2014 Spring Qualifying Exam in Complex Analysis

Proof.

(a) Since , we have . Thus

has a removable singularity at and extends to an entire function. For ,

By Liouville's theorem, is constant. Hence

for some constant with .

Thus the functions are exactly

(b) The condition says that as . We claim must be constant.

For any , there is such that

whenever . By Cauchy's estimate on the circle ,

Letting gives . Applying the same estimate to for gives

as . Thus all derivatives of positive order vanish at , and is constant.

Conversely, every constant function satisfies . Hence the functions are exactly the constant entire functions.

Proof.

(a) The Laurent expansion is

There is no coefficient. Hence the residue at is , and therefore

(b) For ,

Taking gives

Thus

Proof.

(a) By the Schwarz-Pick lemma,

Since ,

Thus the sharp upper bound is

(b) Equality is attained by a disk automorphism mapping to , for example

Then and

Proof.

For ,

This limit function is holomorphic and nonzero on .

Let

The series converges uniformly on the closed disk , so

uniformly on .

Since has no zeros on the compact set , there is such that

on . For all sufficiently large ,

on . Hence cannot vanish there. Therefore there exists such that for ,

for all .

Proof.

(a) For a holomorphic function , the function is subharmonic for every . This follows, for example, because is subharmonic and is increasing and convex.

A finite sum of subharmonic functions is subharmonic. Hence

is subharmonic in .

(b) Let

By part (a), is subharmonic. The hypothesis says attains a maximum at the interior point . By the maximum principle for subharmonic functions, is constant on .

Since each is subharmonic and nonnegative, the distributional Laplacian of each term is a positive measure. The sum has Laplacian , because is constant. Therefore each has Laplacian , so each is harmonic.

If a holomorphic function is nonconstant, then is strictly subharmonic away from its zeros. Thus cannot be harmonic unless is constant. Hence every is constant.

Proof.

The two boundary circles

are tangent at . The map

sends circles through to lines.

For , we have

Dividing by gives

so

For , we have

Dividing by gives

so

Thus maps onto the vertical strip

Translate and scale the strip by

Then

The exponential map

maps this strip conformally onto the upper half-plane. Finally,

maps the upper half-plane onto the unit disk.

Therefore one required conformal map is

Proof.

Let

Then ,

and

It is enough to prove .

If , then every automorphism has the form

The conditions and give and , so is the identity.

If , then by the Riemann mapping theorem there is a conformal map

The map

is an automorphism of the unit disk. It fixes and has derivative of modulus there. Conjugating once more by a disk automorphism that sends to , we get a disk automorphism fixing with derivative at .

By the Schwarz lemma, such a map must be the identity. Therefore is the identity, hence is the identity, and so

Proof.

The singularity at is isolated. If it is removable, there is nothing to prove. Otherwise, suppose first that is a pole of order . Then near ,

for some . In polar coordinates,

has the same convergence behavior as

This integral is finite exactly when . Since is a positive integer, we must have . Thus a pole can only be simple.

It remains to rule out an essential singularity. A standard form of the removable-singularity theorem says: if a holomorphic function on a punctured disk is locally integrable and has an isolated singularity at the puncture, then the singularity is either removable or a pole of order strictly less than . Applying this theorem with gives precisely that the singularity is removable or a simple pole.

Therefore is either removable or a simple pole.

Proof.

Since , the function

is positive and harmonic on .

By Harnack's inequality, for ,

Take . Then

Hence, for ,

Equivalently,

Raising both sides to the third power,

Since

the Weierstrass -test shows that

converges uniformly on .