={︁x+iy : (x+ 1)2+y2>4(x2+y2)}︁=
={︁x+iy : 3x2+ 3y2−2x−1<0}︁=
={︃x+iy :x2+y2−2
3x−1
3<0}︃=
={︄x+iy :(︃x−1
3)︃2
+y2<4
9}︄.
Ω
Ω
1
34
−1
3
−1
3
1
31
Ωis open and connec ed se , and he e o e Ωis a domain.
e) Ω = {z∈C: 1 <|z|<2}.
Ω
Ω
1 2
i
2i
i
2i
21
Ωis open and connec ed se , and he e o e Ωis a domain.
) Ω = {︁z∈C:|z|<1∧a g z∈(−π, π⟩
∖
{0}}︁.
Ω
Ω
1
1
Ωis open and connec ed se , and he e o e Ωis a domain.
34
g)
Ω = {z∈C:|2z|<|1 + z2|} =
={︁x+iy : 4(x2+y2)<(1 + x2−y2)2+ 4x2y2}︁=
={︁x+iy : 4x2+ 4y2<1 + x4+y4+ 2x2−2y2−2x2y2+ 4x2y2}︁=
={︁x+iy : 0 <1 + x4+y4−2x2−6y2+ 2x2y2}︁=
={︁x+iy : (x2+y2−1)2−4y2>0}︁=
={︁x+iy : (x2+y2−1+2y)(x2+y2−1−2y)>0}︁=
={︁x+iy : [x2+ (y+ 1)2−2][x2+ (y−1)2−2] >0}︁.
i
−i1
Ωis open, bu no connec ed se , and he e o e Ωis no a domain.
Exe cise 23.
Find all o he poin s whe e he unc ion has a de i a i e and he poin s whe e i is
holomo phic, i
a) (z) := Re z;
b) (z) := |z2|;
c) (z) := zez;
d) (z) := z|z|;
e) (z) := Re z
z;
) (z) := z2z;
g) (z) := z2+ 2z−1.
Solu ion:
a)
(x+iy) = x
⏞⏟⏟⏞
=:u(x,y)
+ 0
⏞⏟⏟⏞
=: (x,y)·i.
Fo e e y (x, y)∈R2we ha e ha
∂u
∂x(x, y) = 1 =
/0 = ∂
∂y(x, y),
and because o ha i ollows ha he unc ion does no ha e a de i a i e anywhe e
and he unc ion is no holomo phic a any poin .
35
b) (x+iy) = |(x+iy)2|= (|x+iy|)2=x2+y2. So =u+i , whe e u(x, y) := x2+y2
and (x, y) := 0.
∂u
∂x(x, y) = 2x=∂
∂y(x, y) = 0
∂u
∂y (x, y) = 2y=−∂
∂x(x, y)=0
⎫
⎪
⎪
⎬
⎪
⎪
⎭⇔(x, y) = (0,0),
and a he same ime he unc ions uand a e di e en iable in R2, and he e o e
has a de i a i e (only) in he poin 0an i is no holomo phic anywhe e.
c)
(x+iy) = (x+iy)ex(cos y+isin y) =
=xexcos y−yexsin y
⏞ ⏟⏟ ⏞
=:u(x,y)
+i(xexsin y+yexcos y)
⏞ ⏟⏟ ⏞
=: (x,y)
.
Func ions uand a e di e en iable in R2,
∂u
∂x(x, y) = excos y+xexcos y−yexsin y,
∂
∂y(x, y) = xexcos y+excos y−yexsin y,
and
∂u
∂y (x, y) = −xexsin y−exsin y−yexcos y,
−∂
∂x(x, y) = −(exsin y+xexsin y+yexcos y).
So ∂u
∂x =∂
∂y and ∂u
∂y =−∂
∂x in R2, and he e o e is holomo phic e e ywhe e in Cand
′(z)exis s a e e y z∈C.
(︃ ′(z) = ′(x+iy) = (︃∂u
∂x +i∂
∂x)︃(x, y) = ··· =ez+zez.)︃
d)
(x+iy)=(x−iy)√︁x2+y2=x√︁x2+y2
⏞ ⏟⏟ ⏞
=:u(x,y)
+i(−y√︁x2+y2)
⏞ ⏟⏟ ⏞
=: (x,y)
.
F om his, i ollows ha o e e y (x, y)∈R2
∖
{(0,0)}we ha e ha
∂u
∂x(x, y) = √︁x2+y2+x2
√︁x2+y2>0,
∂
∂y(x, y) = −√︁x2+y2−y2
√︁x2+y2<0,
and he e o e: i z=
/0, hen ′(z)does no exis .
36
I emains o p o e o disp o e he exis ence o he de i a i e a he poin 0:
′(0) = lim
z→0
(z)− (0)
z−0= lim
z→0
z·|z|
z=
= lim
z→0|z|(cos(a g z)−isin(a g z)) ·|z|
|z|(cos (a g z) + isin (a g z)) =
= lim
z→0[|z|·(cos (−2 a g z) + isin (−2 a g z))] = 0,
because ∀z=
/0: |cos (−2 a g z) + isin (−2 a g z)|= 1.
Summa y: he unc ion has a de i a i e only a he poin 0, and he e o e is no
holomo phic a any poin .
e)
(x+iy) = x
x+iy =x(x−iy)
x2+y2=x2
x2+y2
⏞ ⏟⏟ ⏞
=:u(x,y)
+i(︃−xy
x2+y2)︃
⏞ ⏟⏟ ⏞
=: (x,y)
.
Fo e e y (x, y)∈R2
∖
{(0,0)}we ha e ha
∂u
∂x(x, y) = 2x(x2+y2)−x22x
(x2+y2)2=2xy2
(x2+y2)2,
∂
∂y(x, y) = −x(x2+y2) + xy2y
(x2+y2)2=−x3+xy2
(x2+y2)2,
∂u
∂y (x, y) = −x22y
(x2+y2)2,
∂
∂x(x, y) = −y(x2+y2) + xy2x
(x2+y2)2=x2y−y3
(x2+y2)2,
and he e o e he de i a i e can exis only in he poin s x+iy whe e
(︃2xy2
(x2+y2)2=x(−x2+y2)
(x2+y2)2)︃∧(︃2x2y
(x2+y2)2=y(x2−y2)
(x2+y2)2)︃,
ha is (︃xy2
(x2+y2)2=−x3
(x2+y2)2)︃∧(︃x2y
(x2+y2)2=−y3
(x2+y2)2)︃.
I is easy o obse e ha his sys em o equa ions has no solu ion.
Summa y: he unc ion does no ha e a de i a i e a any poin , and he e o e i is no
holomo phic a any poin .
) (x+iy) = (x2−y2+ 2ixy)(x−iy) =
=x3−xy2+ 2xy2+i(−x2y+y3+ 2x2y) =
=x3+xy2
⏞ ⏟⏟ ⏞
=:u(x,y)
+i(y3+x2y)
⏞ ⏟⏟ ⏞
=: (x,y)
.
37
A he same ime he unc ion uis con inuous on R2
∖
{(0,0)}(a e e y poin R2
∖
{(0,0)}
i mus be di e en iable), and he e o e
lim
x→0−u(x, 1) = u(0,1) = lim
x→0+ u(x, 1).
=
=
π+c2−π+c1
F om his, i ollows ha
2π=c1−c2.
Analogously
lim
x→0−u(x, −1) = u(0,−1) = lim
x→0+ u(x, −1),
=
=
−π+c2π+c1
and he e o e
2π=c2−c1.
This leads us o he ac ha
2π=c1−c2=−(c2−c1) = −2π,
which is a con adic ion. The sough unc ion udoes no exis .
44

Exe cise 30.
Find he o a ional angle and ex ensibili y coe icien o he unc ion a he poin z0, whe e
a) (z) := ez,z0=−1−π
2i;
b) (z) := z3,z0=−3+4i;
c) (z) := z+i
z−i,z0= 2i.
Solu ion:
a)
| ′(z0)|=|ez0|=|e−1−iπ
2|=1
e,
which is he ex ensibili y coe icien o he unc ion a he poin z0(and 1
e<1implies
ha i is a con ac ion).
a g ′(z0) = a g (︃1
e(︂cos π
2−isin π
2)︂)︃= a g (︃−i
e)︃=−π
2,
which is he o a ional angle o he unc ion a he poin z0.
b) z0= 5 (︁cos (︁π
2+ a c an 3
4)︁+isin (︁π
2+ a c an 3
4)︁)︁, and he e o e
′(z0) = 3z2
0= 3 ·25 (︃cos (︃π+ 2 a c an 3
4)︃+isin (︃π+ 2 a c an 3
4)︃)︃.
F om his we ge
| ′(z0)|= 75 . . . ex ensibili y coe icien o he unc ion a z0
(75 >1, he e o e i is a dila a ion),
a g ′(z0) = −π+ 2 a c an 3
4. . . o a ional angle o he unc ion a he poin z0.
c)
′(z) = z−i−(z+i)
(z−i)2=−2i
(z−i)2,
′(z0) = −2i
i2= 2i,
and he e o e
1<| ′(z0)|= 2 . . . ex ensibili y coe icien o a z0(dila a ion),
a g ′(z0) = π
2. . . o a ional angle a z0.
45
Exe cise 31.
De e mine a which poin s o he complex plane is he gi en mapping a con ac ion:
a) (z) := 2
z;
b) (z) := ln(z+ 4).
Solu ion:
a) ′(z) = −2
z2. The e o e o z∈C:
0<| ′(z)|<1⇔−2
z2<1⇔2<|z|2⇔√2<|z|.
The mapping is a con ac ion in e e y poin o he se {︁z∈C:|z|>√2}︁.
√2
√2
b) ′(z)exis s in C
∖
{x+iy :y= 0 ∧x≤ −4}=: Ω. Fo e e y z∈Ωwe ha e ha
| ′(z)|=
1
z+ 4,
0<1
|z+ 4|<1⇔1<|z+ 4|.
The mapping is a con ac ion a e e y poin o he se
{z∈C:|z+ 4|>1}
∖
{x+iy :y= 0 ∧x≤ −4}.
−3−3
−4−3
46
Exe cise 32.
D aw he se s Ωand (Ω) = { (z): z∈Ω}, whe e2
a) Ω = U(1,2), (z) := 1 −2iz;
b) Ω = {z∈C: Re z < 1}, (z) := (1 + i)z+ 1;
c) Ω = U(1,2), (z) := 1
z;
d) Ω = U(1,2), (z) := 2iz
z+3 ;
e) Ω = U(1,2), (z) := z−1
2z−6;
) Ω = {z∈C: Re z < 1}, (z) := 1
z;
g) Ω = {z∈C: Re z < 1}, (z) := z
z−1+i;
h) Ω = {z∈C: Re z < 1}, (z) := z
z−2;
i) Ω = {z∈C: Re z < 0∧Im z < 0}, (z) := 1
z;
j) Ω = {z∈C: Re z > 0∧Im z > 0}, (z) := z−1
z+1;
k) Ω = {z∈C:−1<Re z < 0∧Im z < 0}, (z) := z−i
z+i;
l) Ω = {z∈C:|z|<1∧Re z < 0∧Im z > 0}, (z) := z
z−i.
Solu ion:
a) Ω = U(1,2), (z) := 1 −2iz, (Ω) = U(1 −2i, 4).
Ω
Ω
1
1
−1
−13
3
i
−i
−3i
i
−i
−3i
2i
−2i
−6i
2i
−2i
−6i
−2i
1
1
2i
−2i
−6i
(Ω)
(Ω)
−iz
2z
1 + z
2A hin o some o he ollowing exe cises. Realize (and p o e) ha :
is con o mal in he se Ω⊂C∞,
A, B ⊂Ω}︄⇒ (A∩B) = (A)∩ (B).
47
b) Ω = {z∈C: Re z < 1}, (z) := (1 + i)z+ 1,
(1) = 1 + i+ 1 = 2 + i,
(1 + i) = 2i+ 1,
(0) = 1,
and he e o e ( hink i h ough!) (Ω) = {z∈C: Re z+ Im z < 3}.
1−11−1
Ω
Ω
1 2
i
2i
1 2
i
2i
(Ω)
(Ω)
c) Ω = U(1,2), (z) := 1
z,
(0) = ∞,
(−1) = −1,
(3) = 1
3,
(1 + 2i) = 1
1+2i=1−2i
5,
and he e o e (Ω) = C∞
∖
U(︃−1
3,2
3)︃.
Ω
Ω
1 3−1
−1 1 3
2i
1
3
−11
3
1
5
−1
3
−1
−2
5i
(Ω)
(Ω)
d) Ω = U(1,2), (z) := 2iz
z+3,
(−3) = ∞,
(−1) = −2i
2=−i,
(3) = i,
(1 + 2i) = 2i(1 + 2i)
4+2i=−4−2i
4+2i=−3
5+4
5i,
and he e o e (Ω) = U(0,1).
48
Ω
Ω
1 3−1
−1 1 3
2i
(Ω)
(Ω)
1
1
e) Ω = U(1,2), (z) := z−1
2z−6,
(0) = 1
6,
(3) = ∞,
(−1) = −2
−8=1
4,
(1 + 2i) = 2i
2+4i−6=1
4−1
4i,
and he e o e (Ω) = {︃z∈C: Re z < 1
4}︃.
Ω
Ω
1 3−1
−1 1 3
2i
1
4
−1
4i
1
4
−1
4i
(Ω)
(Ω)
) Ω = {z∈C: Re z < 1}, (z) := 1
z,
(1) = 1,
(0) = ∞,
(1 + i) = 1
1 + i=1−i
2,
(1 −i) = 1
1−i=1 + i
2,
(∞)=0,
and he e o e (Ω) = C∞
∖
U(︃1
2,1
2)︃.
49

1−1
Ω
Ω
1−11
1
2
1
21
(Ω)
(Ω)
g) Ω = {z∈C: Re z < 1}, (z) := z
z−1+i,
(0) = 0,
(1) = 1
i=−i,
(1 −i) = ∞,
(1 + i) = 1 + i
2i=1
2−1
2i,
and he e o e (Ω) = {z∈C: Im z > Re z−1}.
1−1
Ω
Ω
1−1
1
−i
(Ω)
(Ω)
h) Ω = {z∈C: Re z < 1}, (z) := z
z−2,
(2) = ∞,
(1) = −1,
(1 + i) = −2i
2=−i,
(∞) = 1,
and he e o e (Ω) = U(0,1).
1−1
Ω
Ω
1−1
(Ω)
(Ω)
1
1
50
i) Ω = {z∈C: Re z < 0∧Im z < 0}, (z) := 1
z,
(0) = ∞,
(−1) = −1,
(1) = 1,
(i) = −i,
(−i) = i,
and he e o eže Ω = Ω1∩Ω2, whe e Ω1:= {z∈C: Re z < 0},Ω2:= {z∈C: Im z < 0},
je
(Ω) = (Ω1)∩ (Ω2) = {z∈C: Re z < 0∧Im z > 0}.
Ω
Ω
(Ω)
(Ω)
j) Ω = {z∈C: Re z > 0∧Im z > 0}, (z) := z−1
z+1,
(0) = −1,
(1) = 0,
(i) = i−1
i+ 1 =i,
(−i) = −i,
(−1) = ∞,
51
1
1
and he e o e
(Ω) = {z∈C:|z|<1∧Im z > 0}.
Ω
Ω
(Ω)
(Ω)
11
k) Ω = {z∈C:−1<Re z < 0∧Im z < 0}, (z) := z−i
z+i,
(0) = −1,
(i)=0,
(−i) = ∞,
(1) = 1−i
1 + i=(1 −i)2
2=−i,
(−1) = −1−i
−1 + i=(−1−i)2
2=i,
(−1 + i) = −1
−1+2i=1+2i
5,
(−1−i) = 1 + 2i,
−1 1
−1 1
1
3
i
11
i
52
11
and he e o e
(Ω) = {z∈C: Im z > 0∧ |z−(1 + i)|>1∧ |z|>1}.
−1
Ω
Ω
(Ω)
(Ω)
1
i
1
i
l) Ω = {z∈C:|z|<1∧Re z < 0∧Im z > 0}, (z) := z
z−i,
(0) = 0,
(i) = ∞,
(−1) = 1−i
2,
(1) = 1 + i
2,
(−i) = 1
2,
1
1
1
2
−1
2
1
2
−1
2
53
and he e o e
1
z↦→ z2
z↦→ e−iπ
2z
1
Summa y: one o he unc ions wi h he equi ed p ope ies is he unc ion de ined on Ω
(z) := (︃e−iπ
2·z+i
z−i)︃2
=(︃−iz+i
z−i)︃2
=−(︃z+i
z−i)︃2
.
Exe cise 37.
Le
Ω = {z∈C: Re z > 0∧Im z < 0}.
Find he linea ac ional unc ion such ha
(Ω) = {z∈C:|z|<1∧Re z < 0}.
Solu ion:
Ω
Ω
(Ω)
(Ω)
−1
We i s ly ind he linea ac ional unc ion ∗such ha
∗(0) = i,
∗(∞) = −i,
∗(−1) = ∞,
60

ha is he unc ion
∗(z) = ⎧
⎪
⎨
⎪
⎩
−iz +i
z+ 1 , z ∈C,
−i, z =∞.
Then clea ly ei he
∗(Ω)
∗(Ω)
−1
( hen we would de ine := ∗), o
∗(Ω)
∗(Ω)
1
(which would lead us o he de ini ion := − ∗).
Because ∗(i) = 1+i
i+1 = 1 ( he i s possibili y is ealized), we choose
(z) := ∗(z) = ⎧
⎪
⎨
⎪
⎩
−iz +i
z+ 1 , z ∈C,
−i, z =∞.
.
Exe cise 38.
Find he con o mal mapping which maps he domain
Ω = {z∈C: Re z > Im z > 0}
on o U(0,1).
Solu ion:
1
1
61
Le us i s conside he mapping z↦→ z4.
z→z4
Then we ind he linea ac ional unc ion ∗such ha
∗(−1) = −1,
∗(0) = i,
∗(1) = 1,
ha is
∗(z) = ⎧
⎪
⎪
⎨
⎪
⎪
⎩
z+i
iz + 1, z ∈C,
1
i=−i, z =∞.
Clea ly ei he
∗
1
1
(in which case we would ( o z∈Ω) de ine (z) := ∗(z4)), o 3
∗
11
( hen we would de ine (z) := 1
∗(z4)in Ω).
Because ∗(i) = ∞, he second case a ose. We choose ( o z∈Ω)
(z) := 1
∗(z4)=iz4+ 1
z4+i.
3Fo he igh -hand-side image we need o imagine ha ∞= ∗(i).
62
Exe cise 39.
Find he images o he lines pa allel o he eal and imagina y axes by he mapping
(z) := 1
z(conside he lines oge he wi h he poin ∞).
Solu ion:
Fo 0< c ∈Rwe ha e ha
(0) = ∞, (∞) = 0,
(i) = −i, (c) = 1
c,
(−i) = i, (−c) = −1
c,
(1) = 1, (ci) = −1
ci,
(−1) = −1, (−ci) = 1
ci,
and he e o e also
{z∈C: Re z= 0}∪{∞} → {z∈C: Re z= 0}∪{∞},
{z∈C: Im z= 0}∪{∞} → {z∈C: Im z= 0}∪{∞},
{z∈C: Re z=c}∪{∞} → {︃z∈C:z−1
2c=1
2c}︃,
{z∈C: Im z=c}∪{∞} → {︃z∈C:z+1
2ci=1
2c}︃,
{z∈C: Re z=−c}∪{∞} → {︃z∈C:z+1
2c=1
2c}︃,
{z∈C: Im z=−c}∪{∞} → {︃z∈C:z−1
2ci=1
2c}︃.
0
−cc
ci
−ci
1
ci
−1
ci
−1
c01
c
63
Exe cise 40.
Find he images o he se s
Mα={z∈C: a g z=α}and N ={z∈C:|z|= },
whe e α∈(−π, π⟩and ∈R+, by he mapping (z) := ln z.
Solu ion:
ln z= ln |z|+ia g z,
and he e o e
(Mα) = {z∈C: Im z=α},
Mαα
πi
αi
−πi
(Mα)
(N ) = {ln +ik :k∈(−π, π⟩}.
N
πi
−πi
(N )
ln
1
πi
−πi
64
Exe cise 41.
Compu e ∫︂γ|z|dz,
whe e
γ( ) := ⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
3ei , ∈ ⟨0,π
2⟩,
i(︁3 + π
2− )︁, ∈ ⟨π
2,π
2+ 3⟩,
−π
2−3, ∈ ⟨π
2+ 3,π
2+ 6⟩.
Solu ion:
Le us choose
γ1( ) := 3ei , ∈ ⟨0,π
2⟩,
γ2( ) := i, i ∈ ⟨0,3⟩,
γ3( ) := , i ∈ ⟨0,3⟩.
Then
3
3i
⟨γ⟩
⟨γ1⟩
⟨γ2⟩
⟨γ3⟩
and
γ′
1( )=3iei ,
γ′
2( ) = i,
γ′
3( )=1,
and he e o e ∫︂γ|z|dz=∫︂γ1|z|dz−∫︂γ2|z|dz+∫︂γ3|z|dz=
=∫︂π
2
0
3·3iei d −∫︂3
0
i d +∫︂3
0
d =
= 9i∫︂π
2
0
(cos +isin ) d + (1 −i)∫︂3
0
d =
= 9i[sin ]
π
2
0+ 9[cos ]
π
2
0+ (1 −i)[︃ 2
2]︃3
0
=
= 9i−9 + 9
2(1 −i) =
=−9
2+9
2i.
65

Exe cise 42.
Compu e
∫︂γ
z3dz,
whe e
γ( ) := ⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
ei , ∈ ⟨−π
2, π⟩,
3
π −4, ∈ ⟨π, 2π⟩,
−2+i
π + 6 + 2i, ∈ ⟨2π, 3π⟩.
Solu ion:
2−1
−i
⟨γ⟩
I is enough o apply Cauchy’s heo em.
∫︂γ
z3dz= 0,
because (z) := z3is a holomo phic unc ion on he simply connec ed domain Cand γis
piecewise smoo h closed cu e in C.
Exe cise 43.
Compu e
∫︂γ|z|zdz,
whe e γis a simple, closed, piecewise smoo h and posi i ely o ien ed cu e such ha ⟨γ⟩is
he bounda y o he se
{z∈C:|z|<2∧Im z > 0}.
66
Solu ion:
2−2
⟨γ⟩
Le us de ine he cu es
γ1( ) := 2ei , ∈ ⟨0, π⟩,
γ2( ) := , ∈ ⟨−2,2⟩.
Then
γ′
1( )=2iei ,
γ′
2( )=1,
and he e o e
∫︂γ|z|zdz=∫︂γ1|z|zdz+∫︂γ2|z|zdz=
=∫︂π
0
2·2e−i ·2iei d +∫︂2
−2| | d
⏞ ⏟⏟ ⏞
=0
=
= 8i∫︂π
0
1 d = 8πi.
Exe cise 44.
Using he Cauchy’s in eg al o mulas calcula e he gi en in eg als4
a) ∫︂k
z2+i
zdz, whe e k={z∈C:|z−2i|= 1};
b) ∫︂k
sin z
z+idz, whe e k={z∈C:|z+i|= 1};
4Con en ion. By he symbol ∫︁k (z) dz, whe e k⊂C, we mean ∫︁γ (z) dz, whe e γis a simple, closed,
piecewise smoo h and posi i ely o ien ed cu e such ha ⟨γ⟩=k.
67
c) ∫︂k
sin z
z2−7z+ 10 dz, whe e k={z∈C:|z|= 3};
d) ∫︂k
sin z
(z−2i)3dz, whe e k={z∈C:|z|= 3};
e) ∫︂k
cos z
z2−π2dz, whe e k={z∈C:|z|= 4};
) ∫︂k
e1
z
(z2−4)2dz, whe e k={z∈C:|z−2|= 1};
g) ∫︂γ
ezcos(πz)
z2+ 2zdz, whe e γ( ) := 3
2ei , ∈ ⟨0,2π⟩;
h) ∫︂γ
dz
(z2−1)3,whe e γ( ) := −2 + e−4πi
2, ∈ ⟨0,4⟩;
i) ∫︂γ
dz
(1 −z)(z+ 2)(z−i)2,
whe e γis a simple, closed, piecewise smoo h and posi i ely o ien ed cu e such ha
−2∈in γ, i ∈in γ, 1∈ex γ.
Solu ion:
a)
2i
1−1
⟨γ⟩
The unc ion “ z2+i
z” is holomo phic on a simply connec ed domain
Ω := {z∈C: Im z > 0}
and k=⟨γ⟩ ⊂ Ω, and he e o e i ollows om he Cauchy’s heo em ha
∫︂k
z2+i
zdz=∫︂γ
z2+i
z= 0.
68
Bu we we e supposed o use he Cauchy’s in eg al o mulas. Which we can do o
example as
∫︂k
z2+i
zdz=∫︂γ
z2+i
z(z−2i)
z−2idz= 2πi [︃z2+i
z(z−2i)]︃z=2i
= 0.
b)
−i
1
−1
⟨γ⟩
∫︂k
sin z
z+idz=∫︂γ
sin z
z−(−i)dz=
= 2πi [sin z]z=−i=
= 2πi ei(−i)−e−i(−i)
2i=
=π(e−e−1) = 2πsinh 1.
c)
2 5
⟨γ⟩
3
z2−7z+ 10 = (z−5)(z−2), and he e o e
∫︂k
sin z
z2−7z+ 10 dz=∫︂γ
sin z
z−5
z−2dz=
= 2πi [︃sin z
z−5]︃z=2
=
= 2πi sin 2
−3=−(︃2
3πsin 2)︃i.
69
Summa y: he gi en se ies con e ges (absolu ely) o e e y
z∈{︃z∈C:
z+ 1
z−1≤1}︃={z∈C: Re z≤0}.
b) Because o e e y z∈C
∖
{0}
zn+1
(n+1)!
zn
n!=|z|
n+ 1 →0<1,
he se ies ∑︁∞
n=1
zn
n!con e ges absolu ely in C. Because o e e y z∈C
∖
{0}
n
√︄
n2
zn=(n
√n)2
|z|→1
|z|,
he se ies ∑︁∞
n=1
n2
zncon e ges absolu ely o |z|>1and di e ges o |z|<1. I |z|= 1 we
ha e 
n2
zn=n2→ ∞ =
/0,
and he e o e he se ies ∑︁∞
n=1
n2
zndi e ges.
Le us now de ine
sn(z) :=
n
∑︂
k=1 (︃zk
k!+k2
zk)︃,
s∗
n(z) :=
n
∑︂
k=1
zk
k!,
s∗∗
n(z) :=
n
∑︂
k=1
k2
zk.
Then o e e y z∈Cand n∈Nwe ha e ha
sn(z) = s∗
n(z) + s∗∗
n(z),
s∗∗
n=sn(z)−s∗
n(z),
and u he mo e (we al eady know ha ) lim s∗
n(z)∈C o e e y z∈C, and he e o e o
e e y z∈Cwe ha e ha
lim sn(z)∈C⇔lim s∗∗
n(z)∈C.
76

Summa y: he gi en se ies co ne ges (absolu ely) on he se {z∈C:|z|>1}.
11
Exe cise 48.
Find he adius o con e gence Ro he gi en powe se ies
a) ∞
∑︁
n=1
zn
n2011 ;
b) ∞
∑︁
n=1
nn(z−1)n;
c) ∞
∑︁
n=1
3n(z−1)n
√(3n−2)2n;
d) ∞
∑︁
n=0
(z+1+i)n
3n(n−i);
e) ∞
∑︁
n=1
nn
n!zn;
) ∞
∑︁
n=0 (︁cos(in))︁zn;
g) ∞
∑︁
n=0
(n2−n−2)zn;
h) ∞
∑︁
n=0
zn
(n+8)!.
Solu ion:
a)
n
√︃1
n2011 =1
(n
√n)2011 →1,
and he e o e
R= 1.
b)
n
√nn=n→ ∞,
and he e o e
R=1
∞= 0.
c) Because
n
√︄3n
√︁(3n−2)2n=3
√2
1
√︁n
√3n−2→3
√2
(i is enough o ealize ha o n≥3we ha e ha 1≤n
√3n−2≤n
√n·n
√n→1), and
he e o e
R=√2
3.
77
d) 
1
3n+1(n+1−i)
1
3n(n−i)=1
3
n−i
n+ 1 −i=1
3
1−i
n
1 + 1−i
n→1
3,
and he e o e
R= 3.
e)
(n+1)n+1
(n+1)!
nn
n!
=1
n+ 1
(n+ 1)n(n+ 1)
nn=(︃1 + 1
n)︃n
→e,
and he e o e
R=1
e.
)

cos(i(n+ 1))
cos(in)=
ei(i(n+1)) +e−i(i(n+1))
eiin +e−iin =
=e−(n+1) +en+1
e−n+en·
1
en
1
en
=
=
1
enen+1 +e
1
enen+ 1 →e,
and he e o e
R=1
e.
g) 
(n+ 1)2−(n+ 1) −2
n2−n−2→1,
and he e o e
R= 1.
h)
1
(n+9)!
1
(n+8)!
=1
n+ 9 →0,
and he e o e
R=∞.
Exe cise 49.
Find he sum o he powe se ies in he disk o con e gence
a) ∞
∑︁
n=1
nzn; b) ∞
∑︁
n=1
zn
n;
78
c) ∞
∑︁
n=0
z2n+1
2n+1 ;
d) ∞
∑︁
n=1
(−1)n+1 zn
n+1;
e) ∞
∑︁
n=0
(n2−n−2)zn.
Solu ion:
a) n
√n→1, and he e o e he adius o con e gence o a gi en se ies is 1.
Fo e e y z∈C,|z|<1we ha e ha
∞
∑︂
n=1
nzn=z∞
∑︂
n=1
nzn−1=z(︄∞
∑︂
n=1
nzn
n)︄′
=
=z(︄∞
∑︂
n=1
zn)︄′
=z(︃z
1−z)︃′=
=z1−z+z
(1 −z)2=z
(1 −z)2.
b) n
√︂1
n→1, he e o e he adius o con e gence is 1.
Le us de ine he unc ion (z) := ∑︁∞
n=1
zn
n. Then o e e y z∈C,|z|<1we ha e ha
′(z) = ∞
∑︂
n=1
zn−1=1
1−z.
F om his, because
|z|<1⇒1−z∈Ω := {w∈C:|w−1|<1},
ln′w=1
w Ω,
1
1−1
Ω
he e is a c∈Csuch ha o each z∈C,|z|<1we ha e ha
(z) = −ln(1 −z) + c.
Fu he mo e (0) = −ln 1 + c= 0, and he e o e c= 0.
Summa y: o each z∈C,|z|<1we ha e ha
∞
∑︂
n=1
zn
n= (z) = −ln(1 −z).
79
c) 
1
2n+3
1
2n+1 →1,
and he e o e he adius o con e gence is 1.
Le us de ine (z) := ∑︁∞
n=0
z2n+1
2n+1 . Then o e e y z∈C,|z|<1we ha e ha
′(z) = ∞
∑︂
n=0
z2n=1
1−z2=
=1
2−1
z−1+1
2
1
z+ 1.
F om ha i ollows ha he e is c∈C, such ha o each z∈C,|z|<1we ha e ha
(z) = −1
2ln(1 −z) + 1
2ln(1 + z) + c.
And because 0 = (0) = c o each z∈C,|z|<1,we ha e ha
∞
∑︂
n=0
z2n+1
2n+ 1 =−1
2ln(1 −z) + 1
2ln(1 + z).
d) n
√︂1
n+1 →1, and he e o e he adius o con e gence is 1.
Le (z) := ∑︁∞
n=1(−1)n+1 zn+1
n+1 . Then o z∈C,|z|<1,we ha e ha
′(z) = ∞
∑︂
n=1
(−1)n+1zn=−∞
∑︂
n=1
(−z)n=
=z+ 1 −1
1 + z.
F om ha , i ollows ha he e is a c∈Csuch ha
(z) = z−ln(1 + z) + c,
and because 0 = (0) = cwe ha e
∞
∑︂
n=1
(−1)n+1 zn
n+ 1 ={︄1
z (z) = 1 −ln(1+z)
z,0<|z|<1,
0, z = 0.
e) 
(n+ 1)2−(n+ 1) −2
n2−n−2→1,
and he e o e he adius o con e gence is 1.
Fo e e y z∈C,|z|<1,we ha e ha
∞
∑︂
n=0
(n2−n−2)zn=∞
∑︂
n=0
n2zn−∞
∑︂
n=0
nzn−2∞
∑︂
n=0
zn
(i is enough o ealize ha each o he se ies is absolu ely con e gen ).
80
Fu he mo e (|z|<1):
•
∞
∑︂
n=0
zn=1
1−z,
•
∞
∑︂
n=0
nzn=∞
∑︂
n=1
nzn=z∞
∑︂
n=1
nzn−1=
=z(︄∞
∑︂
n=1
nzn
n)︄′
=z(︃z
1−z)︃′=
=z1−z+z
(1 −z)2=z
(1 −z)2,
•
∞
∑︂
n=0
n2zn=∞
∑︂
n=1
n2zn=z∞
∑︂
n=1
n2zn−1=
=z(︄∞
∑︂
n=1
n2zn
n)︄′
=z(︄∞
∑︂
n=1
nzn)︄′
=
=z(︃z
(1 −z)2)︃′=z(1 −z)2+z2 (1 −z)
(1 −z)4=
=zz+ 1
(1 −z)3,
and he e o e o e e y z∈C,|z|<1,we ha e ha
∞
∑︂
n=0
(n2−n−2)zn=z2+z−z(1 −z)−2(1 −z)2
(1 −z)3=2−4z
(z−1)3.
81

Exe cise 50.
Find he sum o he gi en se ies
a) ∞
∑︁
n=1
1
n2n; b) ∞
∑︁
n=1
(−1)n
n2n.
Solu ion:
Le us conside he unc ion
(z) := ∞
∑︂
n=1
zn
n2n.
Because 1
n
√n2n→1
2, he powe se ies in he de ini ion o he unc ion has he adius o
con e gence 2. The e o e o e e y z∈C,0<|z|<2,we ha e ha
′(z) = ∞
∑︂
n=1
zn−1
2n=1
z
∞
∑︂
n=1 (︂z
2)︂n=
=1
z
z
2
1−z
2
=1
2−z.
The e o e he e is a c∈C o which (z) = −ln(2−z)+c. And because (0) = 0 = −ln 2+c,
o e e y z∈C,|z|<2we ha e ha
(z) = −ln(2 −z) + ln 2.
a)
∞
∑︂
n=1
1
n2n= (1) = ln 2,
b)
∞
∑︂
n=1
(−1)n
n2n= (−1) = −ln 3 + ln 2 = ln 2
3.
Exe cise 51.
Find he Taylo se ies o he unc ion cen e ed a z0and ind i s adius o con e gence,
whe e
a) (z) := z+1
z2+4z−5, z0=−1;
b) (z) := z
z2+i, z0= 0;
c) (z) := ln 1+z
1−z, z0= 0;
d) (z) := e3z−2, z0= 1;
e) (z) := sin(3z2+ 2), z0= 0;
) (z) := 1
(z−1)3, z0= 3;
g) (z) := sin2z, z0= 0.
82
Solu ion:
a) (z) = 2
3
1
z+5 +1
3
1
z−1,
−5−1 1
and he e o e he adius o con e gence is 2 and o e e y z∈C,|z+ 1|<2,i holds,
ha
(z) = 2
3·1
4 + z+ 1 +1
3·1
−2 + z+ 1 =2
12 ·1
1 + z+1
4−1
6·1
1−z+1
2
=
=1
6
∞
∑︂
n=0
(−1)n(z+ 1)n
4n−1
6
∞
∑︂
n=0
(z+ 1)n
2n=
=∞
∑︂
n=0 (︃(−1)n
6·4n−1
6·2n)︃(z+ 1)n=
=∞
∑︂
n=1
(−1)n−2n
6·4n(z+ 1)n.
b) z2+i= 0 i and only i z=±(︂√2
2−√2
2i)︂,
11
and he e o e he sough Taylo se ies has he adius o con e gence 1.
Fo e e y z∈C,|z|<1,we ha e ha
(z) = z
i·1
1 + z2
i
=z
i
∞
∑︂
n=0
(−1)n(︃z2
i)︃n
=
=∞
∑︂
n=0
(−1)n
in+1 z2n+1 =
=∞
∑︂
n=0
in−1z2n+1.
83
c) Because clea ly
1
1
z↦→ 1+z
1−z
(0↦→ 0,1↦→ ∞,−1↦→ 0), he adius o con e gence is 1. Fo e e y z∈C,|z|<1we
ha e ha
′(z) = 1−z
1 + z·1−z+ (1 + z)
(1 −z)2=2
(1 + z)(1 −z)=
=2
1−z2=∞
∑︂
n=0
2z2n,
and he e o e he e is a c∈Csuch ha
(z) = ∞
∑︂
n=0
2z2n+1
2n+ 1 +c.
And because (0) = 0 = c, o e e y z∈C,|z|<1we ha e ha
(z) = ∞
∑︂
n=0
2z2n+1
2n+ 1.
d) Clea ly he adius o con e gence is ∞. We know ha o e e y z∈Cwe ha e ez=
∞
∑︁
n=0
zn
n!, and he e o e
(z) = e3z−2=e3(z−1)+1 =e e3(z−1) =
=∞
∑︂
n=0
e·3n
n!(z−1)n.
e) The adius o con e gence is ∞and o any z∈Cwe ha e ha
sin z=∞
∑︂
n=0
(−1)nz2n+1
(2n+ 1)!,
cos z=∞
∑︂
n=0
(−1)nz2n
(2n)!.
84
F om his i ollows ha o e e y z∈Cwe ha e ha
(z) = sin(3z2) cos 2 + cos(3z2) sin 2 =
=∞
∑︂
n=0
cos 2 ·(−1)n32n+1
(2n+ 1)!
⏞ ⏟⏟ ⏞
=:αn
z4n+2 +∞
∑︂
n=0
sin 2 ·(−1)n32n
(2n)!
⏞ ⏟⏟ ⏞
=:βn
z4n=
=∞
∑︂
n=0
anz2n,
whe e a2k:= βkand a2k+1 := αk o e e y k∈N∪{0}.
)
31
Clea ly he adius o con e gence is 2. Fo e e y z∈U(3,2) we ha e ha
1
z−1=1
2 + z−3=1
2·1
1 + z−3
2
=
=1
2
∞
∑︂
n=0 (︃−z−3
2)︃n
=∞
∑︂
n=0
(−1)n
2n+1 (z−3)n
and (︃1
z−1)︃′′ =(︃−1
(z−1)2)︃′= 2 1
(z−1)3.
F om his i easily ollows, ha o each z∈U(3,2) we ha e ha
(z) = 1
2(︃1
z−1)︃′′ =
=1
2(︄∞
∑︂
n=1
(−1)n
2n+1 n(z−3)n−1)︄′
=
=1
2
∞
∑︂
n=2
(−1)n
2n+1 n(n−1)(z−3)n−2=
=∞
∑︂
n=2
(−1)n
2n+2 n(n−1)(z−3)n−2=
=∞
∑︂
n=0
(−1)n
2n+4 (n+ 2)(n+ 1)(z−3)n.
85
α) I z∈C,|z|<1, we ha e
(z) = 1
z+ 1 +1
(z−1)2=1
2·1
1 + z
2
+(︃−1
z−1)︃′=
=1
2
∞
∑︂
n=0
(−1)n
2nzn+(︃1
1−z)︃′=
=∞
∑︂
n=0
(−1)n
2n+1 zn+(︄∞
∑︂
n=0
zn)︄′
=
=∞
∑︂
n=0
(−1)n
2n+1 zn+∞
∑︂
n=1
n(zn−1) =
=∞
∑︂
n=0 (︃(−1)n
2n+1 +n+ 1)︃zn.
β) Fo e e y z∈C,1<|z|<2, we ha e ha
(z) = ∞
∑︂
n=0
(−1)n
2n+1 zn+(︃−1
z
1
1−1
z)︃′=
=∞
∑︂
n=0
(−1)n
2n+1 zn−(︄∞
∑︂
n=0
z−n−1)︄′
=
=∞
∑︂
n=0
(−1)n
2n+1 zn+∞
∑︂
n=0
(n+ 1) 1
zn+2 =
=∞
∑︂
n=0
(−1)n
2n+1 zn+∞
∑︂
n=2
n−1
zn.
γ) Fo e e y z∈Csuch ha |z|>2we ha e ha
(z) = 1
z
1
1 + 2
z
+∞
∑︂
n=2
n−1
zn=
=∞
∑︂
n=0
(−1)n·2n
zn+1 +∞
∑︂
n=2
n−1
zn=
=1
z+∞
∑︂
n=2
(−2)n−1+n−1
zn.
92

b) Because is clea ly holomo phic on C
∖
{0}and |z0−0|=√2, we ha e p ecisely wo
“maximal anluli”:
1
1
i
i
α)P(1 + i, 0,√2),
β)P(1 + i, √2,∞).
α) Fo z∈C,|z−1−i|<√2, we ha e ha
(z) = z+ 1
z2=1
z+1
z2
and u he mo e
1
z=1
1 + i+z−1−i=1
1 + i·1
1 + z−1−i
1+i
=
=∞
∑︂
n=0
(−1)n
(1 + i)n+1 (z−1−i)n,
1
z2=−(︃1
z)︃′=∞
∑︂
n=1
(−1)n+1
(1 + i)n+1 n(z−1−i)n−1,
and he e o e
(z) = ∞
∑︂
n=0 (︃(−1)n
(1 + i)n+1 +(−1)n+2
(1 + i)n+2 (n+ 1))︃(z−1−i)n.
β) Fo e e y z∈C,|z−1−i|>√2we ha e ha
1
z=1
1 + i+z−1−i=1
z−1−i·1
1 + 1+i
z−1−i
=∞
∑︂
n=0
(−1)n(1 + i)n
(z−1−i)n+1 ,
1
z2=−(︃1
z)︃′=∞
∑︂
n=0
(−1)n(1 + i)n(n+ 1)
(z−1−i)n+2 ,
and he e o e
(z) = 1
z−1−i+∞
∑︂
n=1
(−1)n(1 + i)n+ (−1)n−1(1 + i)n−1n
(z−1−i)n+1 .
93
Exe cise 55.
Classi y each o he isola ed singula i ies o he unc ion , whe e
a) (z) := z5+ 4z3−2 + 2
z+3
z2;
b) (z) := z2−4
z−2;
c) (z) := 1
z−z3;
d) (z) := z4
z4+1;
e) (z) := ez
z2+4;
) (z) := z2+4
ez;
g) (z) := 1−ez
2+ez;
h) (z) := e1
z2;
i) (z) := 1
(z−3)2(2−cos z);
j) (z) := z
sin z;
k) (z) := z2sin z
z+1 ;
l) (z) := 1−cos z
sin2z.
Solu ion:
a) The unc ion (z) = z5+ 4z3−2 + 2
z+3
z2has wo isola ed singula i ies: 0and ∞.
Clea ly, we ha e ha
•0is a pole o he o de 3 o ,
•∞is a pole o he o de 5 o .
b) The unc ion (z) = z2−4
z−2has wo isola ed singula i ies: 2and ∞.
•Because
lim
z→2
z2−4
z−2= lim
z→2(z+ 2) = 4,
2is a emo able singula i y o .
•
lim
z→∞
(z)
z= 1 =
/0,
and he e o e ∞is a simple pole o .
c) The unc ion (z) = 1
z−z3=1
z(1 −z)(1 + z)has ou isola ed singula i ies: 0,1,−1
and ∞.
•0,1and −1a e simple poles o .
•Because
lim
z→∞ (z) = lim
z→∞
1
z3(︁1
z2−1)︁=1
−∞ = 0,
∞is a emo able singula i y o .
94
d) (z) = z4
z4+1 and because
z4+ 1 = 0 ⇔z∈{︃1 + i
√2,1−i
√2,−1−i
√2,−1 + i
√2}︃,
he unc ion has i e isola ed singula i ies:
11
•1+i
√2,1−i
√2,−1−i
√2and −1+i
√2a e simple poles o
•and, because lim
z→∞ (z)=1,∞is a emo able singula i y o .
e) The unc ion (z) = ez
z2+4 has h ee isola ed singula i ies: 2i, −2iand ∞.
•2iand −2ia e simple poles o .
•Because
lim
x→∞
x∈R
(x) = lim
x→∞
x∈R
ex
x2+ 4
l’H.
= lim
x→∞
x∈R
ex
2x
l’H.
= lim
x→∞
x∈R
ex
2=∞,
(2nπi) = e2nπi
(2nπi)2+ 4 =1
−4n2π2+ 4 →0,
lim
z→∞ (z)does no exis , and he e o e ∞is an essen ial singula i y o .
) The unc ion (z) = z2+4
ezhas only one isola ed singula i y, which is ∞.
•Because
lim
x→∞
x∈R
(x)=0,
lim
x→−∞
x∈R
(x) = ∞·∞=∞,
he limi lim
z→∞ (z)does no exis . F om his ac , i ollows ha ∞is an essen ial
singula i y o .
95
g) Because (z) = 1−ez
2+ez, and a he same ime
2 + ez= 0 ⇔z= Ln(−2) = ln 2 + (2k+ 1)πi =: zk, k ∈Z,
has isola ed singula i ies p ecisely in he poin s zk.
•Fu he mo e
[(2 + ez)′]z=zk= [ez]z=zk=−2 =
/0,
[1 −ez]z=zk= 3 =
/0,
and he e o e zk= ln 2 + (2k+ 1)πi,k∈Z, a e simple poles o .
Be ca e ul: ∞is no an isola ed singula i y o .
h) (z) = e1
z2and o e e y z∈C
∖
{0}we ha e ha
(z) = ∞
∑︂
n=0
1
n!
1
z2n.
F om his, i ollows ha
•0is an essen ial singula i y o ,
•∞is a emo able singula i y o .
i) (z) = 1
(z−3)2(2−cos z)and because
2 = cos z=eiz +e−iz
2⇔4 = eiz +e−iz ⇔
⇔e2iz −4eiz + 1 = 0 ⇔eiz =4±√16 −4
2= 2 ±√3>0⇔
⇔iz = Ln(2 ±√3) = ln(2 ±√3) + 2kπi, k ∈Z⇔
⇔z=zk:= 2kπ −iln(2 ±√3), k ∈Z,
he unc ion has isola ed singula i ies in he poin s 3and zk,k∈Z.
•Easily we can compu e ha
[(2 −cos z)′]z=zk= [sin z]z=zk=
/0,
and he e o e has in he poin s zk= 2kπ −iln(2 ±√3), whe e k∈Z, simple
poles.
•I is clea ha 3is a pole o he o de 2 o he unc ion .
(∞is no an isola ed singula i y o .)
96
j) The unc ion (z) = z
sin zhas clea ly isola ed singula i ies in he oo s o he unc ion
sinus.
Think h ough he ac ha
•0is a emo able singula i y o ,
•kπ, whe e k∈Z
∖
{0}, a e simple poles o .
(∞is no an isola ed singula i y o .)
k) The unc ion (z) = z2sin z
z+1 has p ecisely wo isola ed singula i ies: −1and ∞.
•−1is an essen ial singula i y o (because lim
z→−1 (z)does no exis ),
•∞is a pole o he o de wo o (because lim
z→∞
(z)
z2= lim
z→∞ sin (︁z
z+1)︁= sin 1 =
/0).
l) The unc ion (z) = 1−cos z
sin2zclea ly has isola ed singula i ies in he oo s o he unc ion
sinus.
•Because5
lim
z→2kπ
1−cos z
sin2z
l’H.
= lim
z→2kπ
sin z
2 sin zcos z=1
2,
we ha e ha he poin s 2kπ, whe e k∈Z, a e emo able singula i ies o .
•Because
lim
z→(2k+1)π(︁z−(2k+ 1)π)︁2·1−cos z
sin2z
l’H.
= 2 lim
z→(2k+1)π
2(︁z−(2k+ 1)π)︁
2 sin zcos z
l’H.
=
l’H.
=−2 lim
z→(2k+1)π
1
cos z=2=
/0,
he poin s (2k+ 1)π, whe e k∈Z, a e poles o he o de 2 o he unc ion .
(∞is no an isola ed singula i y o .)
Exe cise 56.
P o e he L’Hôpi al’s ule:
Le and gbe holomo phic, non-cons an unc ions on some ing neighbo hood o a poin
z0∈Cand le lim
z→z0
(z) = lim
z→z0
g(z)=0.
Then we ha e ha
lim
z→z0
(z)
g(z)= lim
z→z0
′(z)
g′(z).
Solu ion:
F om he assump ions i ollows ha he e a e p, q ∈N, neighbou hood U(z0)o z0and
unc ions 1and g1, which a e holomo phic and non-ze o on U(z0)such ha o e e y
z∈U(z0)
∖
{z0}we ha e ha
(z) = (z−z0)p 1(z),
g(z) = (z−z0)qg1(z).
5We a e using L’Hôpi al’s ule p o en in he ollowing exe cise.
97

The e o e
lim
z→z0
(z)
g(z)= lim
z→z0
(z−z0)p−q 1(z)
g1(z)=⎧
⎪
⎨
⎪
⎩
∞, p < q,
0, p > q,
1(z0)
g1(z0), p =q,
and
lim
z→z0
′(z)
g′(z)= lim
z→z0
p(z−z0)p−1 1(z)+(z−z0)p ′
1(z)
q(z−z0)q−1g1(z)+(z−z0)qg′
1(z)=
= lim
z→z0
(z−z0)p−qp 1(z)+(z−z0) ′
1(z)
qg1(z)+(z−z0)g′
1(z)=
=⎧
⎪
⎨
⎪
⎩
∞, p < q,
0, p > q,
1(z0)
g1(z0), p =q.
The heo em is p o en.
Exe cise 57.
Compu e he esidue o he unc ion in all o i s isola ed singula i ies, whe e
a) (z) := 1
z+z3;
b) (z) := z2
(1+z)3;
c) (z) := 1
(z2+1)3;
d) (z) := z3+1
z−2;
e) (z) := 1
z6(z2+1)2;
) (z) := an z;
g) (z) := 1
sin z;
h) (z) := co g3z;
i) (z) := sin z·sin 1
z;
j) (z) := sin(πz)
(z−1)3.
Solu ion:
a) The unc ion (z) = 1
z+z3=1
z(z−i)(z+i)has clea ly ou isola ed singula i ies:
0, i, −ia∞. Now we will use (as in se e al ollowing exe cises) he [1, Theo em 44, pa
(iii)]:
• es (0) = [︁1
1+3z2]︁z=0 = 1,
• es (i) = [︁1
1+3z2]︁z=i=1
1−3=−1
2,
• es (−i) = [︁1
1+3z2]︁z=−i=1
1−3=−1
2
and [1, Theo em 44, pa ( )]:
• es (∞) = −(︁1−1
2−1
2)︁= 0.
98
b) The unc ion (z) = z2
(1+z)3clea ly has wo isola ed singula i ies.
•−1is a pole o he o de h ee o , and he e o e
es (−1) = 1
2[︁(z2)′′]︁z=−1= 1.
• es (∞) = −1.
c) The unc ion (z) := 1
(z2+1)3has h ee isola ed singula i ies: poles o he o de h ee a
he poin s iand −iand a emo able singula i y a ∞.
• es (±i) = 1
2[︃(︃ 1
(z±i)3)︃′′]︃z=±i
=1
2[︃3·41
(z±i)5]︃z=±i
=
= 6 1
(±2i)5=∓3
16i,
• es (∞) = 0.
d) The unc ion (z) = z3+1
z−2has wo isola ed singula i ies: 2(simple pole) and ∞.
•
es (2) = [︃z3+ 1
1]︃z=2
= 9,
• es (∞) = −9.
e) Because (z) = 1
z6(z2+ 1)2=1
z6(z+i)2(z−i)2, he numbe s ±ia e poles o he o de
wo o ,0is a pole o he o de 6 o and ∞is a emo able singula i y o .
• es (±i) = [︃(︃ 1
z6(z±i)2)︃′]︃z=±i
=
=−[︃6z5(z±i)2+z62(z±i)
z12(z±i)4]︃z=±i
=
=±7
4i.
•Because (z)=1:(z6+ 2z8+z10) = 1
z6+···,6we ha e ha es (∞) = 0,
• es (0) = −7
4i+7
4i−0=0.
) The unc ion (z) = an z=sin z
cos zhas simple poles in he poin s π
2+kπ, whe e k∈Z, and
es (︂π
2+kπ)︂=[︃sin z
−sin z]︃z=π
2+kπ
=−1.
(∞is no an isola ed singula i y o .)
6In he Lau en se ies o he coe icien o 1
zis equal o 0.
99
g) (z) = 1
sin zhas simple poles in he poin s kπ, whe e k∈Z, and
es (kπ) = [︃1
cos z]︃z=kπ
= (−1)k.
h) (z) = co g3z=cos3z
sin3zhas poles o he o de h ee in he poin s kπ, whe e k∈Z,
and
cos z
sin z=(︃1−z2
2+z4
24 −···)︃:(︃z−z3
6+z5
120 −···)︃=1
z−z
3−z3
45 −··· ,
and he e o e
(︂cos z
sin z)︂3=(︃1
z−z
3−z3
45 −···)︃(︃1
z−z
3−z3
45 −···)︃(︃1
z−z
3−z3
45 −···)︃.
es (0) is a “coe icien o 1
z”, he e o e
es (0) = 3 (︃−1
3)︃=−1.
Because he unc ion has a pe iod π, ha is (z) = (z−kπ), we ha e ha
es (kπ) = es (0) = −1 o e e y k∈Z.
i) (z) = sin z·sin 1
zhas an isola ed singula i y a 0and a ∞.
Because o e e y z∈C
∖
{0}we ha e ha
(z) = (︄∞
∑︂
n=0
(−1)nz2n+1
(2n+ 1)!)︄(︄∞
∑︂
k=0
(−1)k1
(2k+ 1)! ·1
z2k+1 )︄=··· ,
we ha e ha 7
es (0) = es (∞)=0.
j) The unc ion (z) = sin(πz)
(z−1)3has wo isola ed singula i ies: 1and ∞.
•Because 1is a pole o he o de wo o ,
es (1) = lim
z→1(︃sin(πz)
(z−1)3(z−1)2)︃′= lim
z→1(︃sin(πz)
z−1)︃′=
= lim
z→1
πcos(πz)(z−1) −sin(πz)
(z−1)2
l’H
=
l’H
= lim
z→1−π2sin(πz)(z−1) + πcos(πz)−πcos(πz)
2(z−1) =
= lim
z→1(︃−π2
2sin(πz))︃= 0.
7The Lau en se ies o has non-ze o coe icien s only o he “e en powe s” o z.
100
O he possible solu ion:
(z) = sin(πz)
(z−1)3=−sin(π(z−1))
(z−1)3=
=−1
(z−1)3
∞
∑︂
n=0
(−1)nπ2n+1
(2n+ 1)!(z−1)2n+1 =
=∞
∑︂
n=0
(−1)n+1 π2n+1
(2n+ 1)!(z−1)2n−2.
The jus compu ed Lau en se ies o he unc ion has non-ze o coe icien s only
o he “e en powe s” o (z−1), and he e o e es (1) = 0.
• es (∞) = 0.
Exe cise 58.
Using he esidue heo em compu e he in eg als
a) ∫︂γ
cos z
z3dz, whe e γ( ) := 3ei , ∈ ⟨0,2π⟩;
b) ∫︂γ
1
z+ 2 cos 1
zdz, whe e γ( ) := 18ei , ∈ ⟨0,2π⟩;
c) ∫︂k
z3
z4−1dz, whe e k={z∈C:|z|= 2};
d) ∫︂k
z3
z+ 1e1
zdz, whe e k={z∈C:|z|= 2};
e) ∫︂γ
zsin z+ 1
z−1dz, whe e γ( ) := 2e−i , ∈ ⟨0,6π⟩;
) ∫︂γ
eπz
2z2−idz,
whe e γis a simple, closed, piecewise smoo h posi i ely o ien ed cu e such ha
in γ={z∈C:|z|<1∧0<a g z < π
2};
g) ∫︂k
dz
z5(z10 −2),whe e k={z∈C:|z|= 2}.
101
c) Because he p oblem
z6+ 1 = 0 ∧Im z≥0
has exac ly h ee solu ions:
z1:= eiπ
6=√3
2+1
2i,
z2:= eiπ
2=i,
z3:= e5
6πi =−√3
2+1
2i,
and he unc ion x4+1
x6+1 is e en, we ha e ha
∫︂∞
0
x4+ 1
x6+ 1 dx=1
2∫︂∞
−∞
x4+ 1
x6+ 1 dx=
=1
2∫︂k
z4+ 1
z6+ 1 dz=
=1
22πi
3
∑︂
j=1
es
z=zj
z4+ 1
z6+ 1,
whe e k⊂Cis he bounda y o he se {z∈C:|z|<2∧Im z > 0}.
1−1
k
The e o e, because
es
z=z1
z4+ 1
z6+ 1 =[︃z4+ 1
6z5]︃z=√3
2+1
2i
=1
6(−i),
es
z=z2
z4+ 1
z6+ 1 =[︃z4+ 1
6z5]︃z=i
=2
6(−i),
es
z=z3
z4+ 1
z6+ 1 =[︃z4+ 1
6z5]︃z=−√3
2+1
2i
=1
6(−i),
we ha e ha ∫︂∞
0
x4+ 1
x6+ 1 dx=2
3π.
108

d) The unc ion x2
(x2+1)3is e en, and he e o e o k⊂C, which is he bounda y o he se
{z∈C:|z|<2∧Im z > 0},
1−1
i
k
we ha e ha
∫︂∞
0
x2dx
(x2+ 1)3=1
2∫︂∞
−∞
x2dx
(x2+ 1)3=1
2∫︂k
z2dz
(z2+ 1)3=
=1
22πi es
z=i(︃z2
(z2+ 1)3)︃=πi1
2[︃(︃ z2
(z+i)3)︃′′]︃z=i
=
=πi
2[︃(︃2z(z+i)3−z23(z+i)2
(z+i)6)︃′]︃z=i
=
=πi
2[︃(︃−z2+ 2zi
(z+i)4)︃′]︃z=i
=
=πi
2[︃(−2z+ 2i)(z+i)4−(−z2+ 2zi)4(z+i)3
(z+i)8]︃z=i
=
=πi
2(︃4(2i)3
(2i)8)︃=2πi
25i5=π
16.
e) Le γ( ) := ei , whe e ∈ ⟨0,2π⟩. Then
∫︂π
−π
cos x
3 + 2 sin xdx=∫︂γ
z+1
z
2
1
3+2z−1
z
2i
1
iz dz=
=1
2∫︂γ
z2+ 1
z
1
z2+ 3iz −1dz.
(We’ e used he change o a iables eix =z, see [1, Chap e 9.3, pa a)]).
Because
z2+ 3iz −1=0 ⇔z=−3±√5
2i,
⟨γ⟩
1
−3−√5
2i
−3+√5
2i
109
we ha e ha
∫︂π
−π
cos x
3 + 2 sin xdx=1
2∫︂γ
z2+ 1
z
1
z2+ 3iz −1
⏞ ⏟⏟ ⏞
=: (z)
dz=
=1
22πi (︄ es (0) + es (︂−3 + √5
2i)︂)︄=
=πi (︄[︃z2+ 1
z2+ 3iz −1]︃z=0
+[︃z2+ 1
z(2z+ 3i)]︃z=−3+√5
2i)︄=
=πi (︄−1 + [︃2−3iz
2−6iz + 3iz ]︃z=−3+√5
2i)︄=πi(−1 + 1) = 0.
) We will use he change o a iables eix =z,
cos x=z+1
z
2,
cos 2x=z2+1
z2
2,
dx=1
iz dz.
Fo γ( ) := ei , whe e ∈ ⟨0,2π⟩, we ha e ha
∫︂2π
0
cos22x
5−4 cos xdx=∫︂γ
1
4(︃z4+ 1
z2)︃21
5−2z2+1
z
1
iz dz=
=∫︂γ
1
4i
(z4+ 1)2
z4
1
5z−2z2−2dz=
=∫︂γ
1
4i
(z4+ 1)2
z4
1
−2(z−2) (︁z−1
2)︁
⏞ ⏟⏟ ⏞
=: (z)
dz=
=2πi
4i(︃ es (0) + es (︃1
2)︃)︃.
⟨γ⟩
1 2
1
2
110
and because
es (0) = 1
3! [︃(︃ (z4+ 1)2
5z−2z2−2)︃′′′]︃z=0
=1
6(︃−255
8)︃=−255
48 ,
es (︃1
2)︃=[︃(z4+ 1)2
z4
1
5−4z]︃z=1
2
=289
48 ,
we ha e ha
∫︂2π
0
cos22x
5−4 cos xdx=17
48π.
g) Because he equa ion z6+ 1 = 0 has, assuming Im z≥0, exac ly h ee solu ions:
z1=eiπ
6=√3
2+1
2i,
z2=eiπ
2=i,
z3=e5
6πi =−√3
2+1
2i,
o he unc ion (z) := 1
1+z6we ha e ha
∫︂∞
−∞
dx
1 + x6= 2πi (︂ es (z1) + es (z2) + es (z3))︂=
= 2πi
3
∑︂
k=1
1
6z5
k
= 2πi
3
∑︂
k=1
zk
6z6
k
=
=−2πi
6(︁z1+z2+z3)︁=
=−π
3i(︄√3
2+1
2i+i−√3
2+1
2i)︄=
=−π
3i2i=2π
3.
111
h) Because
z2+z+ 1 = 0 ⇔z=−1
2±√3
2i,
o k⊂Cde ined as he bounda y o he se {z∈C:|z|<2∧Im z > 0}we ha e ha
∫︂∞
−∞
dx
x2+x+ 1 =∫︂k
dz
z2+z+ 1 =
= 2πi es
z=−1
2+√3
2i(︃1
z2+z+ 1)︃=
= 2πi [︃1
2z+ 1]︃z=−1
2+√3
2i
=
= 2πi 1
−1 + √3i+ 1 =2π
√3.
1−1
k
112
Bibliog aphy
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am. sb.cz/bouchala.
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am. sb.cz/bouchala.
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am. sb.cz/bouchala.
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Ma hema ics, OUP, Ox o d, 2004.
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