2010 Fall Qualifying Exam in Complex Analysis

Problem 1.

Let be distinct complex numbers contained in the disk . Let be analytic in the closed disk . Let

Prove that

is a polynomial of degree at most having the same values as at .

Proof.

For fixed on , the expression

is a polynomial in of degree at most , because vanishes when and is divisible by .

Therefore is an integral of polynomials in of degree at most , so itself is a polynomial of degree at most .

Now evaluate at . Since ,

By Cauchy's integral formula,

Thus interpolates at .

Problem 2.

Show that

is a meromorphic function on .

Proof.

The possible poles occur where

that is,

Let be a compact subset of avoiding all these points. For all sufficiently large , uniformly for ,

Hence

for all large . Since converges, the series converges uniformly on .

Therefore the sum is holomorphic away from the points . At each such point, only one term has a pole, and that pole is simple. Hence the sum is meromorphic on .

Problem 3.

Let be a family of holomorphic functions on the unit disk such that for every ,

Prove that is a normal family.

Proof.

It is enough to prove that is locally uniformly bounded.

Fix a compact set . Choose such that , and choose so that for every , the disk is contained in .

For ,

Since is subharmonic,

Using the lower bound for on ,

Thus

for all and all .

So is locally uniformly bounded. By Montel's theorem, is normal.

Problem 4.

Find an explicit conformal transformation of

onto the unit disk.

Proof.

First use

This maps conformally onto the slit disk

Choose the square-root branch with argument in :

This maps the slit disk onto the upper half unit disk

Now use

which maps the upper half unit disk onto the first quadrant. Squaring maps the first quadrant onto the upper half-plane, and

maps the upper half-plane onto the unit disk.

Therefore one explicit conformal map is

where the branch of is chosen with argument in after the slit is mapped to .

Problem 5.

Find, for ,

Proof.

Use partial fractions:

The standard formula

gives

Thus

Problem 6.

Let be an open subset of , let , and let . Write .

(i) Prove that if is complex differentiable at , then satisfy the Cauchy-Riemann equations at .

(ii) Prove that if is complex differentiable and in , then is an orientation-preserving conformal map.

Proof.

(i) Write . Taking the difference quotient along the real direction gives

Taking it along the imaginary direction gives

Equating real and imaginary parts,

These are the Cauchy-Riemann equations.

(ii) If , then near ,

Multiplication by the nonzero complex number is a rotation followed by a scaling. Such a linear map preserves angles and orientation. Therefore the differential of at each point is angle-preserving and orientation-preserving.

Hence is conformal and orientation-preserving wherever .

Problem 7.

(i) State the mean value theorem for analytic functions and use the Cauchy integral formula to prove it.

(ii) Prove that if is analytic on an open subset , then and are harmonic.

(iii) Let be an open subset of , and let be harmonic. Prove that if there is such that

then is constant.

Proof.

(i) If is analytic on a neighborhood of , then

Indeed, by Cauchy's integral formula,

With , this becomes

(ii) Since is analytic, and satisfy the Cauchy-Riemann equations:

Differentiating,

Since mixed partial derivatives agree,

Similarly,

Thus and are harmonic.

(iii) A harmonic function satisfies the minimum principle. If attains its minimum at an interior point , then attains its maximum at . By the maximum principle for harmonic functions, is constant on the connected component of containing , hence is constant there.

If is connected, this says is constant on all of .

Problem 8.

Let

be a power series with radius of convergence . Determine the region of convergence of the Laurent series

Proof.

Since the original radius of convergence is ,

For the positive-power part,

the root test gives the limiting size

Thus this part converges for

For the negative-power part, write , where . Then

The root test gives the limiting size

Thus this part converges for

Therefore the Laurent series converges in the annulus