2023 Winter Qualifying Exam in Complex Analysis

Proof.

Suppose, toward a contradiction, that is not dense in . Then there are and such that

for all .

Consider

The poles of occur at the zeros of . Since there are no such zeros outside , all poles of lie in the compact set . Hence there are only finitely many poles.

Subtract the principal parts at these finitely many poles. Then

is entire. Also, outside a sufficiently large disk, is bounded by , and each principal part tends to at infinity. Therefore is bounded near infinity. Since is entire, it follows from Liouville's theorem that is constant.

Thus is rational. Hence

is meromorphic on the Riemann sphere. Since is entire, it has no finite poles, so must be a polynomial. This contradicts the assumption that is entire transcendental.

Therefore is dense in .

Proof.

First decompose into partial fractions:

Let

Then , and for ,

Therefore, in the annulus ,

This is one Laurent series centered at .

The example is not unique unless the annulus of convergence is fixed. For example, for ,

Thus on the annulus ,

So different Laurent expansions are possible on different annuli. On any fixed annulus, however, the Laurent series is unique.

Proof.

Let

The function is holomorphic in a neighborhood of . For real sufficiently close to , we are given that

Since the Taylor coefficients satisfy

it is enough to show that every derivative is real. But derivatives at can be computed from real difference quotients along the real axis. Since takes real values on a real interval around , all these real-axis derivatives are real.

Therefore

for every .

Proof.

Let

On the boundary of , we have , so

Also, if , then . Therefore

Thus on ,

By Rouche's theorem, and have the same number of zeros in , counted with multiplicity. Hence has exactly zeros in .

Every point of satisfies , since the disk is tangent to the imaginary axis at and lies in the right half-plane. Thus all these zeros have positive real part.

It remains to prove that the zeros are simple. Suppose were a multiple zero. Then

and

Using the first equation in the second gives

Since , this becomes

so

But does not lie in with positive real part. This is impossible. Therefore all zeros are simple.

Proof.

(a) Suppose could also be analytically continued to an open set containing . By hypothesis, can be analytically continued near every other point of as well. Since is compact, finitely many of these neighborhoods cover the whole boundary circle.

Together with the original disk of convergence, this gives an analytic continuation of to an open neighborhood of the closed disk . In particular, the Taylor series of at would have radius of convergence larger than , contradicting the definition of .

Thus cannot be analytically continued to any open set containing .

(b) No. Here is a counterexample with and :

This series converges normally on compact subsets of

so it defines a holomorphic function in the unit disk. The singularities accumulate at , so the radius of convergence of the Taylor series about is .

Every point of except has a neighborhood avoiding all the singularities , so can be analytically continued near every such point.

However, for every , the punctured disk

contains some point for sufficiently large. Since has a pole there, cannot be analytically continued to the whole punctured disk. Therefore the statement is false.

Proof.

The denominator vanishes at , but also vanishes there, so the singularity is removable.

We use the standard contour integral formula

Taking real parts gives

Since the integrand is even,

Here

so . Therefore

Thus

Proof.

The two boundary circles and are tangent at . Use the Mobius map

which sends the tangency point to infinity. The circle becomes the line

because

The circle becomes

because

The domain maps to the vertical strip

Now set

Then maps to the strip

The map

maps this strip conformally onto the upper half-plane. Finally, the Cayley transform

maps the upper half-plane onto the unit disk.

Therefore one conformal map is

Proof.

Let

for all . Fix . If , then the disk

is contained in . By Cauchy's estimate,

for all .

Since is convex, for any the line segment joining and lies in . Therefore

Thus is Lipschitz, hence uniformly continuous.