Complex Analysis Spring 2001 Homework IV...

Complex Analysis Spring 2001

Homework IV Solutions

1. Conway, chapter 4, section 2, problem 10.

Evaluate

∫γ

z2 + 1

z(z2 + 4)dz where γ(t) = reit for t ∈ [0, 2π] for all possible values of r,

0 < r < 2 and 2 < r < ∞.

Use partial fractions to write

z2 + 1

z(z2 + 4)=

1

4

1

z+

3

8

1

z − 2i+

3

8

1

z + 2i.

Then ∫γ

z2 + 1

z(z2 + 4)dz =

1

4

∫γ

dz

z+

3

8

∫γ

dz

z − 2i+

3

8

∫γ

dz

z + 2i.

Now for γ the boundary of a disk D, oriented in the counterclockwise direction, wehave proved that ∫

γ

dz

z − a=

{2πi, if a inside D

0 if a outside D.

Applying this to the integrals above we have

∫γ

z2 + 1

z(z2 + 4)dz =

2πi

4+ 0 + 0 =

πi

2when

0 < r < 2, and

∫γ

z2 + 1

z(z2 + 4)dz =

2πi

4+

3

82πi +

3

82πi = 2πi when 2 < r < ∞.

Before working on the next problem, be sure to review Prop. 1.6 in Chapter 3.


This is a marathon problem – just one step after the other, but for a very long time.Since ez =

∑∞n=0

zn

n!, ez − 1 =

∑∞n=1

zn

n!so

ez − 1

z=

∞∑n=1

zn−1

n!=

∞∑k=0

zk

(k + 1)!

for z 6= 0. But the series on the right is convergent at z = 0 also, and so defines ananalytic function. Since this formula shows that the series on the right is convergentfor z with arbitrarily large modulus, the radius of convergence is ∞. The reciprocal,f(z) − z

ez−1, is analytic for |z| < 2π, since the denominator is zero when z = ±2πi.

(Note that by the power series above the extended analytic function we denote byez−1

zhas the value 1 at z = 0 so its reciprocal is also 1 at z = 0. By theorem 2.8 and

corollary 2.11, f(z) has a convergent power series expansion about z = 0 with radiusof convergence equal to 2π. Now by Proposition 1.6, for |z| < 2π,

1 =ez − 1

zf(z)

=∞∑

k=0

zj

(j + 1)!

∞∑k=0

ak

k!zk

=∞∑

n=0

cnzn

where

cn =n∑

k=0

ak

k!

1

(n + 1− k)!

=1

(n + 1)!

n∑k=0

ak(n + 1)!

k!(n + 1− k)!

=1

(n + 1)!

n∑k=0

ak

(n + 1

k

)Comparing the coefficients of the two series, we see that

0 =n∑

k=0

ak

(n + 1

k

)for n ≥ 1

which is one of the relations to be proved. Now

f(z) +1

2z =

z

2

(2 + ez − 1

ez − 1

)=

z

2

ez + 1

ez − 1

ande−z + 1

e−z − 1=

1 + ez

1− ez= −ez + 1

ez − 1

and so f(z) + z/2 si the product of two odd functions and so is even. Then

0 = f(z) +1

2z − (f(−z) +

1

2z)

= z +∑ ak

k!(1− (−1)k)zk

= z +∑kodd

2ak

k!zk

and so 0 = 1 + 2a1 and 0 = ak for k > 1 and odd. There now follows the solution ofmany equations for the a2j and from that the Bernoulli numbers B2n = (−1)n−1a2n,but that algebra will not be reproduced here.


Let G be a region and suppose that f : G → C is analytic and there is a ∈ G suchthat |f(a)| ≤ |f(z)| for all z ∈ G. Prove that either f(a) = 0 or f is constant.

If f(a) = 0 we are done, so we may suppose f(a) 6= 0. Then |f(z)| ≥ |f(a)| > 0 so 1f(z)

is analytic, and | 1f(z)| achieves its maxmimum at z = a. By the maximum principle,

1f(z)

is constant, and so f is constant.

4. Let f be an entire function, and suppose that there exist positive numbers K and msuch that

<f(z) ≤ K + m log(1 + |z|) for all z ∈ C.

Prove that f is constant.

Let g(z) = ef(z). Then g is also entire, and

|g(z)| = e<f(z) ≤ eK+m log(1+|z|) = C(1 + |z|)m.

By a variation on Liouville’s theorem from class, an entire funtion bounded by (1+|z|)m

must be a polynomial of degree at most the greatest integer in m. Furthermore, g(z) isnever zero, since the range of the exponential omits 0. But by the fundamental theoremof algebra, every non-constant polynomial has a root, and so g must be constant. Thus

0 = g′(z) = f ′(z)ef(z)

so f ′(z) = 0 and f is constant.


Suppose that f is analytic on the disk, |f(z)| ≤ 1 for |z| < 1 and that f is non-constant.

Let g(z) = f(z)−a1−af(z)

where f(0) = a. Then g(0) = 0 and g maps the unit disk to the unit

disk, so the Schwarz lemma implies that |g(z)| ≤ |z|. In terms of f , this is

|f(z)− a||1− af(z)|

≤ |z|

or|f(z)− a| ≤ |z||1− af(z)| ≤ |z|(1 + |a||f(z)|).

This gives the two equations

|f(z)| − |a| ≤ |z|+ |z||a||f(z)|

and|a| − |f(z)| ≤ |z|+ |z||a||f(z)|,

which yield the desired inequalities when solved for |f(z)|.


Suppose that f : D → C is analytic and not constant, and satisfies <f(z) ≥ 0 for allz ∈ D.

(a) Show <f(z) > 0 for all z ∈ D.

This can be approached several ways. One way is to look at e−f(z) and use themaximum principle, similar to what was done in problem 4. However, this doesn’tseem to help in the subsequent parts of this problem. Thus we compose f with aMobius transformation which maps the right half plane onto the unit disk. Sucha transformation is Sz = z−1

z+1so let us consider

g(z) =f(z)− 1

f(z) + 1.

Thus g maps the unit disk into the closed unit disk. Since f is not constant, gis not constant, and so |g| does not have an interior maximum, and so |g(z)| < 1for |z| < 1.

(b) If f(0) = 1, then g(0) = 0 so g satisfies the hypotheses of the Schwarz lemma,and hence |g(z)| ≤ |z| for z ∈ D. Expressing this in terms of f gives

|f(z)− 1||f(z)− 1|

≤ |z|.

The rest of the algebra is the same as in the preceding problem, and gives con-clusions (b) and (c). In the case that f(0) 6= 1, then g(0) = a 6= 0 and problem 1from this section applies (with this g replacing f in that problem).

Complex Analysis Spring 2001 Homework IV...

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