Complex Analysis : Exercises With Solutions

Complex analysis Exercises with solutions Jiří Bouchala (and Ondřej Bouchala) 17. listopadu 2172/15 708 00 Ostrava-Poruba Czech Republic univ[email protected] www.vsb.cz Transformation of the structure and content of higher education at VŠB-TUO NPO_VŠB-TUO_MSMT-16605/2022 This work is licenced under CC BY-SA 4.0 cba. The author of the painting Imaginární džungle on the cover page is Jiří Bouchala (and it is owned by Ondřej Bouchala). Preface This text contains the solutions to all of the practice problems in the 10th chapter of the lecture notes “An Introduction to Complex Analysis” [1]. It is a translation of the Czech text [3]. The typesetting and all of the pictures are the work of my son Ondřej. 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 proofreading. We are grateful 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, 2024 Jiří Bouchala (and Ondřej Bouchala) 1Please send all of the remarks (notes, recommendations, threats and gifts) to my e-mail address [email protected]. Exercise 1. Find the real and imaginary part of the complex number a) z= (1 + i)(3 −2i) ; b) z=2−3i 3+4i; c) z=1+i 1−i; d) z= 2i−2−4i 2. Solution: 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. Exercise 2. Write the given complex number in the trigonometric form a) z=−1 + √3i; b) z=i; c) z=−8 ; d) z=−1−√3i; e) z=2+i 3−2i; f) z=3−i 2+i. Solution: 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 ,tan φ= 7 13 4 13 =7 4, φ = arctan 7 4; z=√65 13 (︃cos (︃arctan 7 4)︃+isin (︃arctan 7 4)︃)︃. f) 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 Exercise 3. Prove the de Moivre’s theorem (∀n∈N) (∀φ∈R) : (︁cos φ+isin φ)︁n= cos(nφ) + isin(nφ) using mathematical induction. Solution: 1) We start by checking that the formula holds for n= 1: (cos φ+isin φ)1= cos (1 ·φ) + isin (1 ·φ). 2) Now we prove the implication (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 it suffices 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 φand cos φ. Solution: 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 therefore (it is enough to compare the real and imaginary parts) cos(4φ) = cos4φ−6 sin2φcos2φ+ sin4φ, sin(4φ) = 4 sin φcos3φ−4 sin3φcos φ. 3 Exercise 5. Find Re zand Im zfor z=(︂1−i 1+√3i)︂24 . Solution: 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. Exercise 6. Find Arg zand arg zfor a) z=(︁√3 + i)︁126 ; b) z= (1 + i)137 ; c) z=−1−5i. Solution: a) z= (√3 + i)126 =(︁2(cos π 6+isin π 6))︁126 = 2126 (cos(21π) + isin(21π)) = −2126; Arg z={π+ 2kπ :k∈Z},arg z=π. b) z= 2137 2(︁cos (︁137π 4)︁+isin (︁137π 4)︁)︁= 2137 2(︁cos π 4+isin π 4)︁; Arg z={︂π 4+ 2kπ :k∈Z}︂,arg z=π 4. 4 c) −5i −1 α tan α=5 1, α = arctan 5; Arg z={−π+ arctan 5 + 2kπ :k∈Z},arg z=−π+ arctan 5. Exercise 7. Draw in the complex plane the set 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}; f) {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≤arg(z+ 2i)≤π 2}; l) {z∈C:|z|+ Re z≤1∧ ∧−π 2≤argz≤π 4}. Solution: a) {z∈C: Re z≤1}: 1 1 2 5 b) From the assumptions it follows that for all sufficiently large nwe have that zn=|zn|(cos (arg zn) + isin (arg zn)) , and the claim follows directly from the continuity of cosine and sine and the theorem of the limit of a product. As a counterexample disproving the reverse inequality, we can use the sequence zn:= cos (︃π+(−1)n n)︃+isin (︃π+(−1)n n)︃ and the choice r= 1, φ =π. Exercise 11. Find all z∈Csuch that a) z3= 1; b) z2=i; c) z2= 24i−7; d) (︁z−1 z+1)︁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+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 therefore 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 therefore 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) Let 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 if and only if z=x+iy = 3 + 4ior z=−3−4i. 13 d) After the change of variables z−1 z+1 =: u=|u|(cos φ+isin φ)we firstly solve the equation u2= 2i, that is |u|2(cos (2φ) + isin (2φ)) = 2 (︂cos (︂π 2)︂+isin (︂π 2)︂)︂. The solution is u=±√2(︂cos (︂π 4)︂+isin (︂π 4)︂)︂=±(1 + i), and then easily z−1 z+1 = 1 + iif 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 similarly z−1 z+1 =−1−iif 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: (︃z−1 z+ 1)︃2 = 2i⇔(︃z=−1+2i∨z=−1 5−2 5i)︃. e) |z|4(cos (4φ) + isin (4φ)) = cos π+isin πif and only if z=zk= cos (︂π 4+kπ 2)︂+isin (︂π 4+kπ 2)︂, k ∈Z,that is z4=−1⇔z∈{︃1 + i √2,−1 + i √2,−1−i √2,1−i √2}︃. 11 f) |z|3(cos (3φ) + isin (3φ)) = √2(︃cos (︃3π 4)︃+isin (︃3π 4)︃)︃ if and only if (︃|z|=3 √︂√2)︃∧(︃3φ=3π 4+ 2kπ, k ∈Z)︃. From this, it easily follows that z3=i−1if and only if 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. . . not possible, −11 + √3721 2=−11 + 61 2= 25, and therefore z2=−11 + 60i⇔z=±(5 + 6i). i) Let 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. . . not possible, 4, and therefore z2= 3 + 4i⇔z=±(2 + i). 15 Exercise 12. Find and draw the set M={︁1 z:z∈Ω}︁, if a) Ω = {z∈C:argz=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}. Solution: a) α∈(−π, π)⇒M={z∈C: arg z=−α}; Ω α M −α α=π⇒M=Ω={z∈C: arg z=π}. ΩM 16 b) M={︃u+iv :1 u+iv ∈Ω}︃∪{∞} = ={︃u+iv : 1 u+iv −1= 1}︃∪{∞} = ={u+iv :|1−u−iv|=|u+iv|}∪{∞} = ={︁u+iv : (1 −u)2+v2=u2+v2}︁∪{∞} = ={u+iv : 1 −2u= 0}∪{∞} = ={︃u+iv :u=1 2}︃∪{∞}. 1 i 2 2 Ω 1 i 1 2 1 2 M c) M={︃u+iv :1 u+iv ∈Ω}︃∪{∞} = ={︃u+iv :u−iv u2+v2∈Ω}︃∪{∞}, and because u u2+v2=−v u2+v2⇔(u= /0∧u=−v), we have that M={u+iv :u= /0∧u=−v}∪{∞}. Ω M 17 d) M={︃u+iv :1 u+iv ∈Ω}︃= ={︃u+iv :u u2+v2= 1}︃= ={︄u+iv :(︃u−1 2)︃2 +v2=1 4}︄ ∖ {0}. 1 i 1 1 Ω i 1 2 1 2 M e) M={︃u+iv :1 u+iv ∈Ω}︃∪{∞} = ={︃u+iv :−v u2+v2= 0}︃∪{∞} = ={u+iv :v= 0 = /u}∪{∞}. ΩM Exercise 13. Find and draw the set M={f(z): z∈Ω}, if a) Ω = {z∈C:|argz| ≤ π 6}, f(z) := z2; b) Ω = {z∈C:|Im 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. 18 Solution: a) M={︂z∈C:|arg 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 Exercise 14. Compute a) sin(2 −3i); b) cos i; c) cosh i; d) Ln(−5+3i)aln(−5+3i); e) Ln(−4−√3i)aln(−4−√3i); f) Ln(ie2). Solution: 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+ arctan 5 3)︃+isin (︃π 2+ arctan 5 3)︃)︃, and therefore Ln(−5+3i) = ln √34 + i(︃π 2+ arctan 5 3)︃+ 2kπi, k ∈Z; ln(−5+3i) = ln √34 + i(︃π 2+ arctan 5 3)︃. e) −4−√3i=√19 (︄cos (︄−π+ arctan √3 4)︄+isin (︄−π+ arctan √3 4)︄)︄, and therefore Ln(−4−√3i) = ln √19 + i(︄−π+ arctan √3 4)︄+ 2kπi, k ∈Z; ln(−4−√3i) = ln √19 + i(︄−π+ arctan √3 4)︄. f) Ln(ie2) = ln(e2) + iπ 2+ 2kπi = = 2 + π 2i+ 2kπi, k ∈Z. Exercise 15. Find all z∈C, for which we have that 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 where 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). From that, it follows that (we are using the assumption φ1, φ2∈⟨︁0,π 4)︁): z1, z2∈Ω z3 1=z3 2⇒|z1|=|z2| φ1=φ2⇒z1=z2, therefore the function fis injective on Ω. Exercise 19. Decide if the given limit exists, and if it does compute it 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; f) lim z→i z2+z(2−i)−2i z2+1 ; g) lim z→0 Re z 1+|z|. Solution: a) lim z→0 Re z zdoes not exist, because 0 = /1 n→0∧Re (︁1 n+ 0i)︁ 1 n = 1 →1 and at the same time 0 = /i1 n→0∧Re (︁i1 n)︁ 1 n = 0 →0. b) lim z→0 Im z2 z·zdoes not exist, because for 0 = /z=x+iy we have that 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 not exist, because for 0 = /z=x+iy we have that 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. f) 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. Alternatively we can use the continuity of the function f(z) := z+2 z+iat the point 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 Exercise 20. Draw the set ⟨φ⟩:= {φ(t): t∈Dφ}, if a) φ(t) := 1 −it, Dφ =⟨0,2⟩; b) φ(t) := t−it2, Dφ =⟨−1,2⟩; c) φ(t) := 1 + e−it, Dφ =⟨0,2π⟩; d) φ(t) := e2it −1, Dφ =⟨0,2π⟩; e) φ(t) := {︄eiπt, t ∈ ⟨0,1), t−2, t ∈ ⟨1,3⟩; f) φ(t) := {︄eit, t ∈ ⟨−π 2, π), 3t π−4, t ∈ ⟨π, 2π⟩. Solution: a) φ(t) := 1 −it, Dφ =⟨0,2⟩. 1 i −2i⟨φ⟩ b) φ(t) := t−it2, Dφ =⟨−1,2⟩. −4i −12 ⟨φ⟩ −i −i c) φ(t) := 1 + e−it, Dφ =⟨0,2π⟩. 1⟨φ⟩ d) φ(t) := e2it −1, Dφ =⟨0,2π⟩. −1 ⟨φ⟩ “2×oběhnutᔓ2×around” 30 e) φ(t) := {︄eiπt, t ∈ ⟨0,1), t−2, t ∈ ⟨1,3⟩. −1 1 i i ⟨φ⟩ f) φ(t) := {︄eit, t ∈ ⟨−π 2, π), 3t π−4, t ∈ ⟨π, 2π⟩. −i −1 2 i i 1 1 ⟨φ⟩ Exercise 21. Find a parametrization of the set Ω(i.e. find a curve φsuch that ⟨φ⟩= Ω), if a) Ω = {z∈C:|z−2+3i|= 2}; b) Ωis a line segment with the endpoints a, b ∈C,a= /b; c) Ω = {z∈C: Re z= 2 Im z}; d) Ω = {z∈C: Re (︁1 z)︁= 2}. Solution: a) Ω = {z∈C:|z−2+3i|= 2};φ(t) := 2 −3i+ 2eit, t ∈ ⟨0,2π⟩. −3i 2 Ω = ⟨φ⟩ 31 b) Ωis a line segment with the endpoints a, b ∈C,a= /b;φ(t) := a+ (b−a)t, t ∈ ⟨0,1⟩. Ω = ⟨φ⟩ a b c) Ω = {z∈C: Re z= 2 Im z};φ(t) := t+t 2i, t ∈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(t) := 1 4+1 4eit, t ∈(−π, π). 1 4 Ω = ⟨φ⟩ 1 2 1 2 32 Exercise 22. Draw the set Ω, and decide if Ωis a domain and if it is an open set, where 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}; f) Ω = {︁z∈C:|z|<1∧arg z∈(−π, π⟩ ∖ {0}}︁; g) Ω = {z∈C:|2z|<|1 + z2|}. Solution: a) Ω = {z∈C:|z−i|<1∨ |z+i|<1}. Ω Ω i i −i −i Ωis open, but not connected, and therefore Ωis not a domain. b) Ω = {z∈C:|z−1|<1∧ |z−2|<2}. Ω Ω 1 14 4 2 2 Ωis open and connected set, and therefore Ωis a domain. c) Ω = {z∈C:|z−1|<|z+ 1|}. Ω Ω −11 1 Ωis open and connected set, and therefore 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 connected set, and therefore Ωis a domain. e) Ω = {z∈C: 1 <|z|<2}. Ω Ω 1 2 i 2i i 2i 21 Ωis open and connected set, and therefore Ωis a domain. f) Ω = {︁z∈C:|z|<1∧arg z∈(−π, π⟩ ∖ {0}}︁. Ω Ω 1 1 Ωis open and connected set, and therefore 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, but not connected set, and therefore Ωis not a domain. Exercise 23. Find all of the points where the function fhas a derivative and the points where it is holomorphic, if a) f(z) := Re z; b) f(z) := |z2|; c) f(z) := zez; d) f(z) := z|z|; e) f(z) := Re z z; f) f(z) := z2z; g) f(z) := z2+ 2z−1. Solution: a) f(x+iy) = x ⏞⏟⏟⏞ =:u(x,y) + 0 ⏞⏟⏟⏞ =:v(x,y)·i. For every (x, y)∈R2we have that ∂u ∂x(x, y) = 1 = /0 = ∂v ∂y(x, y), and because of that it follows that the function fdoes not have a derivative anywhere and the function fis not holomorphic at any point. 35 b) f(x+iy) = |(x+iy)2|= (|x+iy|)2=x2+y2. So f=u+iv, where u(x, y) := x2+y2 and v(x, y) := 0. ∂u ∂x(x, y) = 2x=∂v ∂y(x, y) = 0 ∂u ∂y (x, y) = 2y=−∂v ∂x(x, y)=0 ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭⇔(x, y) = (0,0), and at the same time the functions uand vare differentiable in R2, and therefore fhas a derivative (only) in the point 0ant it is not holomorphic anywhere. c) f(x+iy) = (x+iy)ex(cos y+isin y) = =xexcos y−yexsin y ⏞ ⏟⏟ ⏞ =:u(x,y) +i(xexsin y+yexcos y) ⏞ ⏟⏟ ⏞ =:v(x,y) . Functions uand vare differentiable in R2, ∂u ∂x(x, y) = excos y+xexcos y−yexsin y, ∂v ∂y(x, y) = xexcos y+excos y−yexsin y, and ∂u ∂y (x, y) = −xexsin y−exsin y−yexcos y, −∂v ∂x(x, y) = −(exsin y+xexsin y+yexcos y). So ∂u ∂x =∂v ∂y and ∂u ∂y =−∂v ∂x in R2, and therefore fis holomorphic everywhere in Cand f′(z)exists at every z∈C. (︃f′(z) = f′(x+iy) = (︃∂u ∂x +i∂v ∂x)︃(x, y) = ··· =ez+zez.)︃ d) f(x+iy)=(x−iy)√︁x2+y2=x√︁x2+y2 ⏞ ⏟⏟ ⏞ =:u(x,y) +i(−y√︁x2+y2) ⏞ ⏟⏟ ⏞ =:v(x,y) . From this, it follows that for every (x, y)∈R2 ∖ {(0,0)}we have that ∂u ∂x(x, y) = √︁x2+y2+x2 √︁x2+y2>0, ∂v ∂y(x, y) = −√︁x2+y2−y2 √︁x2+y2<0, and therefore: if z= /0, then f′(z)does not exist. 36 It remains to prove or disprove the existence of the derivative at the point 0: f′(0) = lim z→0 f(z)−f(0) z−0= lim z→0 z·|z| z= = lim z→0|z|(cos(arg z)−isin(arg z)) ·|z| |z|(cos (arg z) + isin (arg z)) = = lim z→0[|z|·(cos (−2 arg z) + isin (−2 arg z))] = 0, because ∀z= /0: |cos (−2 arg z) + isin (−2 arg z)|= 1. Summary: the function fhas a derivative only at the point 0, and therefore fis not holomorphic at any point. e) f(x+iy) = x x+iy =x(x−iy) x2+y2=x2 x2+y2 ⏞ ⏟⏟ ⏞ =:u(x,y) +i(︃−xy x2+y2)︃ ⏞ ⏟⏟ ⏞ =:v(x,y) . For every (x, y)∈R2 ∖ {(0,0)}we have that ∂u ∂x(x, y) = 2x(x2+y2)−x22x (x2+y2)2=2xy2 (x2+y2)2, ∂v ∂y(x, y) = −x(x2+y2) + xy2y (x2+y2)2=−x3+xy2 (x2+y2)2, ∂u ∂y (x, y) = −x22y (x2+y2)2, ∂v ∂x(x, y) = −y(x2+y2) + xy2x (x2+y2)2=x2y−y3 (x2+y2)2, and therefore the derivative can exist only in the points x+iy where (︃2xy2 (x2+y2)2=x(−x2+y2) (x2+y2)2)︃∧(︃2x2y (x2+y2)2=y(x2−y2) (x2+y2)2)︃, that is (︃xy2 (x2+y2)2=−x3 (x2+y2)2)︃∧(︃x2y (x2+y2)2=−y3 (x2+y2)2)︃. It is easy to observe that this system of equations has no solution. Summary: the function fdoes not have a derivative at any point, and therefore it is not holomorphic at any point. f) f(x+iy) = (x2−y2+ 2ixy)(x−iy) = =x3−xy2+ 2xy2+i(−x2y+y3+ 2x2y) = =x3+xy2 ⏞ ⏟⏟ ⏞ =:u(x,y) +i(y3+x2y) ⏞ ⏟⏟ ⏞ =:v(x,y) . 37 At the same time the function uis continuous on R2 ∖ {(0,0)}(at every point R2 ∖ {(0,0)} it must be differentiable), and therefore lim x→0−u(x, 1) = u(0,1) = lim x→0+ u(x, 1). = = π+c2−π+c1 From this, it follows that 2π=c1−c2. Analogously lim x→0−u(x, −1) = u(0,−1) = lim x→0+ u(x, −1), = = −π+c2π+c1 and therefore 2π=c2−c1. This leads us to the fact that 2π=c1−c2=−(c2−c1) = −2π, which is a contradiction. The sought function udoes not exist. 44 Exercise 30. Find the rotational angle and extensibility coefficient of the function fat the point z0, where a) f(z) := ez,z0=−1−π 2i; b) f(z) := z3,z0=−3+4i; c) f(z) := z+i z−i,z0= 2i. Solution: a) |f′(z0)|=|ez0|=|e−1−iπ 2|=1 e, which is the extensibility coefficient of the function fat the point z0(and 1 e<1implies that it is a contraction). arg f′(z0) = arg (︃1 e(︂cos π 2−isin π 2)︂)︃= arg (︃−i e)︃=−π 2, which is the rotational angle of the function fat the point z0. b) z0= 5 (︁cos (︁π 2+ arctan 3 4)︁+isin (︁π 2+ arctan 3 4)︁)︁, and therefore f′(z0) = 3z2 0= 3 ·25 (︃cos (︃π+ 2 arctan 3 4)︃+isin (︃π+ 2 arctan 3 4)︃)︃. From this we get |f′(z0)|= 75 . . . extensibility coefficient of the function fat z0 (75 >1, therefore it is a dilatation), arg f′(z0) = −π+ 2 arctan 3 4. . . rotational angle of the function fat the point z0. c) f′(z) = z−i−(z+i) (z−i)2=−2i (z−i)2, f′(z0) = −2i i2= 2i, and therefore 1<|f′(z0)|= 2 . . . extensibility coefficient of fat z0(dilatation), arg f′(z0) = π 2. . . rotational angle fat z0. 45 Exercise 31. Determine at which points of the complex plane is the given mapping a contraction: a) f(z) := 2 z; b) f(z) := ln(z+ 4). Solution: a) f′(z) = −2 z2. Therefore for z∈C: 0<|f′(z)|<1⇔−2 z2<1⇔2<|z|2⇔√2<|z|. The mapping fis a contraction in every point of the set {︁z∈C:|z|>√2}︁. √2 √2 b) f′(z)exists in C ∖ {x+iy :y= 0 ∧x≤ −4}=: Ω. For every z∈Ωwe have that |f′(z)|= 1 z+ 4, 0<1 |z+ 4|<1⇔1<|z+ 4|. The mapping fis a contraction at every point of the set {z∈C:|z+ 4|>1} ∖ {x+iy :y= 0 ∧x≤ −4}. −3−3 −4−3 46 Exercise 32. Draw the sets Ωand f(Ω) = {f(z): z∈Ω}, where2 a) Ω = U(1,2), f(z) := 1 −2iz; b) Ω = {z∈C: Re z < 1}, f(z) := (1 + i)z+ 1; c) Ω = U(1,2), f(z) := 1 z; d) Ω = U(1,2), f(z) := 2iz 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). 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 f(Ω) f(Ω) −iz 2z 1 + z 2A hint for some of the following exercises. Realize (and prove) that: fis conformal in the set Ω⊂C∞, A, B ⊂Ω}︄⇒f(A∩B) = f(A)∩f(B). 47 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}. 1−11−1 Ω Ω 1 2 i 2i 1 2 i 2i f(Ω) f(Ω) c) Ω = U(1,2), f(z) := 1 z, f(0) = ∞, f(−1) = −1, f(3) = 1 3, f(1 + 2i) = 1 1+2i=1−2i 5, and therefore f(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 f(Ω) f(Ω) d) Ω = U(1,2), f(z) := 2iz z+3, f(−3) = ∞, f(−1) = −2i 2=−i, f(3) = i, f(1 + 2i) = 2i(1 + 2i) 4+2i=−4−2i 4+2i=−3 5+4 5i, and therefore f(Ω) = U(0,1). 48 Ω Ω 1 3−1 −1 1 3 2i f(Ω) f(Ω) 1 1 e) Ω = U(1,2), f(z) := z−1 2z−6, f(0) = 1 6, f(3) = ∞, f(−1) = −2 −8=1 4, f(1 + 2i) = 2i 2+4i−6=1 4−1 4i, and therefore f(Ω) = {︃z∈C: Re z < 1 4}︃. Ω Ω 1 3−1 −1 1 3 2i 1 4 −1 4i 1 4 −1 4i f(Ω) f(Ω) f) Ω = {z∈C: Re z < 1}, f(z) := 1 z, f(1) = 1, f(0) = ∞, f(1 + i) = 1 1 + i=1−i 2, f(1 −i) = 1 1−i=1 + i 2, f(∞)=0, and therefore f(Ω) = C∞ ∖ U(︃1 2,1 2)︃. 49 1−1 Ω Ω 1−11 1 2 1 21 f(Ω) f(Ω) g) Ω = {z∈C: Re z < 1}, f(z) := z z−1+i, f(0) = 0, f(1) = 1 i=−i, f(1 −i) = ∞, f(1 + i) = 1 + i 2i=1 2−1 2i, and therefore f(Ω) = {z∈C: Im z > Re z−1}. 1−1 Ω Ω 1−1 1 −i f(Ω) f(Ω) h) Ω = {z∈C: Re z < 1}, f(z) := z z−2, f(2) = ∞, f(1) = −1, f(1 + i) = −2i 2=−i, f(∞) = 1, and therefore f(Ω) = U(0,1). 1−1 Ω Ω 1−1 f(Ω) f(Ω) 1 1 50 i) Ω = {z∈C: Re z < 0∧Im z < 0}, f(z) := 1 z, f(0) = ∞, f(−1) = −1, f(1) = 1, f(i) = −i, f(−i) = i, and thereforeže Ω = Ω1∩Ω2, where Ω1:= {z∈C: Re z < 0},Ω2:= {z∈C: Im z < 0}, f f je f(Ω) = f(Ω1)∩f(Ω2) = {z∈C: Re z < 0∧Im z > 0}. Ω Ω f(Ω) f(Ω) j) Ω = {z∈C: Re z > 0∧Im z > 0}, f(z) := z−1 z+1, f(0) = −1, f(1) = 0, f(i) = i−1 i+ 1 =i, f(−i) = −i, f(−1) = ∞, 51 f 1 1 f and therefore f(Ω) = {z∈C:|z|<1∧Im z > 0}. Ω Ω f(Ω) f(Ω) 11 k) Ω = {z∈C:−1<Re z < 0∧Im z < 0}, f(z) := z−i z+i, f(0) = −1, f(i)=0, f(−i) = ∞, f(1) = 1−i 1 + i=(1 −i)2 2=−i, f(−1) = −1−i −1 + i=(−1−i)2 2=i, f(−1 + i) = −1 −1+2i=1+2i 5, f(−1−i) = 1 + 2i, −1 1 −1 1 f 1 3 i 11 i 52 f f 11 and therefore f(Ω) = {z∈C: Im z > 0∧ |z−(1 + i)|>1∧ |z|>1}. −1 Ω Ω f(Ω) f(Ω) 1 i 1 i l) Ω = {z∈C:|z|<1∧Re z < 0∧Im z > 0}, f(z) := z z−i, f(0) = 0, f(i) = ∞, f(−1) = 1−i 2, f(1) = 1 + i 2, f(−i) = 1 2, 1 1 f 1 2 −1 2 1 2 −1 2 53 and therefore 1 z↦→ z2 z↦→ e−iπ 2z f1 Summary: one of the functions with the required properties is the function defined on Ω f(z) := (︃e−iπ 2·z+i z−i)︃2 =(︃−iz+i z−i)︃2 =−(︃z+i z−i)︃2 . Exercise 37. Let Ω = {z∈C: Re z > 0∧Im z < 0}. Find the linear fractional function fsuch that f(Ω) = {z∈C:|z|<1∧Re z < 0}. Solution: Ω Ω ff(Ω) f(Ω) −1 We firstly find the linear fractional function f∗such that f∗(0) = i, f∗(∞) = −i, f∗(−1) = ∞, 60 that is the function f∗(z) = ⎧ ⎪ ⎨ ⎪ ⎩ −iz +i z+ 1 , z ∈C, −i, z =∞. Then clearly either f∗(Ω) f∗(Ω) −1 (then we would define f:= f∗), or f∗(Ω) f∗(Ω) 1 (which would lead us to the definition f:= −f∗). Because f∗(i) = 1+i i+1 = 1 (the first possibility is realized), we choose f(z) := f∗(z) = ⎧ ⎪ ⎨ ⎪ ⎩ −iz +i z+ 1 , z ∈C, −i, z =∞. . Exercise 38. Find the conformal mapping which maps the domain Ω = {z∈C: Re z > Im z > 0} onto U(0,1). Solution: 1 1 61 Let us first consider the mapping z↦→ z4. z→z4 Then we find the linear fractional function f∗such that f∗(−1) = −1, f∗(0) = i, f∗(1) = 1, that is f∗(z) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ z+i iz + 1, z ∈C, 1 i=−i, z =∞. Clearly either f∗ 1 1 (in which case we would (for z∈Ω) define f(z) := f∗(z4)), or3 f∗ 11 (then we would define f(z) := 1 f∗(z4)in Ω). Because f∗(i) = ∞, the second case arose. We choose (for z∈Ω) f(z) := 1 f∗(z4)=iz4+ 1 z4+i. 3For the right-hand-side image we need to imagine that ∞=f∗(i). 62 Exercise 39. Find the images of the lines parallel to the real and imaginary axes by the mapping f(z) := 1 z(consider the lines together with the point ∞). Solution: For 0< c ∈Rwe have that f(0) = ∞, f(∞) = 0, f(i) = −i, f(c) = 1 c, f(−i) = i, f(−c) = −1 c, f(1) = 1, f(ci) = −1 ci, f(−1) = −1, f(−ci) = 1 ci, and therefore 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 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. Solution: ln z= ln |z|+iarg z, and therefore f(Mα) = {z∈C: Im z=α}, Mααf πi αi −πi f(Mα) f(Nr) = {ln r+ik :k∈(−π, π⟩}. Nr r f πi −πi f(Nr) ln r 1 f πi −πi 64 Exercise 41. Compute ∫︂γ|z|dz, where γ(t) := ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ 3eit, t ∈ ⟨0,π 2⟩, i(︁3 + π 2−t)︁, t ∈ ⟨π 2,π 2+ 3⟩, t−π 2−3, t ∈ ⟨π 2+ 3,π 2+ 6⟩. Solution: Let us choose γ1(t) := 3eit, t ∈ ⟨0,π 2⟩, γ2(t) := ti, i ∈ ⟨0,3⟩, γ3(t) := t, i ∈ ⟨0,3⟩. Then 3 3i ⟨γ⟩ ⟨γ1⟩ ⟨γ2⟩ ⟨γ3⟩ and γ′ 1(t)=3ieit, γ′ 2(t) = i, γ′ 3(t)=1, and therefore ∫︂γ|z|dz=∫︂γ1|z|dz−∫︂γ2|z|dz+∫︂γ3|z|dz= =∫︂π 2 0 3·3ieit dt−∫︂3 0 ti dt+∫︂3 0 tdt= = 9i∫︂π 2 0 (cos t+isin t) dt+ (1 −i)∫︂3 0 tdt= = 9i[sin t] π 2 0+ 9[cos t] π 2 0+ (1 −i)[︃t2 2]︃3 0 = = 9i−9 + 9 2(1 −i) = =−9 2+9 2i. 65 Exercise 42. Compute ∫︂γ z3dz, where γ(t) := ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ eit, t ∈ ⟨−π 2, π⟩, 3 πt−4, t ∈ ⟨π, 2π⟩, −2+i πt+ 6 + 2i, t ∈ ⟨2π, 3π⟩. Solution: 2−1 −i ⟨γ⟩ It is enough to apply Cauchy’s theorem. ∫︂γ z3dz= 0, because f(z) := z3is a holomorphic function on the simply connected domain Cand γis piecewise smooth closed curve in C. Exercise 43. Compute ∫︂γ|z|zdz, where γis a simple, closed, piecewise smooth and positively oriented curve such that ⟨γ⟩is the boundary of the set {z∈C:|z|<2∧Im z > 0}. 66 Solution: 2−2 ⟨γ⟩ Let us define the curves γ1(t) := 2eit, t ∈ ⟨0, π⟩, γ2(t) := t, t ∈ ⟨−2,2⟩. Then γ′ 1(t)=2ieit, γ′ 2(t)=1, and therefore ∫︂γ|z|zdz=∫︂γ1|z|zdz+∫︂γ2|z|zdz= =∫︂π 0 2·2e−it ·2ieit dt+∫︂2 −2|t|tdt ⏞ ⏟⏟ ⏞ =0 = = 8i∫︂π 0 1 dt= 8πi. Exercise 44. Using the Cauchy’s integral formulas calculate the given integrals4 a) ∫︂k z2+i zdz, where k={z∈C:|z−2i|= 1}; b) ∫︂k sin z z+idz, where k={z∈C:|z+i|= 1}; 4Convention. By the symbol ∫︁kf(z) dz, where k⊂C, we mean ∫︁γf(z) dz, where γis a simple, closed, piecewise smooth and positively oriented curve such that ⟨γ⟩=k. 67 c) ∫︂k sin z z2−7z+ 10 dz, where k={z∈C:|z|= 3}; d) ∫︂k sin z (z−2i)3dz, where k={z∈C:|z|= 3}; e) ∫︂k cos z z2−π2dz, where k={z∈C:|z|= 4}; f) ∫︂k e1 z (z2−4)2dz, where k={z∈C:|z−2|= 1}; g) ∫︂γ ezcos(πz) z2+ 2zdz, where γ(t) := 3 2eit, t ∈ ⟨0,2π⟩; h) ∫︂γ dz (z2−1)3,where γ(t) := −2 + e−4πit 2, t ∈ ⟨0,4⟩; i) ∫︂γ dz (1 −z)(z+ 2)(z−i)2, where γis a simple, closed, piecewise smooth and positively oriented curve such that −2∈int γ, i ∈int γ, 1∈ext γ. Solution: a) 2i 1−1 ⟨γ⟩ The function “ z2+i z” is holomorphic on a simply connected domain Ω := {z∈C: Im z > 0} and k=⟨γ⟩ ⊂ Ω, and therefore it follows from the Cauchy’s theorem that ∫︂k z2+i zdz=∫︂γ z2+i z= 0. 68 But we were supposed to use the Cauchy’s integral formulas. Which we can do for 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 therefore ∫︂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 Summary: the given series converges (absolutely) for every z∈{︃z∈C: z+ 1 z−1≤1}︃={z∈C: Re z≤0}. b) Because for every z∈C ∖ {0} zn+1 (n+1)! zn n!=|z| n+ 1 →0<1, the series ∑︁∞ n=1 zn n!converges absolutely in C. Because for every z∈C ∖ {0} n √︄ n2 zn=(n √n)2 |z|→1 |z|, the series ∑︁∞ n=1 n2 znconverges absolutely for |z|>1and diverges for |z|<1. If |z|= 1 we have  n2 zn=n2→ ∞ = /0, and therefore the series ∑︁∞ n=1 n2 zndiverges. Let us now define 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 for every z∈Cand n∈Nwe have that sn(z) = s∗ n(z) + s∗∗ n(z), s∗∗ n=sn(z)−s∗ n(z), and furthermore (we already know that) lim s∗ n(z)∈Cfor every z∈C, and therefore for every z∈Cwe have that lim sn(z)∈C⇔lim s∗∗ n(z)∈C. 76 Summary: the given series covnerges (absolutely) on the set {z∈C:|z|>1}. 11 Exercise 48. Find the radius of convergence Rof the given power series 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; f) ∞ ∑︁ n=0 (︁cos(in))︁zn; g) ∞ ∑︁ n=0 (n2−n−2)zn; h) ∞ ∑︁ n=0 zn (n+8)!. Solution: a) n √︃1 n2011 =1 (n √n)2011 →1, and therefore R= 1. b) n √nn=n→ ∞, and therefore R=1 ∞= 0. c) Because n √︄3n √︁(3n−2)2n=3 √2 1 √︁n √3n−2→3 √2 (it is enough to realize that for n≥3we have that 1≤n √3n−2≤n √n·n √n→1), and therefore 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 therefore 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 therefore R=1 e. f)  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 therefore R=1 e. g)  (n+ 1)2−(n+ 1) −2 n2−n−2→1, and therefore R= 1. h) 1 (n+9)! 1 (n+8)! =1 n+ 9 →0, and therefore R=∞. Exercise 49. Find the sum of the power series in the disk of convergence 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. Solution: a) n √n→1, and therefore the radius of convergence of a given series is 1. For every z∈C,|z|<1we have that ∞ ∑︂ 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, therefore the radius of convergence is 1. Let us define the function f(z) := ∑︁∞ n=1 zn n. Then for every z∈C,|z|<1we have that f′(z) = ∞ ∑︂ n=1 zn−1=1 1−z. From this, because |z|<1⇒1−z∈Ω := {w∈C:|w−1|<1}, ln′w=1 wvΩ, 1 1−1 Ω there is a c∈Csuch that for each z∈C,|z|<1we 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|<1we have that ∞ ∑︂ n=1 zn n=f(z) = −ln(1 −z). 79 c)  1 2n+3 1 2n+1 →1, and therefore the radius of convergence is 1. Let us define f(z) := ∑︁∞ n=0 z2n+1 2n+1 . Then for every z∈C,|z|<1we have that f′(z) = ∞ ∑︂ n=0 z2n=1 1−z2= =1 2−1 z−1+1 2 1 z+ 1. From that it follows that there is c∈C, such that for each z∈C,|z|<1we have that f(z) = −1 2ln(1 −z) + 1 2ln(1 + z) + c. And because 0 = f(0) = cfor each z∈C,|z|<1,we have that ∞ ∑︂ n=0 z2n+1 2n+ 1 =−1 2ln(1 −z) + 1 2ln(1 + z). d) n √︂1 n+1 →1, and therefore the radius of convergence is 1. Let f(z) := ∑︁∞ n=1(−1)n+1 zn+1 n+1 . Then for z∈C,|z|<1,we have that f′(z) = ∞ ∑︂ n=1 (−1)n+1zn=−∞ ∑︂ n=1 (−z)n= =z+ 1 −1 1 + z. From that, it follows that there is a c∈Csuch that f(z) = z−ln(1 + z) + c, and because 0 = f(0) = cwe have ∞ ∑︂ n=1 (−1)n+1 zn n+ 1 ={︄1 zf(z) = 1 −ln(1+z) z,0<|z|<1, 0, z = 0. e)  (n+ 1)2−(n+ 1) −2 n2−n−2→1, and therefore the radius of convergence is 1. For every z∈C,|z|<1,we have that ∞ ∑︂ n=0 (n2−n−2)zn=∞ ∑︂ n=0 n2zn−∞ ∑︂ n=0 nzn−2∞ ∑︂ n=0 zn (it is enough to realize that each of the series is absolutely convergent). 80 Furthermore (|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 therefore for every z∈C,|z|<1,we have that ∞ ∑︂ n=0 (n2−n−2)zn=z2+z−z(1 −z)−2(1 −z)2 (1 −z)3=2−4z (z−1)3. 81 Exercise 50. Find the sum of the given series a) ∞ ∑︁ n=1 1 n2n; b) ∞ ∑︁ n=1 (−1)n n2n. Solution: Let us consider the function f(z) := ∞ ∑︂ n=1 zn n2n. Because 1 n √n2n→1 2, the power series in the definition of the function fhas the radius of convergence 2. Therefore for every z∈C,0<|z|<2,we have that f′(z) = ∞ ∑︂ n=1 zn−1 2n=1 z ∞ ∑︂ n=1 (︂z 2)︂n= =1 z z 2 1−z 2 =1 2−z. Therefore there is a c∈Cfor which f(z) = −ln(2−z)+c. And because f(0) = 0 = −ln 2+c, for every z∈C,|z|<2we have that f(z) = −ln(2 −z) + ln 2. a) ∞ ∑︂ n=1 1 n2n=f(1) = ln 2, b) ∞ ∑︂ n=1 (−1)n n2n=f(−1) = −ln 3 + ln 2 = ln 2 3. Exercise 51. Find the Taylor series of the function fcentered at z0and find its radius of convergence, where a) f(z) := z+1 z2+4z−5, z0=−1; b) f(z) := z z2+i, z0= 0; c) f(z) := ln 1+z 1−z, z0= 0; d) f(z) := e3z−2, z0= 1; e) f(z) := sin(3z2+ 2), z0= 0; f) f(z) := 1 (z−1)3, z0= 3; g) f(z) := sin2z, z0= 0. 82 Solution: a) f(z) = 2 3 1 z+5 +1 3 1 z−1, −5−1 1 and therefore the radius of convergence is 2 and for every z∈C,|z+ 1|<2,it holds, that f(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 if and only if z=±(︂√2 2−√2 2i)︂, 11 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 + 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 clearly 1 1 z↦→ 1+z 1−z (0↦→ 0,1↦→ ∞,−1↦→ 0), the radius of convergence is 1. For every z∈C,|z|<1we have that f′(z) = 1−z 1 + z·1−z+ (1 + z) (1 −z)2=2 (1 + z)(1 −z)= =2 1−z2=∞ ∑︂ n=0 2z2n, and therefore there is a c∈Csuch that f(z) = ∞ ∑︂ n=0 2z2n+1 2n+ 1 +c. And because f(0) = 0 = c, for every z∈C,|z|<1we have that f(z) = ∞ ∑︂ n=0 2z2n+1 2n+ 1. d) Clearly the radius of convergence is ∞. We know that for every z∈Cwe have ez= ∞ ∑︁ n=0 zn n!, and therefore f(z) = e3z−2=e3(z−1)+1 =e e3(z−1) = =∞ ∑︂ n=0 e·3n n!(z−1)n. e) The radius of convergence is ∞and for any z∈Cwe have that sin z=∞ ∑︂ n=0 (−1)nz2n+1 (2n+ 1)!, cos z=∞ ∑︂ n=0 (−1)nz2n (2n)!. 84 From this it follows that for every z∈Cwe have that f(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, where a2k:= βkand a2k+1 := αkfor every k∈N∪{0}. f) 31 Clearly the radius of convergence is 2. For every z∈U(3,2) we have that 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. From this it easily follows, that for each z∈U(3,2) we have that f(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) If z∈C,|z|<1, we have f(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) For every z∈C,1<|z|<2, we have that f(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) For every z∈Csuch that |z|>2we have that f(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 fis clearly holomorphic on C ∖ {0}and |z0−0|=√2, we have precisely two “maximal anluli”: 1 1 i i α)P(1 + i, 0,√2), β)P(1 + i, √2,∞). α) For z∈C,|z−1−i|<√2, we have that f(z) = z+ 1 z2=1 z+1 z2 and furthermore 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 therefore f(z) = ∞ ∑︂ n=0 (︃(−1)n (1 + i)n+1 +(−1)n+2 (1 + i)n+2 (n+ 1))︃(z−1−i)n. β) For every z∈C,|z−1−i|>√2we have that 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 therefore f(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 Exercise 55. Classify each of the isolated singularities of the function f, where a) f(z) := z5+ 4z3−2 + 2 z+3 z2; b) f(z) := z2−4 z−2; c) f(z) := 1 z−z3; d) f(z) := z4 z4+1; e) f(z) := ez z2+4; f) f(z) := z2+4 ez; g) f(z) := 1−ez 2+ez; h) f(z) := e1 z2; i) f(z) := 1 (z−3)2(2−cos z); j) f(z) := z sin z; k) f(z) := z2sin z z+1 ; l) f(z) := 1−cos z sin2z. Solution: a) The function f(z) = z5+ 4z3−2 + 2 z+3 z2has two isolated singularities: 0and ∞. Clearly, we have that •0is a pole of the order 3 of f, •∞is a pole of the order 5 of f. b) The function f(z) = z2−4 z−2has two isolated singularities: 2and ∞. •Because lim z→2 z2−4 z−2= lim z→2(z+ 2) = 4, 2is a removable singularity of f. • lim z→∞ f(z) z= 1 = /0, and therefore ∞is a simple pole of f. c) The function f(z) = 1 z−z3=1 z(1 −z)(1 + z)has four isolated singularities: 0,1,−1 and ∞. •0,1and −1are simple poles of f. •Because lim z→∞ f(z) = lim z→∞ 1 z3(︁1 z2−1)︁=1 −∞ = 0, ∞is a removable singularity of f. 94 d) f(z) = z4 z4+1 and because z4+ 1 = 0 ⇔z∈{︃1 + i √2,1−i √2,−1−i √2,−1 + i √2}︃, the function fhas five isolated singularities: 11 •1+i √2,1−i √2,−1−i √2and −1+i √2are simple poles of f •and, because lim z→∞ f(z)=1,∞is a removable singularity of f. e) The function f(z) = ez z2+4 has three isolated singularities: 2i, −2iand ∞. •2iand −2iare simple poles of f. •Because lim x→∞ x∈R f(x) = lim x→∞ x∈R ex x2+ 4 l’H. = lim x→∞ x∈R ex 2x l’H. = lim x→∞ x∈R ex 2=∞, f(2nπi) = e2nπi (2nπi)2+ 4 =1 −4n2π2+ 4 →0, lim z→∞ f(z)does not exist, and therefore ∞is an essential singularity of f. f) The function f(z) = z2+4 ezhas only one isolated singularity, which is ∞. •Because lim x→∞ x∈R f(x)=0, lim x→−∞ x∈R f(x) = ∞·∞=∞, the limit lim z→∞ f(z)does not exist. From this fact, it follows that ∞is an essential singularity of f. 95 g) Because f(z) = 1−ez 2+ez, and at the same time 2 + ez= 0 ⇔z= Ln(−2) = ln 2 + (2k+ 1)πi =: zk, k ∈Z, fhas isolated singularities precisely in the points zk. •Furthermore [(2 + ez)′]z=zk= [ez]z=zk=−2 = /0, [1 −ez]z=zk= 3 = /0, and therefore zk= ln 2 + (2k+ 1)πi,k∈Z, are simple poles of f. Be careful: ∞is not an isolated singularity of f. h) f(z) = e1 z2and for every z∈C ∖ {0}we have that f(z) = ∞ ∑︂ n=0 1 n! 1 z2n. From this, it follows that •0is an essential singularity of f, •∞is a removable singularity of f. i) f(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, the function fhas isolated singularities in the points 3and zk,k∈Z. •Easily we can compute that [(2 −cos z)′]z=zk= [sin z]z=zk= /0, and therefore fhas in the points zk= 2kπ −iln(2 ±√3), where k∈Z, simple poles. •It is clear that 3is a pole of the order 2 of the function f. (∞is not an isolated singularity of f.) 96 j) The function f(z) = z sin zhas clearly isolated singularities in the roots of the function sinus. Think through the fact that •0is a removable singularity of f, •kπ, where k∈Z ∖ {0}, are simple poles of f. (∞is not an isolated singularity of f.) k) The function f(z) = z2sin z z+1 has precisely two isolated singularities: −1and ∞. •−1is an essential singularity of f(because lim z→−1f(z)does not exist), •∞is a pole of the order two of f(because lim z→∞ f(z) z2= lim z→∞ sin (︁z z+1)︁= sin 1 = /0). l) The function f(z) = 1−cos z sin2zclearly has isolated singularities in the roots of the function sinus. •Because5 lim z→2kπ 1−cos z sin2z l’H. = lim z→2kπ sin z 2 sin zcos z=1 2, we have that the points 2kπ, where k∈Z, are removable singularities of f. •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, the points (2k+ 1)π, where k∈Z, are poles of the order 2 of the function f. (∞is not an isolated singularity of f.) Exercise 56. Prove the L’Hôpital’s rule: Let fand gbe holomorphic, non-constant functions on some ring neighborhood of a point z0∈Cand let lim z→z0 f(z) = lim z→z0 g(z)=0. Then we have that lim z→z0 f(z) g(z)= lim z→z0 f′(z) g′(z). Solution: From the assumptions it follows that there are p, q ∈N, neighbourhood U(z0)of z0and functions f1and g1, which are holomorphic and non-zero on U(z0)such that for every z∈U(z0) ∖ {z0}we have that f(z) = (z−z0)pf1(z), g(z) = (z−z0)qg1(z). 5We are using L’Hôpital’s rule proven in the following exercise. 97 Therefore lim z→z0 f(z) g(z)= lim z→z0 (z−z0)p−qf1(z) g1(z)=⎧ ⎪ ⎨ ⎪ ⎩ ∞, p < q, 0, p > q, f1(z0) g1(z0), p =q, and lim z→z0 f′(z) g′(z)= lim z→z0 p(z−z0)p−1f1(z)+(z−z0)pf′ 1(z) q(z−z0)q−1g1(z)+(z−z0)qg′ 1(z)= = lim z→z0 (z−z0)p−qpf1(z)+(z−z0)f′ 1(z) qg1(z)+(z−z0)g′ 1(z)= =⎧ ⎪ ⎨ ⎪ ⎩ ∞, p < q, 0, p > q, f1(z0) g1(z0), p =q. The theorem is proven. Exercise 57. Compute the residue of the function fin all of its isolated singularities, where a) f(z) := 1 z+z3; b) f(z) := z2 (1+z)3; c) f(z) := 1 (z2+1)3; d) f(z) := z3+1 z−2; e) f(z) := 1 z6(z2+1)2; f) f(z) := tan z; g) f(z) := 1 sin z; h) f(z) := cotg3z; i) f(z) := sin z·sin 1 z; j) f(z) := sin(πz) (z−1)3. Solution: a) The function f(z) = 1 z+z3=1 z(z−i)(z+i)has clearly four isolated singularities: 0, i, −ia∞. Now we will use (as in several following exercises) the [1, Theorem 44, part (iii)]: •res f(0) = [︁1 1+3z2]︁z=0 = 1, •res f(i) = [︁1 1+3z2]︁z=i=1 1−3=−1 2, •res f(−i) = [︁1 1+3z2]︁z=−i=1 1−3=−1 2 and [1, Theorem 44, part (v)]: •res f(∞) = −(︁1−1 2−1 2)︁= 0. 98 b) The function f(z) = z2 (1+z)3clearly has two isolated singularities. •−1is a pole of the order three of f, and therefore res f(−1) = 1 2[︁(z2)′′]︁z=−1= 1. •res f(∞) = −1. c) The function f(z) := 1 (z2+1)3has three isolated singularities: poles of the order three at the points iand −iand a removable singularity at ∞. •res f(±i) = 1 2[︃(︃ 1 (z±i)3)︃′′]︃z=±i =1 2[︃3·41 (z±i)5]︃z=±i = = 6 1 (±2i)5=∓3 16i, •res f(∞) = 0. d) The function f(z) = z3+1 z−2has two isolated singularities: 2(simple pole) and ∞. • res f(2) = [︃z3+ 1 1]︃z=2 = 9, •res f(∞) = −9. e) Because f(z) = 1 z6(z2+ 1)2=1 z6(z+i)2(z−i)2, the numbers ±iare poles of the order two of f,0is a pole of the order 6 of fand ∞is a removable singularity of f. •res f(±i) = [︃(︃ 1 z6(z±i)2)︃′]︃z=±i = =−[︃6z5(z±i)2+z62(z±i) z12(z±i)4]︃z=±i = =±7 4i. •Because f(z)=1:(z6+ 2z8+z10) = 1 z6+···,6we have that res f(∞) = 0, •res f(0) = −7 4i+7 4i−0=0. f) The function f(z) = tan z=sin z cos zhas simple poles in the points π 2+kπ, where k∈Z, and res f(︂π 2+kπ)︂=[︃sin z −sin z]︃z=π 2+kπ =−1. (∞is not an isolated singularity of f.) 6In the Laurent series of fthe coefficient of 1 zis equal to 0. 99 g) f(z) = 1 sin zhas simple poles in the points kπ, where k∈Z, and res f(kπ) = [︃1 cos z]︃z=kπ = (−1)k. h) f(z) = cotg3z=cos3z sin3zhas poles of the order three in the points kπ, where k∈Z, and cos z sin z=(︃1−z2 2+z4 24 −···)︃:(︃z−z3 6+z5 120 −···)︃=1 z−z 3−z3 45 −··· , and therefore (︂cos z sin z)︂3=(︃1 z−z 3−z3 45 −···)︃(︃1 z−z 3−z3 45 −···)︃(︃1 z−z 3−z3 45 −···)︃. res f(0) is a “coefficient of 1 z”, therefore res f(0) = 3 (︃−1 3)︃=−1. Because the function fhas a period π, that is f(z) = f(z−kπ), we have that res f(kπ) = res f(0) = −1for every k∈Z. i) f(z) = sin z·sin 1 zhas an isolated singularity at 0and at ∞. Because for every z∈C ∖ {0}we have that f(z) = (︄∞ ∑︂ n=0 (−1)nz2n+1 (2n+ 1)!)︄(︄∞ ∑︂ k=0 (−1)k1 (2k+ 1)! ·1 z2k+1 )︄=··· , we have that7 res f(0) = res f(∞)=0. j) The function f(z) = sin(πz) (z−1)3has two isolated singularities: 1and ∞. •Because 1is a pole of the order two of f, res f(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 Laurent series of fhas non-zero coefficients only for the “even powers” of z. 100 Other possible solution: f(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 just computed Laurent series of the function fhas non-zero coefficients only for the “even powers” of (z−1), and therefore res f(1) = 0. •res f(∞) = 0. Exercise 58. Using the residue theorem compute the integrals a) ∫︂γ cos z z3dz, where γ(t) := 3eit, t ∈ ⟨0,2π⟩; b) ∫︂γ 1 z+ 2 cos 1 zdz, where γ(t) := 18eit, t ∈ ⟨0,2π⟩; c) ∫︂k z3 z4−1dz, where k={z∈C:|z|= 2}; d) ∫︂k z3 z+ 1e1 zdz, where k={z∈C:|z|= 2}; e) ∫︂γ zsin z+ 1 z−1dz, where γ(t) := 2e−it, t ∈ ⟨0,6π⟩; f) ∫︂γ eπz 2z2−idz, where γis a simple, closed, piecewise smooth positively oriented curve such that intγ={z∈C:|z|<1∧0<arg z < π 2}; g) ∫︂k dz z5(z10 −2),where k={z∈C:|z|= 2}. 101 c) Because the problem z6+ 1 = 0 ∧Im z≥0 has exactly three solutions: z1:= eiπ 6=√3 2+1 2i, z2:= eiπ 2=i, z3:= e5 6πi =−√3 2+1 2i, and the function x4+1 x6+1 is even, we have that ∫︂∞ 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 res z=zj z4+ 1 z6+ 1, where k⊂Cis the boundary of the set {z∈C:|z|<2∧Im z > 0}. 1−1 k Therefore, because res z=z1 z4+ 1 z6+ 1 =[︃z4+ 1 6z5]︃z=√3 2+1 2i =1 6(−i), res z=z2 z4+ 1 z6+ 1 =[︃z4+ 1 6z5]︃z=i =2 6(−i), res z=z3 z4+ 1 z6+ 1 =[︃z4+ 1 6z5]︃z=−√3 2+1 2i =1 6(−i), we have that ∫︂∞ 0 x4+ 1 x6+ 1 dx=2 3π. 108 d) The function x2 (x2+1)3is even, and therefore for k⊂C, which is the boundary of the set {z∈C:|z|<2∧Im z > 0}, 1−1 i k we have that ∫︂∞ 0 x2dx (x2+ 1)3=1 2∫︂∞ −∞ x2dx (x2+ 1)3=1 2∫︂k z2dz (z2+ 1)3= =1 22πi res 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) Let γ(t) := eit, where t∈ ⟨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’ve used the change of variables eix =z, see [1, Chapter 9.3, part a)]). Because z2+ 3iz −1=0 ⇔z=−3±√5 2i, ⟨γ⟩ 1 −3−√5 2i −3+√5 2i 109 we have that ∫︂π −π cos x 3 + 2 sin xdx=1 2∫︂γ z2+ 1 z 1 z2+ 3iz −1 ⏞ ⏟⏟ ⏞ =:f(z) dz= =1 22πi (︄res f(0) + res f(︂−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. f) We will use the change of variables eix =z, cos x=z+1 z 2, cos 2x=z2+1 z2 2, dx=1 iz dz. For γ(t) := eit, where t∈ ⟨0,2π⟩, we have that ∫︂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)︁ ⏞ ⏟⏟ ⏞ =:f(z) dz= =2πi 4i(︃res f(0) + res f(︃1 2)︃)︃. ⟨γ⟩ 1 2 1 2 110 and because res f(0) = 1 3! [︃(︃ (z4+ 1)2 5z−2z2−2)︃′′′]︃z=0 =1 6(︃−255 8)︃=−255 48 , res f(︃1 2)︃=[︃(z4+ 1)2 z4 1 5−4z]︃z=1 2 =289 48 , we have that ∫︂2π 0 cos22x 5−4 cos xdx=17 48π. g) Because the equation z6+ 1 = 0 has, assuming Im z≥0, exactly three solutions: z1=eiπ 6=√3 2+1 2i, z2=eiπ 2=i, z3=e5 6πi =−√3 2+1 2i, for the function f(z) := 1 1+z6we have that ∫︂∞ −∞ dx 1 + x6= 2πi (︂res f(z1) + res f(z2) + res f(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, for k⊂Cdefined as the boundary of the set {z∈C:|z|<2∧Im z > 0}we have that ∫︂∞ −∞ dx x2+x+ 1 =∫︂k dz z2+z+ 1 = = 2πi res 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 Bibliography [1] J. Bouchala, M. Lampart: An Introduction to Complex Analysis, 2024, am.vsb.cz/bouchala. [2] J. Bouchala: Funkce komplexní proměnné, 2012, am.vsb.cz/bouchala. [3] J. Bouchala (and O. 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