Title See discussions, stats, and author profiles for this publication at https www researchgate netpublication280722238 Complex Analysis Problems with solutions Book August 2016 CITATIONS 0 READS.
See discussions, stats, and author profiles for this publication at: https://www.researchgate.net/publication/280722238 Complex Analysis: Problems with solutions Book · August 2016 CITATIONS READS 102,190 1 author: Juan Carlos Ponce Campuzano The University of Queensland 35 PUBLICATIONS 16 CITATIONS SEE PROFILE Some of the authors of this publication are also working on these related projects: Dynamic, interactive simulations for enhancing student learning View project All content following this page was uploaded by Juan Carlos Ponce Campuzano on 15 December 2016 The user has requested enhancement of the downloaded file Complex Analysis Problems with solutions Juan Carlos Ponce Campuzano Copyright c 2016 Juan Carlos Ponce Campuzano P UBLISHED BY J UAN C ARLOS P ONCE C AMPUZANO This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License License at: https://creativecommons.org/licenses/by-nc-sa/4.0/ First e-book version, August 2015 Contents Foreword Complex Numbers 1.1 Basic algebraic and geometric properties 1.2 Modulus 13 1.3 Exponential and Polar Form, Complex roots 16 Functions 21 2.1 Basic notions 21 2.2 Limits, Continuity and Differentiation 35 2.3 Analytic functions 39 2.3.1 Harmonic functions 45 Complex Integrals 49 3.1 Contour integrals 49 3.2 Cauchy Integral Theorem and Cauchy Integral Formula 55 3.3 Improper integrals 71 Series 75 4.1 Taylor and Laurent series 75 4.2 Classification of singularities 85 4.3 Applications of residues 91 4.3.1 Improper integrals 96 Bibliography 100 Foreword This text constitutes a collection of problems for using as an additional learning resource for those who are taking an introductory course in complex analysis The problems are numbered and allocated in four chapters corresponding to different subject areas: Complex Numbers, Functions, Complex Integrals and Series The majority of problems are provided with answers, detailed procedures and hints (sometimes incomplete solutions) Of course, no project such as this can be free from errors and incompleteness I will be grateful to everyone who points out any typos, incorrect solutions, or sends any other suggestion for improving this manuscript Contact: j.ponce@uq.edu.au 2016 Complex Numbers 1.1 Basic algebraic and geometric properties Verify that √ √ − i − i − 2i = −2i (a) (b) (2 − 3i) (−2 + i) = −1 + 8i Solution We have √ √ √ √ − i − i − 2i = − i − i + = −2i, and (2 − 3i) (−2 + i) = −4 + 2i + 6i − 3i2 = −4 + + 8i = −1 + 8i Reduce the quantity 5i (1 − i)(2 − i)(3 − i) to a real number Solution We have 5i 5i i i = = = = (1 − i)(2 − i)(3 − i) (1 − i)(5 − 5i) (1 − i) −2i Chapter Complex Numbers Show that (a) Re(iz) = − Im(z); (b) Im(iz) = Re(z) Proof Let z = x + yi with x = Re(z) and y = Im(z) Then Re(iz) = Re(−y + xi) = −y = − Im(z) and Im(iz) = Im(−y + xi) = x = Re(z) Verify the associative law for multiplication of complex numbers That is, show that (z1 z2 )z3 = z1 (z2 z3 ) for all z1 , z2 , z3 ∈ C Proof Let zk = xk + iyk for k = 1, 2, Then (z1 z2 )z3 = ((x1 + y1 i)(x2 + y2 i))(x3 + y3 i) = ((x1 x2 − y1 y2 ) + i(x2 y1 + x1 y2 ))(x3 + y3 i) = (x1 x2 x3 − x3 y1 y2 − x2 y1 y3 − x1 y2 y3 ) + i(x2 x3 y1 + x1 x3 y2 + x1 x2 y3 − y1 y2 y3 ) and z1 (z2 z3 ) = (x1 + y1 i)((x2 + y2 i))(x3 + y3 i)) = (x1 + y1 i)((x2 x3 − y2 y3 ) + i(x2 y3 + x3 y2 )) = (x1 x2 x3 − x3 y1 y2 − x2 y1 y3 − x1 y2 y3 ) + i(x2 x3 y1 + x1 x3 y2 + x1 x2 y3 − y1 y2 y3 ) Therefore, (z1 z2 )z3 = z1 (z2 z3 ) Compute 2+i ; (a) 2−i (b) (1 − 2i)4 Answer: (a) (3 + 4i)/5, (b) −7 + 24i 1.1 Basic algebraic and geometric properties Let f be the map sending each complex number z = x + yi → − x y −y x Show that f (z1 z2 ) = f (z1 ) f (z2 ) for all z1 , z2 ∈ C Proof Let zk = xk + yk i for k = 1, Then z1 z2 = (x1 + y1 i)(x2 + y2 i) = (x1 x2 − y1 y2 ) + i(x2 y1 + x1 y2 ) and hence f (z1 z2 ) = x1 x2 − y1 y2 x2 y1 + x1 y2 −x2 y1 − x1 y2 x1 x2 − y1 y2 On the other hand, x2 y2 x1 x2 − y1 y2 x2 y1 + x1 y2 = −y2 x2 −x2 y1 − x1 y2 x1 x2 − y1 y2 x1 y1 −y1 x1 f (z1 ) f (z2 ) = Therefore, f (z1 z2 ) = f (z1 ) f (z2 ) Use binomial theorem n n n n n−1 n n b abn−1 + a b + + a + n n−1 n n n−k k =∑ a b k=0 k (a + b)n = to expand √ (a) (1 + √3i)2011 ; (b) (1 + 3i)−2011 Solution By binomial theorem, 2011 √ 2011 √ k 2011 2011 k/2 k (1 + 3i)2011 = ∑ ( 3i) = ∑ i k k k=0 k=0 Since ik = (−1)m for k = 2m even and ik = (−1)m i for k = 2m + odd, √ (1 + 3i)2011 = ∑ 0≤2m≤2011 +i ∑ 2011 m (−1)m 2m 0≤2m+1≤2011 1005 = ∑ m=0 √ 2011 3m 3(−1)m 2m + 1005 √ 2011 2011 (−3)m + i ∑ (−3)m 2m m=0 2m + 4.2 Classification of singularities 87 ez − z2 Solution Since − z2 = (1 − z)(1 + z), f (z) has poles of order at and −1 Therefore, (d) f (z) = Res z=1 ez ez = − z2 (1 − z2 ) =− z=1 e and ez ez Res = z=−1 − z2 (1 − z2 ) = z=−1 2e And the principal parts of f (z) at z = and z = −1 are − e and 2(z − 1) 2e(z + 1) respectively (e) f (z) = (1 − z2 ) exp z Solution The function has a singularity at where (1 − z2 ) exp z ∞ n n=0 (n!)z = (1 − z2 ) ∑ ∞ ∞ 1 =∑ −∑ n n−2 n=0 (n!)z n=0 (n!)z ∞ ∞ 1 − − (z + z + ) ∑ n n−2 (n!)z (n!)z n=1 n=3 = 1+ ∑ ∞ ∞ 1 = −z2 − z + + ∑ − ∑ n=1 (n!)zn n=1 (n + 2)!zn ∞ 1 = −z2 − z + + ∑ − z−n n=1 n! (n + 2)! So the principal part is ∞ ∑ n=1 1 − z−n n! (n + 2)! the function has an essential singularity at and Res f (z) = z=0 − = 1! 3! Chapter Series 88 (f) f (z) = (sin z)2 Solution The function has singularities at kπ for k ∈ Z At z = kπ, we let w = z − kπ and then (−1)n w2n+1 ∑ n=0 (2n + 1)! ∞ 1 = = (sin z) (sin w)2 −1 = w (−1)n w2n ∑ (2n + 1)! n=0 = w (−1)n+1 w2n 1− ∑ n=1 (2n + 1)! ∞ = w2 −1 ∞ ∞ = ∑ w m=0 −1 m (−1)n+1 w2n ∑ n=1 (2n + 1)! ∞ ∞ + ∑ an wn n=2 So the principal part at kπ is (z − kπ)2 the function has a pole of order at kπ and Res f (z) = z=kπ (g) f (z) = − cos z z2 Solution The function has a singularity at where − cos z = 2 z z (−1)n z2n n=0 (2n)! ∞ 1− ∑ = ∞ (−1)n+1 z2n ∑ (2n)! z2 n=1 (−1)n+1 z2n−2 (2n)! n=1 ∞ = ∑ So the principal part is 0, the function has a removable singularity at and Res f (z) = z=0 4.2 Classification of singularities (h) f (z) = 89 ez z(z − 1)2 Solution The function has two singularities at and At z = 0, ez = z(z − 1) z zn ∑ n! n=0 z ∞ = ∞ ∞ ∑ (n + 1)zn n=0 + ∑ an zn n=1 So the principal part at is 1/z, the function has a pole of order at and Res f (z) = z=0 At z = 1, we let w = z − and then ez e1+w e = = z(z − 1)2 (1 + w)w2 w2 = = wn ∑ n=0 n! ∞ wn n=2 n! ∞ e w2 1+w+ ∑ e w2 + ∑ an wn ∞ ∑ (−1)nwn n=0 ∞ − w + ∑ (−1)n wn n=2 ∞ n=2 So the principal part at is e (z − 1)2 the function has a pole of order at and Res f (z) = z=1 (i) f (z) = tan z Solution The function has singularities at {cos z = 0} = {z = kπ + π/2 : k ∈ Z At z = kπ + π/2, we let w = z − kπ − π/2 and then π π cos w tan z = tan w + kπ + = tan w + =− 2 sin w Since sin w has a zero of multiplicity at w = 0, tan z has a pole of order at z = kπ + π/2 Therefore Res z=kπ+π/2 tan z = Res − w=0 cos w cos w =− sin w (sin w) and the principal part of tan z at z = kπ + π/2 is − 1 =− w z − kπ − π/2 = −1 w=0 Chapter Series 90 (j) f (z) = (1 − z2 ) sin z Solution The function has a singularity at where (−1)n 2n+1 n=0 ((2n + 1)!)z ∞ z (1 − z2 ) sin = (1 − z2 ) ∑ ∞ (−1)n (−1)n − ∑ ∑ 2n+1 2n−1 n=0 ((2n + 1)!)z n=0 ((2n + 1)!)z ∞ = ∞ (−1)n (−1)n − ∑ 2n+1 2n−1 n=0 ((2n + 1)!)z n=1 (2n + 1)!)z ∞ = −z + ∑ ∞ (−1)n (−1)n+1 − ∑ 2n+1 2n+1 n=0 ((2n + 1)!)z n=0 ((2n + 3)!)z ∞ = −z + ∑ ∞ 1 + z−2n−1 (2n + 1)! (2n + 3)! = −z + ∑ (−1)n n=0 So the principal part is ∞ 1 + z−2n−1 (2n + 1)! (2n + 3)! ∑ (−1)n n=0 Therefore, the function has an essential singularity at and Res f (z) = z=0 (k) f (z) = 1 + = 1! 3! ez z2011 Solution The function has a singularity at where ez z2011 = ∞ zn z2011 ∑ n! n=0 n−2011 z ∞ = ∑ n=0 n! = 2010 n−2011 z ∑ n=0 n! zn−2011 ∑ n! n=2011 ∞ + Therefore, the principal part of f (z) at z = is 2010 n−2011 z ∑ n=0 n! and f (z) has a pole of order 2011 and residue Res f (z) = z=0 at z = 2010! 4.3 Applications of residues 91 cos z (l) f (z) = z2 − z3 Solution The function has singularities at {z2 − z3 = 0} = {z = 0, 1} At z = 0, z2 − z3 has a zero of multiplicity and hence f (z) has a pole of order Suppose that the Laurent series of f (z) at z = is given by cos z z2 − z3 = a−2 a−1 + + ∑ an zn z z n≥0 Hence (z2 − z3 ) a−2 a−1 + + ∑ an zn z2 z n≥0 (−1)n z2n n=1 (2n)! ∞ = cos z = + ∑ Comparing the coefficients of and z on both sides, we obtain that a−2 = and a−1 − a−2 = and hence a−1 = a−2 = So the principal part of f (z) at z = is 1 + z2 z with residue Res f (z) = z=0 At z = 1, z2 − z3 has a zero of multiplicity and hence f (z) has a pole of order Hence Res z=1 cos z cos z = z2 − z3 (z2 − z3 ) = − cos(1) z=1 and the principal part of f (z) at z = is − 4.3 cos(1) z−1 Applications of residues Calculate 8−z dz, C z(4 − z) where C is the circle of radius 7, centre 0, negatively oriented Solution Observe that f (z) = 8−z − 2z + z 2 = = + = − z(4 − z) z(4 − z) z (4 − z) z (z − 4) Chapter Series 92 The function f has two singularities on C, z = and z = Both are inside C At z = 4, −1/(z − 4) is analytic and then Res f (z) = z=0 Similarly, since 2/z is analytic at z = 4, Res f (z) = −1 z=4 From Cauchy’s residue theorem, and considering that C is negatively oriented, we have that 8−z dz = −2πi Res f (z) + Res f (z) z=0 z=4 C z(4 − z) = −2πi(2 − 1) = −2πi Compute the integral π dθ − cos θ Solution Since 1/(2 − cos θ ) is even, π dθ = − cos θ π −π dθ − cos θ Let z = eiθ Then cos θ = (z + z−1 )/2 and dθ = −iz−1 dz Hence dθ = − cos θ dθ −π − cos θ −idz = −1 |z|=1 2z(2 − (z + z )/2) dz =i |z|=1 z − 4z + π π The function 1 √ √ = z − 4z + (z − − 3)(z − + 3) √ has a singularity in |z| < at z = − Therefore, dz |z|=1 z2 − 4z + = 2πi Res√ z=2− = 2πi Therefore, π dθ π =√ − cos θ z2 − 4z + 1 (z2 − 4z + 1) √ z=2− πi = −√ 4.3 Applications of residues 93 Let a, b ∈ R such that a2 > b2 Calculate the integral π dθ a + b cos θ π Answer: √ a2 − b2 Evaluate the contour integral of the following functions around the circle |z| = 2011 oriented counterclockwise: (a) ; sin z (b) 2z e − ez Solution (a) f (z) = 1/ sin z is analytic in {z = nπ : n ∈ Z} It has a pole of order one at nπ since (sin z) |z=nπ = cos(nπ) = (−1)n = So Res z=nπ 1 = = (−1)n sin z cos(nπ) Therefore, dz = 2πi ∑ Res z=nπ sin z |z|=2011 sin z |nπ|