Complex Analysis Qualifying Exam Solutions

May 21, 2014

Part 1.

1. Let log z be the principal branch of the logarithm defined on G = z ∈ C | z 6∈(−∞, 0]. Show that if t > 0, then the equation log z = t

zhas exactly one root in G.

Solution. Let z = reiθ where r > 0 and −π < θ < π. Then log z = tz

becomes

log r + iθ =t

rcos θ − i t

rsin θ.

Since θ and − tr

sin θ have opposite signs when θ 6= 0, it follows that if z ∈ G, thenlog z = t

zif and only if z = r where r > 0 satisfies log r = t

r. But log r is strictly

increasing and tr

is strictly decreasing. Hence, the Intermediate Value Theoremimplies that log r = t

rhas exactly one root in r ∈ R | r > 0. Therefore, log z = t

z

has exactly one root in G.

2. Let G = C \ [−1, 1]. Prove that f(z) = z2 − 1 has an analytic square rooton G but does not have an analytic logarithm on G.

Solution. To see that f has a square root, consider the function g defined onG by the formula

g(z) = |f(z)|12 e

12

(arg(z−1)+arg(z+1)

),

where the arguments are chosen in [0, 2π). g is a well defined continuous functionon G satisfying g2 = f . (g is analytic since z2 is locally 1-1 on C \ 0; cf.Proposition 2.20 pg. 39 Conway).

To see that f does not have an analytic logarithm on G, recall that this isequivalent to f ′

fhaving a primitive on G. But if γ is a path in G that winds once

around [−1, 1] in the counterclockwise direction, then∫γ

f ′

fdz =

∫γ

(1

z − 1+

1

z + 1) dz = 4πi 6= 0.

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Therefore, f ′

fcannot have a primitive (cf. Corollary 1.22 pg. 66 Conway).

3. Prove that the zeros of the polynomial p(z) = zn + cn−1zn−1 + . . . c1z + c0

all lie in the open disk with center 0 and radius

R =√

1 + |cn−1|2 + . . .+ |c1|2 + |c0|2.

Solution. Note that R = 1 if and only if p(z) = zn and that in this case theassertion is obviously true. Therefore, we may assume that R > 1. If |z| = R,then

|zn − p(z)| = |cn−1zn−1 + . . . c1z + c0|

≤ (|cn−1|2 + . . .+ |c1|2 + |c0|2)12 (R2(n−1) +R2(n−2) + . . .+ 1)

12

=(R2 − 1

)12

(R2n − 1

R2 − 1

)12

=(R2n − 1

)12

< Rn

= |zn|.

Therefore, by Rouche’s Theorem, the 0’s of p all lie in the open disk with center 0and radius R.

4. Let G be a connected open set in C and let a ∈ G. Prove that if F ⊆ H(G) isa normal family, then D = f ∈ H(G) | f(a) = 0 and f ′ ∈ F is a normal family.

Solution. Assume that F ⊆ H(G) is normal. That D is normal, will follow fromMontel’s Theorem if we can show that D is locally bounded. Accordingly, fixb ∈ G.

Since G is assumed connected, there exists a rectifiable path γ : [0, 1] → Gsatisfying γ(0) = a and γ(1) = b. Since F is normal and γ is compact, thereexists a constant M1 such that

∀g∈F maxt∈[0,1]

|g(γ(t))| ≤M1.

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Also, since F is normal, there exist r > 0 and M2 such that B(b, r) ⊆ G and

∀g∈F supz∈B(b,r)

|g(z)| ≤M2.

It follows that if f ∈ D and w ∈ B(b, r), then

|f(w)| =∣∣∣ ∫

γ

f ′(z) dz +

∫[b,w]

f ′(z) dz∣∣∣

≤∣∣∣ ∫

γ

f ′(z) dz∣∣∣+∣∣∣ ∫

[b,w]

f ′(z) dz∣∣∣

≤M1‖γ‖+M2r.

Thus, D is locally bounded.

5. Show that if G 6= C is a simply connected subset of C, f : G → G is ana-lytic, and f(z) is not identically equal to z, then f has at most one fixed point inG.

Solution. Recall that Problem 2 on the 220B Final Exam was to show that ifg : D → D is analytic and g(z) is not identically equal to z, then g can have atmost one fixed point in D. The simple proof of this fact was based on an applicationof Schwarz’s Lemma.

Since G is assumed simply connected and 6= C, it follows by the Riemann Map-ping Theorem that there exists an analytic bijection φ : G→ D. If we define g byg = φ f φ−1, then g : D→ D is analytic and f(z) is not identically equal to z ifand only if g(z) is not identically equal to z. Furthermore, a ∈ G is a fixed pointfor f if and only if φ(a) is a fixed point in D for g. Therefore, the assertion of thisproblem follows from the result in the previous paragraph.

6. Show that if G is an open subset of C, then there exist ideals in H(G) that arenot finitely generated.

Solution. Construct an infinite sequence of distinct points zk∞k=1 in G that hasno limit points in G. For each n ≥ 1 define

In = f ∈ H(G) | f(zk) = 0 for each k ≥ n.

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It is straightforward to prove that for each n ≥ 1, In is an ideal and also that I,defined by

I =∞⋃n=1

In,

is an ideal.We claim that I is not finitely generated. For if f1, f2, . . . , fm generate I, then,

as there exists N such that fk ∈ IN for k = 1, 2, . . . ,m, it would follow thatI = IN . On the other hand, the Weierstrass Factorization Theorem implies theexistence of an f ∈ IN+1 with f(zN) 6= 0, i.e., I 6= IN .

Part 2.

1. LetG = z ∈ C | |z| < 1 and |z − 1

3| > 2

3,

K = G− = z ∈ C | |z| ≤ 1 and |z − 13| ≥ 2

3,

and A(K) denote the space of continuous functions on K that are analytic onG equipped with the uniform norm on K. For the purposes of this problem, aLaurent polynomial ia a function of the form

∑Nn=−N anz

n and a Laurent seriesis an infinite sum of the form

∑∞n=−∞ anz

n. Determine the truth or falsity of thefollowing assertions. Be sure to justify your answers.

(a) The polynomials are dense in H(G).

(b) The polynomials are dense in A(K).

(c) If f is analytic on a neighborhood of K, then f can be uniformlyapproximated on K by a Laurent polynomial.

(d) If f ∈ H(G) then f can be represented on G by a Laurent series.

Solution to (a). This assertion is true. Fix a compact open exhaustion Kn of Gthat has the property that C \Kn is connected for each n. For example, the setsKn defined by

Kn = z ∈ C | |z| ≤ n

n+ 1and |z − 1

3| ≥ 2

3

n+ 1

n

have this property.To prove that the polynomials are dense in H(G), fix f ∈ H(G). For each n,

since C\Kn is connected, Runge’s Theorem implies that there exists a polynomial

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pn such that |f(z) − pn(z)| ≤ 1n. For such a sequence of polynomials, pn → f in

H(G).

Solution to (b). This assertion is false. Note that the function f(z) = 1z

is inA(K). If pn is a sequence of polynomials and pn → 1

zin A(K), then there exists

n such thatmaxz∈K|1z− pn(z)| < 1.

But if we set f(z) = 1 − zpn(z), as f is analytic on a neighborhood of D−, theMaximum Principle implies that

1 = |f(0)| ≤ max|z|=1|f(z)| = max

|z|=1|1z− pn(z)| < 1.

This contradiction implies that the polynomials are not dense in A(K).

Solution to (c). This assertion is true. C∞ \K has two components,

C1 = z | |z − 13| < 2

3 and C2 = z | |z| > 1 ∪ ∞.

Furthermore, 0 ∈ C1 and∞ ∈ C2. Therefore, by Runge’s Theorem, if f is analyticon a neighborhood of K, f can be uniformly approximated on K by rational func-tions R whose only poles are at 0 and ∞. But if R is a rational function whoseonly poles are at 0 and ∞, then R is a Laurent polynomial.

Solution to (d). This assertion is false. A particularly simple counterexampleis obtained by considering the function f(z) = (z − a)−1 where |a− 1

3| = 2

3. f has

two Laurent series representations,

f(z) =∞∑n=1

an−1

zn, |a| < |z|

and

f(z) = −∞∑n=0

zn

an+1, |z| < |a|.

If a 6= −13, 1, then neither of these series converge on G.

2. (a) Prove that the formula

f(z) = limα→0+

∫ π2

α

tz

sin tdt

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defines an analytic function in z ∈ C | Re z > 0.

(b) Show that if f is defined as in part (a), then f has an analytic continua-tion to z ∈ C | Re z > −1 \ 0 with a simple pole at 0 with residue 1.

Solution. (a) Let G = z ∈ C | Re z > 0. In Conway’s language, the asser-

tion is that the integral∫ π

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tz

sin tdt converges uniformly in G. Fix a compact set

K ⊆ G. Since K is compact, if we let

ρ = minz∈K

Re z,

then ρ > 0. Since tsin t

has a removable singularity at 0, there exists δ ∈ (0, 1) anda constant c such that

0 < t < δ =⇒ | 1

sin t| ≤ c

t.

If α, β ∈ (0, δ) with α < β and z ∈ K, it follows that∣∣∣ ∫ β

α

tz

sin tdt∣∣∣ ≤ ∫ β

α

∣∣ 1

sin t

∣∣ tRe z dt

≤∫ β

α

c

ttρ dt

=c

ρ(βρ − αρ).

Noting that either Morera’s Theorem or Leibniz’s Rule imply that∫ π

2α

tz

sin tdt is

analytic on G when α ∈ (0, π2), this estimate implies that if we choose a sequence

αn in (0, π2), then

∫ π/2αn

tz

sin tdt

is a Cauchy sequence in H(G). Since H(G) iscomplete, it follows that there exists f ∈ H(G) such that∫ π

2

αn

tz

sin tdt→ f in H(G).

Furthermore, the function f does not depend on the choice of sequence αn. Thisproves that there exists an analytic function f on G satisfying

f(z) = limα→0+

∫ π2

α

tz

sin tdt.

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(b) We adapt the trick used in class to analytically continue the Riemann ζ functionto the critical strip z | 0 < Re z < 1. Let

f1(z) =

∫ π2

0

( 1

sin t− 1

t

)tz dt

and

f2(z) =

∫ π2

0

1

ttz dt.

Noting that 1sin z− 1

zhas a removable singularity at 0, it follows by a straightforward

modification of the analysis in part (a) that f1 is a well defined analytic functionon z | Re z > −1. Also, if Re z > 0, then

f2(z) =(π2)z

z

by straightforward calculation. It follows that if Re z > 0, then

f(z) = f1(z) + f2(z) = f1(z) +(π2)z

z.

Hence,

f1(z) +(π2)z

z,

which is analytic in z ∈ C | Re z > −1 \ 0 with a simple pole at 0 gives thedesired continuation of f .

Res(f, 0) = limz→0

z(f1(z) +

(π2)z

z

)= (

π

2)0 = 1.

3. Let G be an open set in C and let (S (G), ρ) denote the sheaf of germs ofanalytic functions on G.

(a) Prove that the sheaf topology on S (G) is Hausdorff.

(b) Let Γ(t) = (γ(t), [ft]γ(t)), 0 ≤ t ≤ 1 be a function from [0, 1] into S (G) andfor each t ∈ [0, 1] choose an open set Dt such that γ(t) ∈ Dt ⊆ G and ft isanalytic on Dt. Prove that Γ is continuous if and only if γ is a path in G and(ft, Dt) | 0 ≤ t ≤ 1 is an analytic continuation along γ.

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Solution to (a). Assume that (a, [f ]a), (b, [f ]b) ∈ S (G) with (a, [f ]a) 6= (b, [f ]b).Either a 6= b, or a = b and [f ]a 6= [f ]b.

If a 6= b, choose disjoint neighborhoods Da, Db ⊆ G of a and b respectively suchthat f is analytic on Da and g is analytic on Db. Then

N(f,Da) = (z, [f ]z) | z ∈ Da and N(g,Db) = (z, [g]z) | z ∈ Db

are disjoint neighborhoods of (a, [f ]a) and (b, [f ]b) respectively.

Solution to (b). First assume that Γ(t) = (γ(t), [ft]γ(t)), 0 ≤ t ≤ 1 is a con-tinuous function from [0, 1] into S (G) and that Dt is an open set such thatγ(t) ∈ Dt ⊆ G and ft is analytic on Dt. We wish to show γ is a path in Gand (ft, Dt) | t ∈ [0, 1] is an analytic continuation along γ. Since γ = ρ Γ, thatγ is a path in G follows immediately follows from the continuity of Γ and ρ. Toprove that (ft, Dt) | t ∈ [0, 1] is an analytic continuation along γ, fix t ∈ [0, 1].Since Γ is continuous and N(ft, Dt) is a neighborhood of Γ(t), there exists δ > 0such that

s ∈ [0, 1] and |s− t| < δ =⇒ Γ(s) ∈ N(ft, Dt).

But Γ(s) = (γ(s), [fs]γ(s)) and N(ft, Dt) = (z, [ft]z) | z ∈ Dt. Therefore,

s ∈ [0, 1] and |s− t| < δ =⇒ γ(s) ∈ Dt and [fs]γ(s) = [ft]γ(s).

This proves that (ft, Dt) | t ∈ [0, 1] is an analytic continuation along γ.Conversely, assume that γ is a path in G, Γ(t) = (γ(t), [ft]γ(t)), 0 ≤ t ≤ 1

is a function from [0, 1] into S (G), and that (ft, Dt) | t ∈ [0, 1] is an analyticcontinuation along γ. We wish to show that Γ is continuous. Accordingly, fixt ∈ [0, 1] and an open set Ω in S (G) such that Γ(t) ∈ Ω.

By the definition of the sheaf topology, there exists an open set U in G suchthat Γ(t) ∈ N(ft, U) ⊆ Ω. Since (ft, Dt) | t ∈ [0, 1] is an analytic continuationalong γ, there exists δ1 > 0 such that

s ∈ [0, 1] and |s− t| < δ1 =⇒ γ(s) ∈ Dt and [fs]γ(s) = [ft]γ(s).

Since γ is a path, there exists δ2 > 0 such that

s ∈ [0, 1] and |s− t| < δ2 =⇒ γ(s) ∈ U.

Therefore, if δ = minδ1, δ2, then

s ∈ [0, 1] and |s− t| < δ =⇒ γ(s) ∈ Dt ∩ U and [fs]γ(s) = [ft]γ(s).

But γ(s) ∈ Dt ∩ U and [fs]γ(s) = [ft]γ(s) imply that

Γ(s) = (γ(s), [fs]γ(s)) ∈ (z, [ft]z) | z ∈ Dt ∩ U = N(ft, U) ⊆ Ω.

Summarizing, we have shown that if t ∈ [0, 1] and Ω is a neighborhood of Γ(t)in S (G), then there exists δ > 0 such that Γ(s) ∈ Ω whenever s ∈ [0, 1] and|s− t| < δ. Therefore, Γ is continuous.

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