COMPLEX ANALYSIS -Solved Problems 1

COMPLEX ANALYSIS - Solved Problems 1 Nicolae Cotfas, version 08 June 2024 (for future updates see https://unibuc.ro/user/nicolae.cotfas/ ) T 1 Theorem (f ± g)′ = f ′ ± g ′ (z n )′ = nz n−1 Problem 1 (f g)′ = f ′ g + f g ′ (ez )′ = ez  ′ Compute: ′ g′ f (sin z)′ = cos z = f g−f 2+i 10 2 g g2 (2+3i) +(1−i)+ 1+i +i +|3−4i|. ′ ′ ′ (f (ϕ(z)) = f (ϕ(z)) ϕ (z) (cos z)′ = − sin z. π ei 6 +e3+πi +cos(3i). T 2 Theorem (z 3 + z1 −sin z+z 2 eiz )′ . 1 2 3 1−z = 1+z+z +z + ... for |z| < 1 √ log(1+ 3 i). 2 3 z + z2! + z3! + ... for any z ez = 1+ 1! Problem 2 D4 Definition Expand each of the functions Let D ⊆ C be an open set, f : C → C, f (z) = eiz f : D −→ C be a continuous function, 1 γ : [a, b] −→ D be a path of class C 1 . g : C\{i} → C, g(z) = z−i R Rb The complex line in a power series around z0 = 0 (series of powers of z), f (z) dz = f (γ(t)) γ ′ (t)dt is integral of f along γ γ a and around z1 = −i (series of powers of (z+i)). T 3 Theorem Problem 3 R Compute the residues of the functions g : D → C is a primitive of f ⇒ γ f (z) dz = g(γ(b))−g(γ(a)) ′ (that is g = f ) 1 f : C\{0, 3i, −3i} → C, f (z) = z2 (z2 +9) R 1 If f : D → C admits a primitive and γ is closed, then f (z) dz = 0. g : C\{−i} → C, g(z) = (z 2 +1)e z+i PROBLEMS 1a 1b 1c 1d 2a 2b 3a 3b γ in the corresponding singular points. Problem 4 Compute the integrals R 4a I0 = z 2 dz, where γ0 : [0, 1] → C, γ0 (t) = t+ti. γ0 R 4b I1 = z2 (z12 +9) dz, where γ1 : [0, 2π] → C, γ1 (t) = i+3eit . γ1 5 Problem 5 Compute by using the theorem of residues 2π R 1 I = 3+sin t dt. 0 D5 Definition Let D be an open set and f : D → C holomorphic. there exists r > 0 such that an isolated z0 ∈ C\D is singular point if { z | 0 < |z−z0 | < r } ⊂ D. f (z0 ) = f ′ (z0 ) = ... = f (k−1) (z0 ) = 0 a zero of if z0 ∈ D is multiplicity k and f (k) (z0 ) 6= 0. a pole of 1 z0 ∈ C\D is order k of f if z0 is a zero of multiplicity k of f . T 4 Theorem Let D be an open set and f : D −→ C holomorphic. z0 is a an isolated singular point and { z | 0 < |z−z0 | < r } ⊂ D The number Resz0 f = a−1 is the residue of f in z0 . T 5 Theorem (x1 +y1 i)(x2 +y2 i) = (x1 x2 −y1 y2 )+(x1 y2 +x2 y1 )i x+yi = x−yi, p |x+yi| = x2 +y 2 , |z1 − z2 | = distance between z1 and z2 , |z| = distance between z and 0, def eit = cos t+i sin t (Euler’s formula), x+yi x x e = e cos y+i e sin y, z = |z| ei arg z , where − π < arg z ≤ π, cos z = 12 (eiz + e−iz ), 1 sin z = 2i (eiz − e−iz ), log z = ln |z|+i arg z, logk z = ln |z|+i(arg z+2kπ). D2 Definition Let D ⊆ C be an open set, and z0 ∈ D. A function f : D −→ C is there exists and is finite def def complex-differentiable ⇐⇒ ′ (z0 ) f (z0 ) = lim f (z)−f . z−z0 z→z0 (C-differentiable) at z0 D3 Definition f : D → C defined on an open set D is called C-differentiable if (or holomorphic function) n=−∞ for z satisfying 0 < |z−z0 | < r . def SOME DEFINITIONS AND THEOREMS D1 Definition there exists a unique Laurent series such that ∞ P ⇒ an (z−z0 )n , f (z) = f is C-differentiable at any point z0 ∈ D. If z0 pole of order k, then around z0 the function f admits an expansion of the form and a−k (z−z0 )k a−1 +a0 +a1 (z−z0 )+ ... +...+ (z−z 0)
(k−1) 1 Resz0 f = (k−1)! lim (z−z0 )k f (z) f (z) = z→z0 D6 Definition (Index of a point z0 with respect to a path). For a closed path γ not passing through z0 , def 1 R how many turns 1 n(z0 , γ) = 2πi dz shows γ z−z0 around z , γ makes. 0 D7 Definition. Let D be an open set . A closed path γ : [a, b] → D is called homotopic to zero in D if γ can be continuosly deformed inside D up to a constant path (a path having as image just a point). T 6 Theorem of Residues. For an open set D ⊆ C: R f : D −→ C holomorphic function f (z)dz = γ S set of isolated singular points ⇒ = 2πi P n(z, γ)Res f z γ path homotopic to zero in D∪S z∈S SOLUTIONS 1a 1b Problem 1 By using D1, we get (2+3i)2 = 4+12i − 9, (1−i) = 1+i, (1−i)(2+i) 3−i 2+i 1+i = (1−i)(1+i) = 2 , 10 8 2 4 2 i =i i √ = (i ) (−1) = −1, |3−4i| = 32 +42 = 5, whence 2+i +i10 +|3−4i| = 3+25i (2+3i)2 +(1−i)+ 1+i 2 . 3b By using D1, we get √ π ei 6 = cos π6 +i sin π6 = 23 + 2i , e3+πi = e3 eπi = e3 (cos π+i sin π) = −e3 , −3 3 cos(3i) = e 2+e , whence √ −3 3 π ei 6 +e3+πi +cos(3i) = 3+e2 −e + 2i . Problem 4 3 4a The function h : C → C, h(z) = z3 is a primitive of z 2 . Therefore, by using T3, we get R 3 = i−1. I0 = z 2 dz = h(γ(1))−h(γ(0)) = (1+i) 3 1c By using D1, we obtain (z 3 + z1 −sin z+z 2 eiz )′ = 3z 2 − z12 −cos z+2zeiz +iz 2 eiz . 1d By using D1, we get √ √ √ log(1+ 3 i) = ln|1+ 3 i|+i arg(1+ 3 i) = ln 2+ πi 3 . γ0 4b Problem 2 2a By using T2, around 0, we get 2 3 4 (iz) (iz) (iz) f (z) = eiz = 1+ iz 1! + 2! + 3! + 4! + ... 2 3 γ1 z iz z = 1+ iz 1! − 2! − 3! + 4! + ... for any z. By using T2, around −i, we get f (z) = eiz = ei(z+i−i) = e ei(z+i) 3 4 (i(z+i))2 = e+e i(z+i) +e (i(z+i)) +e (i(z+i)) + ... 1! +e 2! 3! 4! 3 0 4 (z+i) i(z+i) = e+e i(z+i) +e (z+i) 1! −e 2! −e 3! 4! + ... for any z. 2b By using T2, around 0, we get 1 1 1 1 = −i g(z) = z−i 1−(−iz) = i 1−(−iz) for | − iz| < 1 for |z| < 1. 1 (z+i)−2i = h = 2i 1+ g(z) =
z+i 2i = + i 1 1 1 −2i 1− z+i = 2 1− z+i 2i 2i


z+i 2 z+i 3 z+i 4 + + + 2i 2i 2i = 2i + 14 (z+i)− 8i (z+i)2 + ... ... i for z+i 2i for |z+i| < 2. 1 2 z→3i z (z+3i) 1 = lim = lim (z−3i) z2 (z−3i)(z+3i) z1 = −3i is a pole of order 1, and consequently 1 Res−3i f = 0! lim (z+3i)f (z) z→−3i 1 = lim (z+3i) z2 (z−3i)(z+3i) = lim z→−3i 0 2i 6i+eit −e−it dt 2π R 1 2 eit 6i+eit −e−it or I= (eit )′ dt. is the complex integral R 2 2 z +6iz−1 dz γ z→0 z→3i 2π R 0 <1 z1 = 3i is a pole of order 1, and consequently 1 lim (z−3i)f (z) Res3i f = 0! z→3i 2i This formula, written as 2π R 2 it ′ I = e2it +6ie it −1 (e ) dt Problem 3 3a We use T5 and D1. The singular points are 0, 3i and −3i. z0 = 0 is a pole of order 2, and consequently  ′
′ 1 lim z 2 f (z) = lim z 2 z2 (z12 +9) Res0 f = 1! z→0 z→0  ′ 1 = lim z2 +9 = lim (z−2z 2 +9)2 = 0. z→0 0 or 0 By using T2, around −i, we get 1 z−i can be written as 2π R I = 3+ 1 (eit1 −e−it ) dt I= = i (1−iz+(−iz)2 +(−iz)3 +(−iz)4 + ...) = i+z−iz 2 −z 3 +iz 4 + ... We use the theorem of residues T6. The singular points are 0, 3i and −3i, but only 0, 3i are inside the circular path γ of center i and radius 3. Therefore, by R using the solution of 3a, we obtain π I1 = z2 (z12 +9) dz = 2πi (Res0 f +Res3i f ) = − 27 . Problem 5 By using the definition D1, the expression 2π R 1 I = 3+sin t dt 4 2 We use T2 and T4. The only singular point is −i. It is not a pole. So, we have to use the Laurent series of g. By looking for a representation of z 2 +1 of the form z 2 +1 = α(z+i)2 +β(z+i)+γ we get z 2 +1 = (z+i)2 −2i(z+i). 1 Res−i g is the coefficient of z+i from the Laurent expansion of g around the point −i. By direct computation, we get   1 z+i = (z+i)2 −2i(z+i) g(z) = (z 2 +1)e i h 1 1 1 1 1 1 1 1 1+ 1! + + + + ... 2 3 4 z+i 2! (z+i) 3! (z+i) 4! (z+i)
1 1 2i = ... + 3! − 2! z+i + ... Consequently, Res−i g = 61 −i. i = 54 . 1 2 z→3i z (z−3i) along the circular path γ : [0, 2π] → C, γ(t) = eit . It can be computed by using the theorem T6. Since 2 2 z 2 + 6iz − 1 = (z+3i) √ +2 9 − 1 = (z+3i) + 8 2 =√(z+3i) − (i2 2) = (z−z √ 1 )(z−z2 ), where z1 = (2 2 − 3)i and z2 = (−2 2 − 3)i, the singular points are z1 and z2 , but only z1 is inside the domain with the frontier γ. The point z1 is a pole of order 1, and 2 2 = lim (z − z1 ) (z−z1 )(z−z Resz1 z2 +6iz−1 2) z→z1 2 2 = z1−z 2 z→z1 z−z2 = lim = Therefore, π 2 √ =√ I = 2πi Resz1 z2 +6iz−1 = 2πi 2−i . 2 2 i = − 54 . √2 4 2i √ . = 2−i 2