Worked Examples in Complex Analysis Adrienne Sands April 15, 2017 This document contains sample solutions to the Fall 2014 - Fall 2016 preliminary exams. I hope that it will give you some peace of mind. Good luck! Fall 2016 1 1. Write three terms of the Laurent expansion of f(z) = in the annulus 1 < jzj. z(z − 1)(z + 1) Proof. By partial fraction decomposition, 1 A B C A(z2 − 1) + B(z2 + z) + C(z2 − z) = + + = =) A = −1;B = C = 1=2 z(z − 1)(z + 1) z z − 1 z + 1 z(z − 1)(z + 1) 1 1 1 1 =) = − + + z(z − 1)(z + 1) z 2(z − 1) 2(z + 1) 1 1 1 Since z ; z−1 ; z+1 are analytic in the annulus jzj > 1, their Laurent expansions are just their power 1 series expansions, and we can find them using geometric series since jzj > 1 =) j z j < 1: 1 1 1 1 1 1 X 1 X 1 = = = z − 1 z 1 − 1 z zn zn z n=0 n=1 1 1 1 1 1 1 X 1 X 1 = = (−1)n = (−1)n+1 z + 1 z 1 − (− 1 ) z zn zn z n=0 n=1 Thus 1 1 1 1 1 1 X 1 1 1 1 f(z) = − + + = − + 1 + (−1)n+1 = + + + ::: z 2(z − 1) 2(z + 1) z 2 zn z3 z5 z7 n=1 X zn 2. Show that converges at all points of the unit circle jzj = 1 except z = 1. n n 1 Proof. We apply Dirichlet's test for convergence. The sequence of real numbers f n g is non-increasing and converges to 0. By the triangle inequality, N N+1 N+1 X n z − z jzj + jzj 2 z = ≤ = 1 − z j1 − zj j1 − zj n=1 2 P1 zn Since j1−zj is bounded for fixed z 6= 1 on the unit circle, n=1 n converges. z 3. Determine the radius of convergence of the power series for expanded at 0. sin z 1 z Proof. By Cauchy Theory, the radius of convergence of the power series for f(z) = is the largest sin z disk centered at 0 on which there is a holomorphic function agreeing with f. By inspection of the power series of sin z, the singularity at 0 is removable: z z 1 = = =: F (z) sin z z3 z5 7 z2 z4 6 z − 3! + 5! + O(z ) 1 − 3! + 5! + O(z ) At π, f is unbounded (see PG Product expansion of sin x): π π 1 Y n2 = = = = 1 sin π Q π2 Q (1 − 1 ) n2 − 1 π · n≥1(1 − π2n2 ) n≥1 n2 n≥1 Thus, any function agreeing with f cannot be holomorphic at π, and the radius of convergence of f is strictly less than π. F (z) defined above is holomorphic for jzj < π and clearly agrees with f, so the z radius of convergence for is π. sin z 4. Show that a holomorphic function f with jf(z)j = 1 for all z is constant. Proof. Since f is bounded and entire, it is constant by Liouville. More completely, since f is entire, P (n) n its power series expansion f(z) = n≥0 f (z)z converges for all z 2 C. By Cauchy's formula (integrating counterclockwise around a large circle of radius R > 2jzj centered at 0) n! Z f(w) dw n! Z 2π jf(Reit)j jiReitj dt n! R jf (n)(z)j = ≤ ≤ ! 0 n+1 it n+1 n+1 2πi jwj=R (w − z) 2π 0 jRe − zj (R=2) for n > 0 as R ! 1. Since the power series of f consists only of the constant term, f is constant. p Z 1 3 x dx 5. Evaluate 2 . 0 1 + x Proof. We seek a convenient contour to evaluate this real integral. Notice p p p p Z 0 3 Z 0 p3 Z 1 3 Z 1 3 Z 1 3 x dx (−x)(−dx) πi=3 x dx x dx πi=3 x dx 2 = 2 = e 2 =) 2 = (1 + e ) 2 −∞ 1 + x 1 1 + (−x) 0 1 + x −∞ 1 + x 0 1 + x p 3 z dz As usual, let f(z) = and γ the simple closed path from −R to R along the real line then 1 + z2 R counterclockwise along the arc of the semicircle of radius R from R to −R in the upper half plane H. By construction, Z Z R Z f(z) = f(x) + f(z) γR −R arcR Evaluate the integral over γR using the Residue Theorem and take the limit as R ! 1 to solve for our original integral. By the trivial estimate, p Z Z π 3 Reit iReit dt πR4=3 f(z) dz = ≤ ! 0 it 2 2 arcR 0 1 + (Re ) (R − 1) as R ! 1. Next, note f(z) has poles at ±i where only i 2 H. By residues, Z X 2πi · i1=3 f(z) = 2πi · Res(f; z) = = πeπi=6 2i γR z=i 2 for any R > 1. Taking the limit as R ! 1, we have Z Z R Z ! πeπi=6 = lim f(z) = lim f(x) + f(z) R!1 R!1 γR −R arcR Z 1 Z 1 = f(x) + 0 = (1 + eπi=3) f(x) −∞ 0 πeπi=6 π Thus the value of our original integral is = p . 1 + eπi=3 3 6. Let f be holomorphic, bounded in the upper half plane H, and real-valued on R. Show that f is constant. Proof. By the reflection principle, f extends to a holomorphic function f~ on H [ R [ H = C with f~(z) = f(z) for all z 2 H. By construction f is bounded on the lower half plane, and by continuity near R, it is bounded on the real line. Thus f is bounded, entire, and therefore constant by Liouville. 7. Show that there is a holomorphic function f on the region jzj > 2 such that f(z)4 = (z2 − 1)(z2 − 4). Proof. For any z0 6= 0, we can define a logarithm Z z dw L(z) := log(z0) + z0 w 1 L(z) 4 satisfying (e 4 ) = z and holomorphic for z such that the line segment connecting z0 and z does 1 1 1 not pass through 0. Thus for z0 6= ±1; ± 2 , there are holomorphic logarithms L1 1 − z ;L2 1 + z , 2 2 1 L3 1 − z ;L4 1 + z for z near z0, and 4 1=4(L (1− 1 )+L (1+ 1 )+L (1− 2 )+L (1+ 2 )) 4 1 1 2 2 ze 1 z 2 z 3 z 4 z = z 1 − 1 + 1 − 1 + z z z z 1 4 = z4 1 − 1 − = (z2 − 1)(z2 − 4) z2 z2 1 At z0 = 0, the power series for L1(1− z ) converges absolutely on the largest open disk centered at 0 on 1 1 which L1(1− z ) is holomorphic. Since there is a holomorphic logarithm on the half plane Re(1− z ) < 0, 1 there is a holomorphic logarithm on j z j < 1. A similar argument applies to L2;L3, and L4, so there is a holomorphic 1=4(L (1− 1 )+L (1+ 1 )+L (1− 2 )+L (1+ 2 )) f(z) = ze 1 z 2 z 3 z 4 z 4 2 2 1 such that f(z) = (z − 1)(z − 4) on j z j < 1=2, or equivalently jzj > 2. π2 X 1 8. Show that = . sin2 πz (z − n)2 n2Z 2 Proof. Let f(z) = π and g(z) = P 1 . Since g converges nicely, it gives a holomorphic sin2 πz n2Z (z−n)2 function away from poles. By inspection, both functions have double poles at exactly the integers. The Laurent expansion of g near z 2 Z begins 1 + holomorphic (z − n)2 3 Consider the Laurent expansion of f near 0: π2 π2 1 1 1 1 = = · = · 2 3 2 2 2 2 2 2(πz)2 sin πz (πz) z (πz) z 1 − + ::: πz − 3! + ::: 1 − 3! + ::: 3! 1 2(πz)2 1 = · 1 + + ::: = + holomorphic z2 3! (z − 0)2 By periodicity of f, the Laurent expansions of f and g match for every integer n. Taking their difference cancels all poles, so π2 X 1 h(z) = − sin2 πz (z − n)2 n2Z is entire. We claim h is identically 0. By continuity on R, h is bounded on 0 ≤ z ≤ 1, and since h(z + 1) = h(z), h is bounded on the real line and on every band jyj ≤ N containing R. As jyj ! 1, π2 −4π2 X 1 = ! 0 ! 0 sin2 πz eπ(xi−y) + eπ(y−ix) − 2 (z − n)2 n2Z Thus, h(z) ! 0, and since bounded entire functions are constant by Liouville, h is identically 0. 9. Show that the curve described by w2 = z4 + 1, with points at infinity added as needed, has genus 1, that is, is an elliptic curve. Proof. This ramified covering of P1 by (z; w) ! z is of degree 4. There are 2 distinct local square root functions w above all z 2 C except the four zeros ! = eπi=4;!3;!5;!7 of z4 + 1 since there is no square root function on a neighborhood of 0. We claim these points are totally ramified (i.e. ramification index e = 2). Consider the Newton polygon of w2 − (z4 + 1) with order taken with respect to z.