2016 Spring Qualifying Exam in Complex Analysis

Proof.

The possible poles occur at

On any compact set avoiding these points, for large we have

Thus the series converges uniformly on compact subsets of

Hence it defines a holomorphic function there. At each point , only one summand has a pole, and that pole is simple. Therefore the sum is meromorphic on .

Proof.

Use the standard formula

Here and , so

Proof.

Let be the coefficient of . Then

Therefore the radius of convergence is

Proof.

(a) By Cauchy's estimates,

If , letting gives

Thus all Taylor coefficients above degree vanish, so is a polynomial of degree at most .

(b) Consider

The condition as means

as . Thus has a pole at . Therefore has a pole at infinity, and an entire function with a pole at infinity is a polynomial.

Proof.

Observe that

Thus

If has the same imaginary part as , then

is entire and real-valued. A real-valued holomorphic function is constant. Therefore

Proof.

No such family exists.

If is holomorphic on , then

If uniformly on , then

On , , so

Therefore

a contradiction.

Proof.

Rotate the domain by setting

Then becomes the upper half of the unit disk:

The map

maps this half-disk to the first quadrant. Squaring maps the first quadrant to the upper half-plane:

Finally,

maps the upper half-plane to the exterior of the unit disk.

Thus one conformal map is

Proof.

The Schwarz-Pick lemma for the upper half-plane gives

At , this becomes

Since

we obtain

Equality holds for example when

Then ,