Worked Examples in Complex Analysis

Home , Harmonic function, Holomorphic function

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 < |z|. 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 |z| > 1, their Laurent expansions are just their power 1 series expansions, and we can ﬁnd them using geometric series since |z| > 1 =⇒ | z | < 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 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 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 |z| = 1 except z = 1. n n

1 Proof. We apply Dirichlet’s test for convergence. The sequence of real numbers { n } is non-increasing and converges to 0. By the triangle inequality,

N N+1 N+1 X n z − z |z| + |z| 2 z = ≤ = 1 − z |1 − z| |1 − z| n=1

2 P∞ zn Since |1−z| 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 = = = = ∞ 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 |z| < 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 |f(z)| = 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 ∈ C. By Cauchy’s formula (integrating counterclockwise around a large circle of radius R > 2|z| centered at 0)

Proof. We seek a convenient contour to evaluate this real integral. Notice √ √ √ √ Z 0 3 Z 0 p3 Z ∞ 3 Z ∞ 3 Z ∞ 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 + (−x) 0 1 + x −∞ 1 + x 0 1 + x √ 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 → ∞ to solve for our original integral. By the trivial estimate, √ Z Z π 3 Reit iReit dt πR4/3 f(z) dz = ≤ → 0 it 2 2 arcR 0 1 + (Re ) (R − 1)

as R → ∞. Next, note f(z) has poles at ±i where only i ∈ 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 → ∞, we have

Z Z R Z ! πeπi/6 = lim f(z) = lim f(x) + f(z) R→∞ R→∞ γR −R arcR Z ∞ Z ∞ = f(x) + 0 = (1 + eπi/3) f(x) −∞ 0

πeπi/6 π Thus the value of our original integral is = √ . 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 ∈ 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 |z| > 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 | z | < 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 | z | < 1/2, or equivalently |z| > 2.

π2 X 1 8. Show that = . sin2 πz (z − n)2 n∈Z

2 Proof. Let f(z) = π and g(z) = P 1 . Since g converges nicely, it gives a holomorphic sin2 πz n∈Z (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 ∈ 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 n∈Z 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 |y| ≤ N containing R. As |y| → ∞,

π2 −4π2 X 1 = → 0 → 0 sin2 πz eπ(xi−y) + eπ(y−ix) − 2 (z − n)2 n∈Z 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 ∈ C except the four zeros ω = eπi/4, ω3, ω5, ω7 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. The coefficients have order 0, ∞, 1 at each of the four zeros of z4 + 1. Since these polygons have one segment of slope 1/2, these points have ramification index 2. To determine the ramification above ∞, use coordinates 1/z, 1/w in place of z, w and look near 0:

(1/w)2 = (1/z)4 + 1 =⇒ z4 = w2 + w2z4 =⇒ w2 = z4/(1 + z4)

Near 0, there are two distinct square roots of 1/(1+z4), so there are two distinct holomorphic functions w = iz2/(1 + z4)1/2 and w = −iz2/(1 + z4)1/2 near 0. That is, there is no ramiﬁcation above ∞. By Riemann-Hurwitz, the genus g of this ramiﬁed cover is determined by X 2 − 2g = n · (2 − 2gX ) − (ey0 − 1)

ramiﬁed y0 Thus, X 2 − 2g = 2 · (2 − 2 · 0) + (2 − 1) =⇒ 2 − 2g = 4 − 4 =⇒ g = 1 3 5 7 x0∈{ω,ω ,ω ,ω }

Spring 2016 1 1. Write the ﬁrst three non-zero terms of the Laurent expansion of f(z) = in the annulus |z| > 1. z2 − 1

Proof. By partial fraction decomposition, 1 A B Az − A + Bz + B 1 1 1 1 1 = + = =⇒ A = − ,B = =⇒ = − z2 − 1 z + 1 z − 1 z2 − 1 2 2 z2 − 1 2(z − 1) 2(z + 1)

4 1 1 Since z+1 and z−1 are analytic in the annulus |z| > 1, their Laurent expansions are just their power 1 series expansions, and we can ﬁnd them using geometric series since |z| > 1 =⇒ z < 1:

∞ n ∞ n 1 1 1 1 X 1 X 1 = = = z − 1 z 1 − 1/z z z z n=0 n=1 ∞ n ∞ n −1 1 1 1 X 1 X 1 = − = − − = − z + 1 z 1 − (−1/z) z z z n=0 n=1 Thus

∞ n ∞ n 1 1 1 1 X 1 1 X 1 1 1 1 = − = + . − = + + + ... z2 − 1 2(z − 1) 2(z + 1) 2 z 2 z z2 z4 z6 n=1 n=1

2. Determine the radius of convergence of the power series for log z at −4 + 3i.

Proof. From Cauchy theory, the power series of log z expanded at −4 + 3i converges in the largest disk centered at −4 + 3i on which there is a holomorphic function that agrees with log z. Consider

Z dw Z 1 (z + 4 − 3i) · dt L(z) = log(−4 + 3i) + = log(−4 + 3i) + γ w 0 tz + (1 − t)(−4 + 3i)

where γ is the line segment from −4 + 3i to z. By Morera’s theorem/ inspection of the parametrized integral, L(z) is holomorphic on the slit plane obtained by removing the ray connecting 0 and 4−3i from C. We note that the largest open disk on this half plane centered at −4 + 3i has radius | − 4 + 3i| = 5, and L(z) certainly agrees with log z on this disk. Any disk of holomorphy for a function agreeing with log z cannot contain 0 since log 0 makes no sense. Since we cannot do better with another holomorphic function, we conclude that the radius of convergence for the power series expanded at −4 + 3i is 5.

Z ∞ dx 3. Evaluate 2 4 . −∞ 4 + 5x + x

1 1 Proof. Let f(z) = = for complex z and γ the simple closed path from 4 + 5z2 + z4 (z2 + 1)(z2 + 4) R −R to R along the real line then 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

We will evaluate the integral over γR using the Residue Theorem and take the limit as R → ∞ to solve for our original integral. By the trivial estimate, Z dz 1 1 2 4 ≤ length(arcR) · sup 2 4 ≤ πR · 4 → 0 arcR 4 + 5z + z z∈arcR 4 + 5z + z R as R → ∞. Next note f(z) has poles at ±i, ±2i where only i, 2i ∈ H. By residues,

Z X 1 1 1 1 π f(z) = 2πi · Res(f, z) = 2πi · + = 2πi · − = 4i3 + 10i 4(2i)3 + 10(2i) 6i 12i 6 γR z=i,2i

5 for any R > 1. Taking the limit as R → ∞, we have ! π Z Z R Z = lim f(z) = lim f(x)dx + f(z) R→∞ R→∞ 6 γR −R arcR Z ∞ dx = 2 4 + 0 −∞ 4 + 5x + x π Thus the value of our original integral is 6 . p 4. Show that a holomorphic function f on C satisfying |f(z)| ≤ |z| for all z ∈ C is 0. p Proof. Let f be entire such that |f(z)| ≤ |z| for all z ∈ C. Since f is entire, its power series expansion about 0 converges for all z ∈ C: X f (n)(0) f(z) = zn (for all z ∈ ) n! C n≥0 By Cauchy’s formula, integrating counterclockwise over the circle of radius R > 0 centered at 0, √ √ n! Z f(w) dw n! Z 2π |f(Reit)| · |iReit| dt 1 Z 2π R · R dt R f (n)(0) = ≤ ≤ = n+1 it n+1 n+1 n 2πi |w|=R w 2π 0 |Re | 2π 0 R R For n > 0, this goes to 0 as R → ∞. Since all by the constant term vanish in the power series expansion, f is constant. Furthermore, |f(0)| ≤ 0, so f ≡ f(0) = 0.

5. Let f, g be two holomorphic functions on |z| < 2 with f not identically 0. Show that, for ε > 0 suﬃciently small, f − εg has the same number of zeros inside |z| = 1 as does f.

Proof. Since f is holomorphic and not identically 0, there is some 0 ≤ δ < 1 such that f is nonzero on 1 < |z| ≤ 1 + δ by the identity principle. By continuity of f and g on circles, m = min1<|z|≤1+δ |f| and M = max1<|z|≤1+δ |g| are finite. If g is identically zero on 1 < |z| ≤ 1 + δ, then f − gε = f on |z| < 1 for any ε by the identity principle, and our claim follows immediately. Thus assume M 6= 0 and let m ε = 2M > 0. By choice of δ, f is non-vanishing on |z| = 1 + δ and satisfies m m |f − (f − εg)| = ε|g| ≤ M = < |f| 2M 2 By Rouché’sTheorem, f and f − εg have the same number of zeros inside |z| = 1 + δ, and since εg 6= f by the inequality above, f and f − εg have no zeros on 1 < |z| < 1 + δ. Thus f and f − εg have the same number of zeros inside |z| = 1.

6. Show that there is a holomorphic function f(z) on |z| > 2 so that f(z)2 = (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) 2 satisfying (e 2 ) = 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

2 2 1/2(L (1− 1 )+L (1+ 1 )+L (1− 2 )+L (1+ 2 )) 4 1 1 2 2 z e 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

6 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 | z | < 1. A similar argument applies to L2,L3, and L4, so there is a holomorphic 2 1/2(L (1− 1 )+L (1+ 1 )+L (1− 2 )+L (1+ 2 )) f(z) = z e 1 z 2 z 3 z 4 z 2 2 2 1 such that f(z) = (z − 1)(z − 4) on | z | < 1/2, or equivalently |z| > 2.

7. Describe all complex-valued harmonic functions on the punctured disk 0 < |z| < 1 (PG Example 8.3 ).

Proof. Let u(z) be a harmonic function on 0 < |z| < 1. We can express u in polar coordinates as a Fourier series in θ with coeﬃcients cn(r):

iθ X inθ u(re ) = cn(r)e n∈Z Applying the Laplacian in polar coordinates to the Fourier series and diﬀerentiating termwise,

X ∂2 1 ∂ 1 ∂2 X 1 n2 X ∆u = + + c (r)einθ = c00 + c0 − c · einθ = 0 = 0 · einθ ∂r2 r ∂r r2 ∂θ2 n n r n r2 n n∈Z n∈Z n∈Z since u is harmonic. By uniqueness of Fourier expansions,

1 n2 c00 + c0 − c = 0 n r n r2 n α This ordinary diﬀerential equation is of Euler type, so take cn = r and solve for α:

1 n2 (rα)00 + (rα)0 − rα =⇒ α(α − 1) + α − n2 = 0 (the indicial equation) r r2 with α = ±n. For n 6= 0, the solutions are rn and r−n, and since the root α = 0 is doubled, the solutions for n = 0 are r0 = 1 and r0 · log2−1 r = log r (see PG Example 8.6 ). Taking linear combinations of these solutions and changing back to z, z coordinates,

iθ X n −n inθ u(z) = u(re ) = a0 + b0 log r + (anr + bnr )e 06=n∈Z X n −n = a0 + b0 log |z| + anz + bnz¯ 06=n∈Z

P n −n Thus, for an’s and bn’s with suﬃciently rapid decay, f(z) = a0 + b0 log |z| + anz + bnz¯ 06=n∈Z converges nicely to a harmonic function on the punctured disk.

8. Show that the curve {(z, w) ∈ C2 : z3 + w3 = 1} (with points at inﬁnity added as needed) is an elliptic curve, that is, has genus g = 1.

Proof. This ramified covering of P1 by (z, w) → z is of degree 3. There are 3 distinct local cube root functions w above all z ∈ C except the three zeros 1, ω, ω2 of 1 − z3 since there is no cube root function on a neighborhood of 0. We claim these points are totally ramified (i.e. ramification index e = 3). Consider the Newton polygon of w3 + (z3 − 1) with order taken with respect to z. The coefficients have order 0, ∞, 1 at each of the three zeros of 1 − z3. Since these polygons have one segment of slope 1/3, these points have ramification index 3. To determine the ramification above ∞, use coordinates 1/z, 1/w in place of z, w and look near 0:

(1/w)3 = 1 − (1/z)3 =⇒ z3 = w3z3 − w3 =⇒ w3 = z3/(z3 − 1)

7 Near 0, there are three distinct roots of 1/(z3 − 1), so there are three distinct holomorphic functions w = z/(z3 − 1)1/3, w = ωz/(z3 − 1)1/3, w = ω2z/(z3 − 1)1/3 near 0. That is, there is no ramiﬁcation above ∞. By Riemann-Hurwitz, the genus g of this ramiﬁed cover is determined by X 2 − 2g = n · (2 − 2gX ) − (ey0 − 1)

ramiﬁed y0 Thus, X 2 − 2g = 3 · (2 − 2 · 0) + (3 − 1) =⇒ 2 − 2g = 6 − 6 =⇒ g = 1 2 x0∈{1,ω,ω }

Fall 2015 1 1. Write three terms of the Laurent expansion of f(z) = in the annulus 0 < |z| < 1. z(z − 1)(z + 1)

Proof. By partial fraction decomposition,

1 A B C Az2 − A + Bz2 + Bz + Cz2 − Cz 1 = + + = =⇒ A = −1,B = C = z(z − 1)(z + 1) z z − 1 z + 1 z(z − 1)(z + 1) 2 1 1 1 1 =⇒ = − + + z(z − 1)(z + 2) z 2(z − 1) 2(z + 1)

1 1 1 Since z , z−1 , and z+1 are analytic in the annulus 0 < |z| < 1, their Laurent expansions are just their 1 power series expansions. In particular, z is its own power series expansion, and since |z| < 1, we can use geometric series to ﬁnd the remaining power series expansions:

∞ 1 1 X = − = − zn z − 1 1 − z n=0 ∞ 1 1 X = = (−z)n z + 1 1 − (−z) n=0 Thus ∞ ∞ 1 1 1 1 1 1 X 1 X 1 = − + + = − − zn + (−z)n = − − z − z3 − ... z(z − 1)(z + 2) z 2(z − 1) 2(z + 1) z 2 2 z n=0 n=0

2. Determine the radius of convergence of the power series for log z expanded at −4 + 3i.

Z dw Z 1 (z + 4 − 3i) · dt L(z) = log(−4 + 3i) + = log(−4 + 3i) + γ w 0 tz + (1 − t)(−4 + 3i) where γ is the line segment from −4 + 3i to z. By Morera’s theorem/ inspection of the parametrized integral, L(z) is holomorphic on the slit plane obtained by removing the ray connecting 0 and 4−3i from C. We note that the largest open disk on this half plane centered at −4 + 3i has radius | − 4 + 3i| = 5, and L(z) certainly agrees with log z on this disk. Any disk of holomorphy for a function agreeing with log z cannot contain 0 since log 0 makes no sense. Thus, we cannot do better with another holomorphic function and conclude that the radius of convergence for the power series expanded at −4 + 3i is 5.

8 3. Show that a real-valued holomorphic function is constant (PG Example 4.7 ).

Proof. Assume f is real-valued and holomorphic in a neighborhood of z0 ∈ C. Then f is complex 0 0 diﬀerentiable at z0, and f (z0) is path independent. Consider f (z0) along the real direction and along the purely imaginary direction:

0 f(z0 + ε) − f(z0) f(z0 + iε) − f(z0) f (z0) = lim = lim (ε ∈ R) ε→0 ε ε→0 iε

Since f(z0 + ε), f(z0 + iε), and f(z0) are real, the ﬁrst limit is real, and the second limit is purely imaginary. By equality, both limits are 0 = f 0, so f is constant. √ √ 4. Give an explicit conformal mapping from X = {z : |z − 1| < 2, |z + 1| < 2} to the unit disk {z : |z| < 1}.

iθ π 3π Proof. Let Y the sector {z = re : r > 0, 4 < θ < 4 }, Q the first quadrant, H the upper half plane, and D the unit disk. We exhibit conformal maps f : X → Y , g : Y → Q, h : Q → H, j : H → D whose composition is a conformal map from X to the unit disk. z + i • The Cayley map f : z 7→ is a Möbiustransformation and therefore conformal. We know X iz + 1 √ is the intersection of the circles of radius 2 centered at ±1, and linear fractional transformations √ √ √ preserve circles/lines. Since −i 7→ 0, i 7→ ∞, 1 − 2 7→ 2 (i − 1) ∈ {Re z = − Im z}, 2 − 1 7→ √ 2 2 2 (i + 1) ∈ {Re z = Im z}, and 0 7→ i, we have f(X) = Y . −πi/4 π • g : z 7→ ze is a (clearly conformal) rotation by 4 radians to put the right edge of Y on the positive real axis (i.e. g(Y ) = Q). • h : z 7→ z2 is holomorphic with non-vanishing derivative in Q, so it is conformal. Further, 2 2 2 2 h(Q) = H since iR+ 7→ −R+ (i.e. (ai) = −a ) and R+ 7→ R+ (i.e. (a) = a ). z − i • The inverse Cayley map j : z 7→ is a Möbius transformation and therefore conformal. −iz + 1 Linear fractional transformations preserve circles/lines, and since −1 7→ −1, 0 7→ −i, 1 7→ 1, and i 7→ 0, we have j(H) = D. Thus j ◦ h ◦ g ◦ f given by −i(z+i)2 2 − i z 7→ (iz+1) −(z+i)2 (iz+1)2 + 1 is a conformal map from X to the unit disk. √ Z ∞ x dx 5. Evaluate 2 . 0 1 + x

Proof. To take the usual approach, make the change of variables x 7→ x2: √ Z ∞ x dx Z ∞ x · 2x dx Z ∞ x2 dx 2 = 4 = 4 (by symmetry). 0 1 + x 0 1 + x −∞ 1 + x

z2 Now let f(z) = for complex z and γ the simple closed path from −R to R along the real line 1 + z4 R then 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)dx + f(z) γR −R arcR

9 We will evaluate the integral over γR using the Residue Theorem and take the limit as R → ∞ to solve for our original integral. By the trivial estimate,

Z 2 2 2 z dz z R 4 ≤ length(arcR) · sup 4 ≤ πR · 4 → 0 arcR 1 + z z∈arcR 1 + z (R − 1)

as R → ∞. Next note that f(z) has poles at exactly the eighth roots of unity: ω = eiπ/4, ω3, ω5, and ω7 where only ω, ω3 ∈ H. By residues, √ √ ! Z X ω2 (ω3)2 2πi 2 2 π f(z) = 2πi · Res(f, z) = 2πi · + = · + = √ 4ω3 4(ω3)3 4 i − 1 i + 1 2 γR z=ω,ω3

for any R > 1. Taking the limit as R → ∞, we have ! π Z Z R Z √ = lim f(z) = lim f(x)dx + f(z) R→∞ R→∞ 2 γR −R arcR Z ∞ x2 dx = 4 + 0 −∞ 1 + x

Thus the value of our original integral is √π . 2

6. Let f be an entire function such that f(z + 1) = f(z) = f(z + i) for all z. Show that f is constant (PG Example 4.6 ).

Proof. Given the periodicity relations, f is determined by its values on R = {x+iy : 0 ≤ x, y ≤ 1}. Let z = x+iy any complex number. Given real x, y, there are unique integers m, n such that m ≤ x < m+1, n ≤ y < n + 1. By induction on the periodicity relations,

f(z) = f(x + iy) = f((x − m) + i(y − n))

with 0 ≤ x − m, y − n < 1, so we have a unique representative in R for any z ∈ C. Since R is compact, continuous f is bounded. Thus, bounded entire f is constant by Liouville.

7. Show that z10 − z7 + 4z2 − 1 has exactly two zeros in |z| ≤ 1.

Proof. Let γ the boundary of the unit disk, f(z) = 4z2, and g(z) = z10 − z7 + 4z2 − 1. Clearly f, g are holomorphic on |z| < 2 and f does not vanish on γ. For |z| = 1,

|f(z) − g(z)| = | − z10 − z7 + 1| ≤ |z|10 + |z|7 + 1 = 3 < |f(z)| = 4|z|2 = 4

By Rouch´e’sTheorem, g has the same number of zeros as f inside the unit disk counting multiplicities. Since f has exactly two zeros inside γ, g has exactly 2 zeros inside γ, and by the reverse triangle inequality on |z| = 1,

|4z2 − 1| ≥ |4z2| − 1 = 4|z|2 − 1 = 3 |z7 − 4z2 + 1| ≥ |4z2 − 1| − |z7| ≥ 3 − |z|7 = 2 |g(z)| ≥ |z7 − 4z2 + 1| − |z10| ≥ 2 − |z|10 = 1

Thus g has no zeros on |z| = 1, so there are exactly two zeros in |z| ≤ 1

π2 X 1 8. Show that = (see PG Product expansion of sin x). sin2 πz (z − n)2 n∈Z

10 2 Proof. Let f(z) = π and g(z) = P 1 . Since g converges nicely, it gives a holomorphic sin2 πz n∈Z (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 ∈ Z begins 1 + holomorphic (z − n)2 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 n∈Z 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 |y| ≤ N containing R. As |y| → ∞, π2 −4π2 X 1 = → 0 → 0 sin2 πz eπ(xi−y) + eπ(y−ix) − 2 (z − n)2 n∈Z Thus, h(z) → 0, and since bounded entire functions are constant by Liouville, h is identically 0.

p 9. Let f be a harmonic function on C, such that |f(z)| ≤ 1 + |z| for all z ∈ C. Show f is constant. p Proof. Let f be harmonic on C, such that |f(z)| ≤ 1 + |z| for all z ∈ C. By Poisson’s formula for harmonic functions, integrating counterclockwise over the circle of radius R > 0 centered at 0,

Z Z 2π 2 2 1 1 it R − |z| f(z) = f(w)P (w, z) dw = f(Re ) · it 2 dt (for |z| < R) 2πR |w|=R 2πR 0 |z − Re | For |z| < R/2,

Z 2π 2 2 1 it R − |z| |f(z) − f(0)| = f(Re ) it 2 − 1 dt 2πR 0 |z − Re | 1 Z 2π R2 − |z|2 ≤ |f(Reit)| · dt 2πR |z − Reit|2 0 √ √ 1 Z 2π 1 + R · R2 4 1 + R ≤ 2 dt = 2πR 0 (R/2) R

This goes to 0 as R → ∞, so f(z) = f(0) for all z ∈ C. Thus, f is constant.

Spring 2015 √ 1. Describe all values of (−1)i, where i = −1.

2 Proof. (−1)i = ei (π+2nπ) = e−π−2nπ (n ∈ Z)

1 2. Write three terms of the Laurent expansion of f(z) = in the annulus 1 < |z| < 2. z(z − 1)(z − 2)

11 Proof. By partial fraction decomposition,

1 A B C Az2 − 3Az + 2A + Bz2 − 2Bz + Cz2 − Cz = + + = z(z − 1)(z − 2) z z − 1 z − 2 z(z − 1)(z − 2) 1 1 =⇒ A = ,B = −1,C = 2 2 1 1 1 1 =⇒ = − + z(z − 1)(z − 2) 2z z − 1 2(z − 2)

1 1 1 Since z , z−1 , and z−2 are analytic in the annulus 1 < |z| < 2, their Laurent expansions are just their 1 power series expansions. In particular, 2z is its own power series expansion, and since |z| > 1 =⇒ 1 z z < 1 and |z| < 2 =⇒ 2 < 1, we can use geometric series to ﬁnd the remaining power series expansions: ∞ n ∞ n 1 1 1 1 X 1 X 1 = = = z − 1 z 1 − 1/z z z z n=0 n=1 ∞ 1 1 1 1 X z n = − = − z − 2 2 1 − z/2 2 2 n=0 Thus

∞ n ∞ 1 1 1 1 1 X 1 1 X z n 1 1 z = − + = − − = − − − + ... z(z − 1)(z − 2) 2z z − 1 2(z − 2) 2z z 4 2 2z 4 8 n=1 n=0

3. Give an explicit conformal mapping from the half-disk {z : |z| < 1, Re(z) > 0} to the unit disk {z : |z| < 1}.

Proof. Let X the given half-disk, Q the first quadrant, H the upper half plane, and D the unit disk. We exhibit conformal maps f : X → Q, g : Q → H, h : H → D whose composition is a conformal map from X to the unit disk. z + i • The Cayley map f : z 7→ is a Möbiustransformation and therefore conformal. Linear iz + 1 fractional transformations preserve circles/lines, and since −i 7→ 0, 0 7→ i, i 7→ ∞, and 1 7→ 1 we have f(X) = Q. • g : z 7→ z2 is holomorphic with non-vanishing derivative in Q, so it is conformal. Further, 2 2 2 2 g(Q) = H since iR+ 7→ −R+ (i.e. (ai) = −a ) and R+ 7→ R+ (i.e. (a) = a ). z − i • The inverse Cayley map h : z 7→ is a Möbiustransformation and therefore conformal. −iz + 1 Linear fractional transformations preserve circles/lines, and since −1 7→ −1, 0 7→ −i, 1 7→ 1, and i 7→ 0, we have h(H) = D. Thus h ◦ g ◦ f given by 2 z+i iz+1 − i z2 + 2z − 1 z 7→ 2 = 2 z+i −z + 2z + 1 −i iz+1 + 1 is a conformal map from X to the unit disk. √ 4. Determine the radius of convergence of the power series for z expanded at −4 + 3i.

12 √ Proof. From Cauchy theory, the power series of z expanded at −4 + 3i converges√ in the largest disk centered at −4 + 3i on which there is a holomorphic function that agrees with z. First we construct a holomorphic logarithm Z dw Z 1 (z + 4 − 3i) · dt L(z) = log(−4 + 3i) + = log(−4 + 3i) + γ w 0 tz + (1 − t)(−4 + 3i) where γ is the line segment from −4 + 3i to z. By Morera’s theorem/ inspection of the parametrized integral, L(z) is a holomorphic logarithm on the slit plane obtained by removing the ray connecting 0 and 4 − 3i from C. Next let S(z) = e1/2L(z) By construction, S(z) is a holomorphic square root where L(z) is holomorphic, namely on the slit plane described above. The largest open√ disk on this half plane centered at −4 + 3i has radius | − 4 + 3i| = 5, and S(z) certainly agrees with z on this disk. We will√ show that we can do no better with another holomorphic function. Suppose there was a square root z holomorphic at the origin. Then there is also a power series expansion at the origin: √ 2 2 z = c0 + c1z + c2z + ... =⇒ z = c0 + 2c0c1z + ...

Equating coeﬃcients, we must have c0 = 0 and 2c0c1 = 0 6= 1, a contradiction. Since any larger disk of holomorphy centered at −4 + 3i would contain the origin, we cannot do better with another holomorphic function. Thus the radius of convergence for the power series expanded at −4 + 3i is 5.

Z ∞ eiξxdx 5. Evaluate 2 for real ξ. −∞ 1 + x

eiξz Proof. We adjust the standard method to avoid blowups from the exponential. Let f(z) = for 1 + z2 complex z and γR the simple closed path from −R to R along the real line then from R to −R along the arc of the circle of radius R. If ξ ≥ 0, trace the semicircle counterclockwise in the upper half plane, and if ξ < 0, trace the semicircle clockwise in the lower half plane. By construction, Z Z R Z f(z) = f(x)dx + f(z) γR −R arcR

We will evaluate the integral over γR using the Residue Theorem and take the limit as R → ∞ to solve for our original integral. First we estimate the integral over the arc by the trivial estimate: Z iξz iξz −ξ Im z e dz e e 2 ≤ length(arcR) · sup 2 ≤ πR · 2 arcR 1 + z z∈arcR 1 + z R − 1 By careful construction, ξ · Im z > 0 on arcR, so this integral vanishes as R → ∞. By residues,  eiξ·i  2πi · Res(f, i) = 2πi = πe−ξ ξ ≥ 0 Z  2i f(z) = eiξ·−i γR −2πi · Res(f, −i) = −2πi = πeξ ξ < 0  −2i for any R > 1. Accommodating both signs and taking the limit as R → ∞, we have Z Z R Z ! πe−|ξ| = lim f(z) = lim f(x)dx + f(z) R→∞ R→∞ γR −R arcR Z ∞ eiξxdx = 2 + 0 −∞ 1 + x for any real ξ. Thus the value of our original integral is πe−|ξ|.

13 p 6. Show that a holomorphic function f on C satisfying |f(z)| ≤ 1 + |z| for all z ∈ C is a constant.

p Proof. Let f be entire such that |f(z)| ≤ 1 + |z| for all z ∈ C. Since f is entire, its power series expansion about 0 converges for all z ∈ C:

X f (n)(0) f(z) = zn (for all z ∈ ) n! C n≥0

By Cauchy’s formula, integrating counterclockwise over the circle of radius R > 0 centered at 0,

7. Show that 4z5 − z + 2 has all its zeros in the unit disk.

Proof. Let γ the boundary of the unit disk, f(z) = 4z5, and g(z) = 4z5 − z + 2. Clearly f, g are holomorphic on |z| < 2 and f does not vanish on γ. For |z| = 1 we have

|f(z) − g(z)| = |z − 2| ≤ |z| + 2 = 3 < |f(z)| = |4z5| = 4|z|5 = 4

By Rouch´e’sTheorem, g has the same number of zeros as f inside the unit disk counting multiplicities. Since f has exactly ﬁve zeros inside γ, g has exactly 5 zeros inside γ; furthermore, g is a 5th degree polynomial, so it has at most 5 zeros. Thus all zeros of g are in the unit disk.

sin z 8. Show that there is a holomorphic function f(z) on a neighborhood of 0 so that f(z)2 = , and z determine the radius of convergence of the power series at 0.

Proof. By Cauchy Theory, the radius of convergence of the power series for f(z) is the largest disk centered at 0 on which there is a holomorphic function agreeing with f. First observe that the singularity sin z of z at 0 is removable:

z2 z · Q 1 − 2 sin z n≥1 π2n2 Y z = = 1 − = g(z) z z π2n2 n≥1

sin z The product above deﬁnes an entire function, so the power series of z converges for all z ∈ C. For z0 6= 0, we can deﬁne a logarithm Z z dw L(z) := log(z0) + z0 w 1 L(z) 2 satisfying (e 2 ) = z and holomorphic for z such that the line segment connecting z0 and z does not pass through 0. Thus, for z0 away from nonzero multiples of π, there are holomorphic logarithms z2 Ln(1 − π2n2 ) for z near z0 such that

2 2 2 1/2 P L (1− z ) Y z e n≥1 n π2n2 = 1 − = g(z) π2n2 n≥1

14 z2 At z0 = 0, the power series for each Ln(1 − π2 ) converges absolutely on the largest open disk centered z2 at 0 on which Ln(1 − π2n2 ) is holomorphic. Since there are holomorphic logarithms on the half planes z2 Re(1− π2n2 ) > 0, there are suitable holomorphic logarithms Ln on |z| < π. Thus, there is a holomorphic

2 1/2 P L (1− z ) f(z) = e n≥1 n π2n2

2 sin z such that f(z) = z on |z| < π. Since any function agreeing with f(z) has a branch point at ±π, we can do no better with another holomorphic function. Thus, the power series of f(z) converges on |z| < π.

9. Describe all complex-valued harmonic functions on the annulus 1 < |z| < 2 which extend continuously to the circle |z| = 2 and take value 0 on that circle.

Proof. Let f(z) be a harmonic function on 1 < |z| < 2. By applying the Laplacian to its Fourier series in polar coordinates and solving an Euler type ordinary diﬀerential equation for the coeﬃcients cn(r), we can describe all complex-valued harmonic functions on an annulus (see PG Exercise 8.3 ):

iθ X inθ X n −n inθ f(re ) = cn(r)e = a0 + b0 log r + anr + bnr e n∈Z 06=n∈Z Not every harmonic function on the punctured disk satisﬁes the given boundary conditions, so we must restrict the Fourier coeﬃcients. We make an immediate restriction so f=0 on |z| = r = 2:

iθ X n −n inθ X inθ f(2 · e ) = a0 + b0 log 2 + an2 + bn2 e = 0 = 0 · e 06=n∈Z n∈Z n −n By uniqueness of Fourier expansions, a0 + b0 log 2 = an2 + bn2 = 0 for 0 6= n ∈ Z. Thus, r X z X f(z) = f(reiθ) = b log + a rn − 4nr−n einθ = b log + a zn − 4nz¯−n 0 2 n 0 2 n 06=n∈Z 06=n∈Z Since harmonic functions are continuous and therefore L2 on 0 ≤ |z| ≤ 2, their Fourier coefficients must have finite sum of squares of absolute values 1 < |z| < 2 by Parseval-Plancherel. Thus for an’s with z P n n −n sufficiently rapid decay, f(z) = b0 log + an (z − 4 z¯ ) converges nicely to a harmonic 2 06=n∈Z function on the annulus satisfying the given boundary conditions.

Fall 2014 √ 1. Describe all the values of ii, where i = −1.

i i2(π/2+2nπ) − π −2nπ Proof. i = e = e 2 (n ∈ Z)

2. Give an explicit conformal mapping that gives a bijection of the ﬁrst quadrant with the unit disk.

Proof. Let Q the ﬁrst quadrant, H the upper half plane, and D the unit disk. We exhibit bijective conformal maps f : Q → H, g : H → D whose composition g ◦ f : Q → D is a bijective conformal map from the ﬁrst quadrant to the unit disk.

• f : z 7→ z2 is bijective and holomorphic with non-vanishing derivative in Q, so it is conformal. 2 2 2 2 Further, f(Q) = H since iR+ 7→ −R+ (i.e. (ai) = −a ) and R+ 7→ R+ (i.e. (a) = a ). z − i • The inverse Cayley map g : z 7→ is a M¨obius transformation and therefore conformal. −iz + 1 Linear fractional transformations preserve circles/lines, and since −1 7→ −1, 0 7→ −i, 1 7→ 1, and i 7→ 0, we have g(H) = D.

15 z2−i Thus the conformal map g ◦ f given by z 7→ −iz2+1 gives a bijection of the ﬁrst quadrant with D.

3. What is the radius of convergence of the power series for log z expanded at −4 + 3i?

Z dw Z 1 (z + 4 − 3i) · dt L(z) = log(−4 + 3i) + = log(−4 + 3i) + γ w 0 tz + (1 − t)(−4 + 3i)

where γ is the line segment from −4 + 3i to z. By Morera’s theorem/ inspection of the parametrized integral, L(z) is holomorphic on the slit plane obtained by removing the ray connecting 0 and 4−3i from C. We note that the largest open disk on this half plane centered at −4+3i has radius |−4+3i| = 5, and L(z) certainly agrees with log z on this disk. Now, any disk of holomorphy for a function agreeing with log z cannot contain 0 since log 0 makes no sense. Thus, we cannot do better with another holomorphic function and conclude that the radius of convergence for the power series expanded at −4 + 3i is 5.

1 4. Write four terms of the Laurent expansion of f(z) = in the annulus 1 < |z|. z(z − 1)

Proof. By partial fraction decomposition, 1 A B Az − A + Bz 1 −1 1 = + = =⇒ A = −1,B = 1 =⇒ = + z(z − 1) z z − 1 z(z − 1) z(z − 1) z z − 1

1 −1 Since z−1 and z are analytic in the annulus |z| > 1, their Laurent expansions are just their power −1 1 series expansions. In particular, z is its own power series expansion, and since |z| > 1 =⇒ z < 1, 1 we can use a geometric series to ﬁnd the power series expansion of z−1 :

∞ n ∞ n 1 1 1 1 X 1 X 1 = = = z − 1 z 1 − 1/z z z z n=0 n=1 Thus

∞ n ∞ n 1 −1 1 −1 X 1 X 1 1 1 1 1 = + = + = = + + + + ... z(z − 1) z z − 1 z z z z2 z3 z4 z5 n=1 n=2

R ∞ dx 5. Evaluate −∞ 1+x4 .

1 Proof. Let f(z) = for complex z and γ the simple closed path from −R to R along the real 1 + z4 R line then 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)dx + f(z) γR −R arcR

We will evaluate the integral over γR using the Residue Theorem and take the limit as R → ∞ to solve for our original integral. First estimate the integral over the arc by the trivial estimate: Z dz 1 1 4 ≤ length(arcR) · sup 4 ≤ πR · 4 → 0 arcR z + 1 z∈arcR z + 1 (R − 1)

16 as R → ∞. Next note that f(z) has poles at exactly the primitive eighth roots of unity: ω = eiπ/4, ω3, ω5, and ω7 where only ω, ω3 ∈ H. By residues, √ √ ! Z X 1 1 2πi 2 2 π f(z) = 2πi · Res(f, z) = 2πi · + = · + = √ 4ω3 4ω9 4 i − 1 i + 1 2 γR z=ω,ω3

for any R > 1. Taking the limit as R → ∞, we have ! π Z Z R Z √ = lim f(z) = lim f(x)dx + f(z) R→∞ R→∞ 2 γR −R arcR Z ∞ dx = 4 + 0 −∞ 1 + x

Thus the value of our original integral is √π . 2

6. Classify the holomorphic functions f on C which satisfy |f(z)| ≤ (1 + |z|)2 for all z ∈ C.

Proof. Let f be entire such that |f(z)| ≤ (1 + |z|)2 for all z ∈ C. Since f is entire, its power series expansion about 0 converges for all z ∈ C:

X f (n)(0) f(z) = zn (for all z ∈ ) n! C n≥0

By Cauchy’s formula, integrating counterclockwise over the circle of radius R > 0 centered at 0,

7. Show that 3z5 − z + 1 has all its zeros in the unit disk.

Proof. Let γ the boundary of the unit disk, f(z) = 3z5, and g(z) = 3z5 − z + 1. Clearly f, g are holomorphic on |z| < 2 and f does not vanish on γ. For |z| = 1 we have

|f(z) − g(z)| = |1 − z| ≤ |z| + 1 = 2 < 3 = 3|z|5 = |f(z)|

8. Describe all complex-valued harmonic functions on the punctured disk 0 < |z| < 1 which extend continuously to the circle |z| = 1 and take value 0 on that circle.

Proof. Let f(z) be a harmonic function on 0 < |z| < 1. By applying the Laplacian to its Fourier series in polar coordinates and solving an Euler type ordinary diﬀerential equation for the coeﬃcients cn(r), we can describe all complex-valued harmonic functions on an annulus (see PG Exercise 8.3 ):

iθ X inθ X n −n inθ f(re ) = cn(r)e = a0 + b0 log r + anr + bnr e n∈Z 06=n∈Z

17 Not every harmonic function on the punctured disk satisﬁes the given boundary conditions, so we must restrict the Fourier coeﬃcients. We make an immediate restriction so f=0 on |z| = r = 1:

iθ X inθ X inθ f(1 · e ) = a0 + (an + bn) e = 0 = 0 · e 06=n∈Z n∈Z

By uniqueness of Fourier expansions, a0 = an + bn = 0 for 0 6= n ∈ Z. Thus,

iθ X n −n inθ X n −n f(z) = f(re ) = b0 log r + an(r − r )e = b0 log |z| + an(z − z¯ ) 06=n∈Z 06=n∈Z Since harmonic functions are continuous and therefore L2 on 0 ≤ |z| ≤ 1, their Fourier coefficients must have finite sum of squares of absolute values 0 < |z| < 1 by Parseval-Plancherel. Thus for an’s P n −n with sufficiently rapid decay, f(z) = b0 log |z| + an(z − z¯ ) converges nicely to a harmonic 06=n∈Z function on the punctured disk satisfying the given boundary conditions.

2 ez −1 9. Show that there is a holomorphic function f(z) on a neighborhood of 0 so that f(z) = z . What is the radius of convergence of the power series at 0 for this function?

n z P z n 2 e − 1 n≥1 X z z z = n! = = 1 + + + ... = g(z) z z (n + 1)! 2! 3! n≥0

ez −1 The product above deﬁnes an entire function, so the power series of z converges for all z ∈ C. For z0 6= 0, we can deﬁne a logarithm Z z dw L(z) := log(z0) + z0 w 1 L(z) 2 satisfying (e 2 ) = z and holomorphic for z such that the line segment connecting z0 and z does not pass through 0. Thus, for z0 away from nonzero multiples of 2πi, there is a holomorphic logarithm L(g) for z near z0 such that 2 e1/2L(g) = g(z)

At z0 = 0, the power series for L(g) converges absolutely on the largest open disk centered at 0 on which L(g) is holomorphic. Since there is a holomorphic logarithm on the half plane Re(z − 2π) > 0, there is a suitable holomorphic logarithms L(g) on |z| < 2π. Thus, there is a holomorphic

f(z) = e1/2L(g)

2 ez −1 such that f(z) = z on |z| < 2π. Since any function agreeing with f(z) has a branch point at ±2πi, we can do no better with another function. Thus, the power series of f converges on |z| < 2π.