UMD Analysis Qualifying Exam/Jan08 Complex

From Wikibooks, open books for an open world
< UMD Analysis Qualifying Exam
Jump to navigation Jump to search

Problem 2[edit]

Prove there is an entire function so that for any branch of

for all in the domain of definition of

Solution 2[edit]

Key steps

  • ratio test

Problem 4[edit]

Let be the domain . Prove that is a 1:1 conformal mapping of onto a domain . What is ?

Solution 4[edit]

Showing G 1:1 conformal mapping[edit]

First note that

Also, applying a trigonometric identity, we have for all ,

Hence if , then


The latter cannot happen in since so


Note that the zeros of occur at . Similary the zeros of occur at .

Therefore from and , is a conformal mapping.

Finding the domain D[edit]

To find , we only need to consider the image of the boundaries.

Consider the right hand boundary,

Since ,

Now, consider the left hand boundary .

Since ,

Now consider the bottom boundary .

Since ,

Hence, the boundary of maps to the real line. Using the test point , we find

We then conclude

Problem 6[edit]

Suppose that for a sequence and any , the series

is convergent. Show that is analytic on and has analytic continuation to

Solution 6[edit]

Summation a_n Convergent[edit]

We want to show that is convergent. Assume for the sake of contradiction that is divergent i.e.

Since is convergent in the upper half plane, choose as a testing point.

Since converges in the upper half plane, so does its imaginary part and real part.

The sequence is increasing () since and e.g. the gap between and is grows as grows. Hence,

This contradicts that is convergent on the upper half plane.

Show that h is analytic[edit]

In order to prove that is analytic, let us cite the following theorem

Theorem Let be a sequence of holomorphic functions on an open set .   Assume that for each compact subset of the sequence converges uniformly on , and let the limit function be .   Then is holomorphic.

Proof See Theorem 1.1 in Chapter V, Complex Analysis Fourth Edition, Serge Lang.

Now, define .   Let be a compact set of .   Since is continuous