2023 Fall Qualifying Exam in Complex Analysis

Proof.

The integrand is entire, so the integral is independent of path. A primitive is

because

Therefore

Now

and

Thus

Proof.

(a) The domain

is an open half-plane. In particular, it is convex. Every convex domain is simply connected, and every holomorphic function on a simply connected domain has a primitive.

Thus, if is holomorphic on , there exists a holomorphic function on such that

For every closed piecewise curve in ,

Therefore is holomorphically simply connected.

(b) Let

The function

is holomorphic on . Let for . Then is a simple closed piecewise curve in , and

Therefore is not holomorphically simply connected.

Proof.

(a) This is a geometric series with ratio

It converges precisely when

Writing , we have

so

Thus the series converges when

or equivalently .

Since satisfies , the point lies in this half-plane. Hence the series converges uniformly on compact subsets of a neighborhood of , and its sum is holomorphic there.

(b) Where the geometric series converges,

This meromorphic function has singularities when

Equivalently,

so

The singularity nearest to occurs at , whose distance from is

All other singularities are farther away. Therefore the radius of convergence of the Taylor series about is

Proof.

The boundary circles

meet at and . Use the Mobius map

which sends to and to . Therefore the two boundary circles, both passing through , are mapped to two straight lines through the origin.

One checks that these lines are

The point maps to , so is mapped to the sector

Rotate this sector by defining

Then the sector becomes

Squaring maps this sector conformally onto the upper half-plane:

Finally, the Cayley transform

maps the upper half-plane onto the unit disk.

Thus an explicit conformal map is

Proof.

The statement is false.

Take

Then

up to the harmless indexing convention. Consider the family

on .

This family is not normal. For example, by Marty's criterion, a family of holomorphic functions is normal only if the spherical derivatives are locally bounded. The spherical derivative of at is

which is unbounded as .

Therefore the iterates of an arbitrary entire function need not form a normal family on the unit disk.

Proof.

(a) No. Take

This function is harmonic on , and on the real axis , so

But is not identically zero. Therefore the condition does not imply .

(b) Yes. Since is harmonic on , it is harmonic on the closed unit disk. If

on , then by uniqueness for the Dirichlet problem, or by the maximum and minimum principles,

throughout .

Harmonic functions are real analytic. Since vanishes on the open set , unique continuation implies that vanishes everywhere on . Thus

Proof.

Since is a polynomial of degree , there is a constant such that

for all sufficiently large . Because

we get

for some constant .

Thus is an entire function of polynomial growth. We now use Cauchy's estimates. Write

For any ,

If , then letting gives

Therefore all sufficiently high Taylor coefficients vanish. Hence is a polynomial in .

Proof.

Since is holomorphic in the punctured disk, it has a Laurent expansion

for .

We will show that all negative coefficients vanish. For ,

Put . Then , and

This holds for every . Letting gives

Thus all negative Laurent coefficients vanish, so the singularity at is removable.