2019 Fall Qualifying Exam in Complex Analysis

Proof.

(a) Write . Then

The Wirtinger derivative is

Thus the Cauchy-Riemann equations hold only when

that is, at .

However, being analytic at a point means holomorphic in a neighborhood of that point. Since the Cauchy-Riemann equations fail in every punctured neighborhood of , the function is holomorphic on no nonempty open set. Thus it is analytic nowhere.

(b) Since is , Green's theorem gives

The left-hand side is zero for every sufficiently small disk. Therefore

for every such disk.

Since is continuous, this implies

throughout . Hence satisfies the Cauchy-Riemann equations and is holomorphic in .

Proof.

Let be the zeros of in , counted with multiplicity. Jensen's formula and the bound imply

Since ,

But

Therefore

so

Thus

for the open disk .

If the intended disk is the closed disk , then the sharp bound is , achieved by a finite Blaschke product with twenty zeros at .

Proof.

Let

Then , and is not a root of unity. Put

The terms have modulus

Thus the series converges absolutely exactly when

This is equivalent to

which is the right half-plane

If , the terms do not tend to zero, so the series diverges. If , then the terms are

which do not tend to zero in general; in fact the phase is a nonconstant quadratic rotation and has no limit to zero. Hence the series does not converge on the boundary.

Therefore both absolute convergence and convergence occur exactly for

Proof.

For , define

The standard beta-integral gives

Differentiate twice under the integral sign:

At ,

Since

a direct differentiation gives

Hence

Proof.

This is impossible.

If converges uniformly on a nonempty open set to , then by the Weierstrass theorem the limit must be holomorphic on . But

is not holomorphic on any nonempty open set. Indeed,

which cannot vanish on a nonempty open set.

Therefore no such sequence exists.

Proof.

The inequality implies that has no zeros outside . Hence has only finitely many zeros in .

Let be the polynomial whose zeros, with multiplicity, are exactly the zeros of . Then

is entire. Outside a large disk,

Thus has polynomial growth. By Cauchy's estimates, is a polynomial.

Since

is entire and rational, it must be a polynomial.

Proof.

A proper holomorphic self-map of the disk is a finite Blaschke product.

Indeed, properness implies that each value in has finitely many preimages, counted with multiplicity. In particular, has finitely many zeros in . Also, continuity on the closed disk and properness imply

on the boundary.

Let

with the zeros repeated according to multiplicity. Then is holomorphic and zero-free in the disk, continuous on the boundary, and has modulus on the boundary. By the maximum modulus principle applied to and its reciprocal,

is constant of modulus .

Thus

This is rational.

Proof.

The series

converges normally on compact subsets of , since the terms are uniformly on compact sets away from the integers. Hence it defines a meromorphic function with double poles at the integers.

At each integer, the principal part of

is the same as the principal part of the corresponding term in the series, namely

Therefore the difference

has removable singularities at every integer. Hence is entire.

Both terms are periodic with period , so is entire and periodic. On a vertical strip, the two terms are bounded away from the removable singularities, and after removability is bounded on the whole strip. By periodicity, is bounded on . Liouville's theorem implies that is constant.

Finally, as , both

and

tend to . Therefore the constant is , and