Math 185: Complex Analysis Fall 2009/10 Problem Set 10 Solutions

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MATH 185: COMPLEX ANALYSIS FALL 2009/10 PROBLEM SET 10 SOLUTIONS

1. (a) Show that if f has a pole or an essential singularity at a, then ef has an essential singularity at a. Solution. If f has a pole of order m at a, then there exists ε > 0 and g : D(a, ε) → C analytic, g(a) 6= 0, such that g(z) f(z) = (z − a)m for all z ∈ D∗(a, ε). Let the power series representation of g on D(a, ε) be ∞ X n g(z) = an(z − a) . n=0 Let ∞ X n h(z) := an+m(z − a) . n=0 Then1 m−1 X an f(z) = h(z) + (z − a)m−n n=0 and so m−1 Y an ef(z) = eh(z) e (z−a)m−n =: eh(z)F (z). n=0 Note that eh(z) is analytic and non-zero. If ef(z) has a pole or a removable singularity at a, then F (z) = ef(z)e−h(z) will have a pole or a removable singularity at a. So since F (z) has an essential singularity at a, ef(z) must have an essential singularity at a. ∗ If f has an essential singularity at a, then f(D (a, ε)) is dense in C for all ε > 0, ie. f(D∗(a, ε)) = C (here S denotes the closure of the set S). Recall that if g is any continuous ∗ function, then g(S) ⊇ g(S). Since exp : C → C is a continuous function, let S = f(D (a, ε)) and we have × exp(S) ⊇ exp(S) = exp(C) = C . ∗ Hence, (exp ◦f)(D (a, ε)) is dense in C. By part (a), a is an essential singularity of exp ◦f. (b) Let Ω ⊆ C be a region. Let a ∈ Ω and f :Ω\{a} → C be a function with an isolated singularity at a. Suppose for some m ∈ N and ε > 0, Re f(z) ≤ −m log|z − a| for all z ∈ D∗(a, ε). Show that a is a removable singularity of f. Solution. Note that the condition implies that 1 |ef(z)| = eRe f(z) ≤ e−m log|z−a| = . |z − a|m

Date: December 5, 2009. 1Now that you have learnt about Laurent series and its relation with poles, you could of course just write this down without going through the preceding arguments.

1 Hence |(z − a)mef(z)| ≤ 1 for all z ∈ D∗(a, ε) and thus lim|(z − a)m+1ef(z)| ≤ lim|z − a| = 0. z→a z→a So a must either be a pole (of order k ≤ m) or a removable singularity of ef . By part (a), a cannot be a pole nor an essential singularity of f (otherwise a will be an essential singularity of ef , contradicting the previous statement). Hence a must be a removable singularity of f.

∗ 2. (a) Let f : D (0, 1) → C be analytic. Show that if 1 |f(z)| ≤ log |z| for all z ∈ D∗(0, 1), then f ≡ 0. Solution. Clearly f has an isolated singularity (which may be removable) at 0. Let the Laurent expansion of f in D∗(0, 1) be ∞ X n f(z) = anz . n=−∞ Then by the integral formula for Laurent coeﬃcients, 1 Z f(z) an = n+1 dz 2πi Γr z it where Γr is given by z : [0, 2π] → C, z(t) = re , and 0 < r < 1. Hence 1 Z 2π |f(reit)ireit| |an| ≤ it n+1 dt 2π 0 |re | 1 Z 2π |f(reit)| = n dt 2π 0 r 1 1 1 Z 2π ≤ n log it dt r |re | 2π 0 1 1 = log . rn r Since this true for all 0 < r < 1, for n = −1, −2, −3,..., we may take limit as r → 0+ to get −n 1 |an| ≤ lim r log = 0 r→0+ r (note that this only works when −n is positive). Hence an = 0 for all n < 0 and so ∞ X n f(z) = anz n=0

must have a removable singularity at 0. Upon deﬁning f(0) = a0, we may assume that f : D(0, 1) → C is an analytic function. Now we may apply maximum modulus theorem to see that for all 0 < r < 1, 1 1 max|f(z)| = max|f(z)| ≤ max log = log . |z|≤r |z|=r |z|=r |z| r Hence 1 max|f(z)| ≤ lim log = 0, |z|<1 r→1− r

2 and so f ≡ 0 on D(0, 1). × × (b) Let f : C → C be analytic on C with a pole of order 1 at 0. Show that if f(z) ∈ R for × all |z| = 1, then for some α ∈ C and β ∈ R, 1 f(z) = αz + α + β z × for all z ∈ C . Solution. Since f has a pole of order 1 at 0 and is analytic otherwise, the Laurent expansion of f takes the form ∞ X n f(z) = anz . (2.1) n=−1

iθ iθ Let Γ be the closed curve z : [0, 2π] → C, z(θ) = e . Since f(e ) ∈ R for all θ ∈ [0, 2π], we must have f(eiθ) = f(eiθ).

The integral formula for Laurent coeﬃcients yields, for any n ∈ Z, 1 Z f(z) an = n+1 dz 2πi Γ z 1 Z 2π = f(eiθ)e−inθ dz 2π 0 1 Z 2π = f(eiθ)e−inθ dz 2π 0 1 Z 2π = f(eiθ)einθ dz. 2π 0 Note that Z 2π 1 iθ inθ a−n = f(e )e dz 2π 0 and so we get

an = a−n

for all n ∈ Z. By (2.1), a−n = 0 for all n > 1. So an = 0 for all n > 1. Let a1 = α and so a−1 = a1 = α. Let a0 = β. Then (2.1) becomes 1 f(z) = αz + α + β. z We know that α 6= 0 since f has a simple pole at 0. We also know that β ∈ R since β = a−0 = a0 = β. 3. Evaluate the integral Z fi Γi for i = a, b. × (a) fa : C → C is given by 1 e z fa(z) = e

and Γa is the boundary ∂D(0, 2) traversed once counter-clockwise.

3 × Solution. fa is analytic in C and its Laurent expansion about z = 0 may be obtained as follows: 1 e z 1 1 2 1 n e = 1 + e z + e z + ··· + e z + ··· 2! n! " # 1 1 1 1 2 1 22 = 1 + 1 + + + ··· + 1 + + + ··· z 2! z2 2! z 2! z 1 n 1 n2 + ··· + 1 + + + ··· + ··· . n! z 2! z Observe that the coeﬃcient of the term z−1 is simply 2 3 n 1 1 1 0 + 1 + + + ··· + + ··· = 1 + + + ··· + + ··· 2! 3! n! 1! 2! (n − 1)! = e. By the residue theorem Z fa = 2πi Res(fa; 0) Ind(Γa; 0) = 2πei. Γa ∗ (b) fb : D (0, π) → C is given by 1 f (z) = b (sin z)3

and Γb is the boundary ∂D(0, 1) traversed once counter-clockwise. ∗ Solution. fb is analytic in D (0, π) and its Laurant expansion about z = 0 may be obtained as follows: 1 1 1 −3 = z − z3 + z5 − · · · (sin z)3 3! 5! 1 1 1 −3 = 1 − z2 − z4 + ··· z3 3! 5! " # 1 1 1 1 1 2 = 1 + 3 z2 − z4 + ··· + 6 z2 − z4 + ··· + ··· . z3 3! 5! 3! 5! Now observe that the terms enclosed in the second parentheses onwards would all have powers at least 4 and so will not contribute to the z−1 term. The only term in the ﬁrst parentheses that contribute to the z−1 term is the z2 term, which has coeﬃcient 3/3! = 1/2. By the residue theorem, Z fb = 2πi Res(fb; 0) Ind(Γb; 0) = πi. Γb 4. (a) Let the Laurent expansion of cot(πz) on A(0; 1, 2) be ∞ X n cot(πz) = anz . n=−∞

Compute an for n < 0. Solution. Let Γ be the circle ∂D(0, r) traversed once counter-clockwise and 1 < r < 2. Note that Ind(Γ; z) = 1 for all z ∈ D(0, r), ie. the bounded component of Γ. By the integral formula for Laurent coeﬃcients, Z Z 1 cot(πz) 1 k−1 a−k = −k+1 dz = z cot(πz) dz 2πi Γ z 2πi Γ

4 for k ∈ N. For k = 1, cos(πz) zk−1 cot(πz) = cot(πz) = sin(πz) has three isolated (non-removable) singularities in the bounded component of Γ, namely, −1, 0, 1. So by the residue theorem2, 1 Z a−1 = cot(πz) dz 2πi Γ = Res(cot(πz); −1) + Res(cot(πz); 0) + Res(cot(πz); 1)

cos(πz) cos(πz) cos(πz) = + + π cos(πz) z=−1 π cos(πz) z=0 π cos(πz) z=1 3 = . π For k ≥ 2, z lim z[zk−1 cot(πz)] = lim [zk−1 cos(πz)] z→0 z→0 sin(πz) 1 πz h i = lim × lim zk−1 cos(πz) π z→0 sin(πz) z→0 = 0 In other words, 0 is a removable singularity of zk−1 cot(πz) for k ≥ 2. Note that the residue about any removable singularity is 0. So by the residue theorem1, Z 1 k−1 a−k = z cot(πz) dz 2πi Γ = Res(zk−1 cot(πz); −1) + Res(zk−1 cot(πz); 0) + Res(zk−1 cot(πz); 1)

zk−1 cos(πz) zk−1 cos(πz) = + 0 + d sin(πz) d cos(πz) dz z=−1 dz z=1 (−1)k−1 1 = + π π  0 if k is odd, = 2 if k is even, π for k ≥ 2. (b) For n = 0, 1, 2,..., compute 1 Z dz 3 2πi Γn z sin z 1 where Γn is the circle ∂D(0, rn) traversed once counter-clockwise and rn = (n + 2 )π. Solution. Note that for m 6= 0, 1 z − mπ lim (z − mπ) = lim z→mπ z3 sin z z→mπ z3 sin(z − mπ) 1 z − mπ = lim m3π3 z→mπ sin(z − mπ) 1 = m3π3

2We use result that Res(ϕ/ψ; a) = ϕ(a)/ψ0(a) if ϕ(a) 6= 0, ψ(a) = 0 and ψ0(a) 6= 0.

5 while for m = 0 1 z lim z4 = lim = 1. z→0 z3 sin z z→0 sin z −3 So the integrand z csc z has a pole of order 3 at 0 and simple poles at mπ for all m ∈ Z, m 6= 0. Since D(0, rn), the bounded component of Γn, contains {mπ | m = −n, , · · · − 1, 0, 1, . . . , n}, the residue theorem yields n 1 Z dz X 1 = Res ; mπ . 2πi z3 sin z z3 sin z Γn m=−n Now for m 6= 0,

1 1/z3 1/m3π3 (−1)m Res ; mπ = = = , z3 sin z d sin z cos(mπ) m3π3 dz z=mπ and for m = 0, 1 Res ; 0 = 0. z3 sin z For the latter, observe that the Laurent expansion of z−3 csc z contains only even powers, and so a−1 = 0. Hence n n 1 Z dz X (−1)m X (−1)m = − = 0. 2πi z3 sin z m3π3 m3π3 Γn m=1 m=1 5. (a) Does the following function have an antiderivative on A(0; 4, ∞)? z (z − 1)(z − 2)(z − 3) Solution. Let f : A(0; 4, ∞) → C be z f(z) := . (z − 1)(z − 2)(z − 3) Let Γ ⊂ A(0; 4, ∞) be the boundary of a rectangle R traversed once counter-clockwise. Since Γ ⊂ A(0; 4, ∞), we must have 1, 2, 3 ∈ R, the bounded component of Γ. So the winding numbers Ind(Γ; n) = 1 for n = 1, 2, 3. By the residue theorem 1 Z f(z) dz = Res(f; 1) + Res(f; 2) + Res(f; 3). 2πi Γ Since f has simple poles at 1, 2, 3, the required residues may be evaluated by z 1 Res(f; 1) = lim(z − 1)f(z) = lim = , z→1 z→1 (z − 2)(z − 3) 2 z Res(f; 2) = lim(z − 2)f(z) = lim = −2, z→2 z→2 (z − 1)(z − 3) z 3 Res(f; 3) = lim(z − 3)f(z) = lim = . z→3 z→3 (z − 1)(z − 2) 2 Hence, 1 Z f(z) dz = 0. 2πi Γ Note that this holds for arbitrary and thus all Γ ⊂ A(0; 4, ∞), Γ = ∂R for some R. Hence by Problem Set 9, Problem 3(a) (which is really Morera’s Theorem), f has an antiderivative on A(0; 4, ∞).

6 (b) Does the following function have an antiderivative on A(0; 4, ∞)? z2 (z − 1)(z − 2)(z − 3) Solution. Let g : A(0; 4, ∞) → C be z2 g(z) := . (z − 1)(z − 2)(z − 3) The same argument above applies but this time z2 1 Res(g; 1) = lim(z − 1)g(z) = lim = , z→1 z→1 (z − 2)(z − 3) 2 z2 Res(g; 2) = lim(z − 2)g(z) = lim = −4, z→2 z→2 (z − 1)(z − 3) z2 9 Res(g; 3) = lim(z − 3)g(z) = lim = . z→3 z→3 (z − 1)(z − 2) 2 So 1 Z g(z) dz = 1 6= 0. 2πi Γ Recall that if g has an antiderivative on A(0; 4, ∞), then the integral about any closed curve in A(0; 4, ∞) is necessarily3 0. Hence g does not have an antiderivative on A(0; 4, ∞).

3This doesn’t depend on having a simply connected region. So it works for annulus too.