2022 Winter Qualifying Exam in Complex Analysis

Proof.

Assume

for all . Define

Then

is a polynomial. On ,

By the maximum modulus principle,

for . But

Thus attains its maximum modulus at an interior point, so is constant. Hence

which implies

Therefore, if is not identically , then there must be some such that

Proof.

Let

First count the roots in . On ,

By Rouche's theorem, and have the same number of zeros in . Thus has exactly one zero in .

Now count the roots in . On , set

A direct elementary minimization on gives

Therefore

on , and by Rouche's theorem, and have the same number of zeros in .

Now

The zero lies in . If and , then

But for ,

with equality only possible at , which is not a zero of . Hence all three roots of lie in .

So has four roots in and one root in . Therefore the number of roots in

is

Proof.

The map

sends the sector conformally onto the right half-plane

The map

sends the right half-plane onto the unit disk and sends to .

Therefore

maps the given sector conformally onto and satisfies .

Moreover,

so

Thus

Hence

Proof.

If , then

has a simple pole at with residue . Hence

Now suppose . The poles inside are and . At , the residue is

At ,

so the coefficient of is . Hence

The sum of residues is , so

Thus

Proof.

(a) The zeros of are , . At , both numerator and denominator vanish simply, and

Thus is removable.

For , the zero of at is simple, while the numerator is nonzero. Hence is a simple pole for every nonzero integer .

At infinity, the singularity is not isolated, because the poles tend to infinity. Therefore infinity is not an isolated singularity of this function.

(b) Factor the denominator:

Partial fractions give

Let

Then

and

Also

Therefore, for ,

Equivalently, the coefficient of is , and the remaining terms are

The residue at is the coefficient of in the partial fraction expansion:

Proof.

The statement is true.

The inequality implies that has no zeros for . Hence has only finitely many zeros in . Let be the polynomial whose zeros, with multiplicities, are exactly the zeros of . Then

is entire.

For ,

for some constants . Thus has polynomial growth. By Cauchy's estimates, must be a polynomial.

Since

is entire and are polynomials, is a rational entire function. Therefore is a polynomial.

Proof.

(a) The series is a lacunary power series with exponents

These satisfy

Also, the radius of convergence is , since

By the Hadamard gap theorem, the unit circle is a natural boundary for . Hence every point of is a singular point.

(b) On ,

Thus the series converges uniformly on .

The formal derivative is

This also converges uniformly on , since

Therefore termwise differentiation is justified up to the closed disk, and is differentiable on .

Proof.

Since , the function

has a removable singularity at and is holomorphic near . Hence there are constants and such that

whenever .

For all sufficiently large ,

Therefore

The series

converges. Hence the given series converges absolutely.

Proof.

The condition says that is locally near . An isolated essential singularity is impossible under such an area-integrability condition; equivalently, the standard Laurent-coefficient estimates for holomorphic functions imply that an isolated singularity can only be removable or a pole of order strictly less than .

Here , so a possible pole order must satisfy

Thus

The only positive integer satisfying this is

Indeed, if had a pole of order , then near ,

and the integral near would behave like

which is finite exactly when .

Therefore the singularity at is either removable or a pole of order .