UMD Analysis Qualifying Exam/Jan08 Complex

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Problem 2Edit

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



for all in the domain of definition of

Solution 2Edit

Key steps

  • ratio test

Problem 4Edit

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

Solution 4Edit

Showing G 1:1 conformal mappingEdit

First note that



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



Hence if , then



or



The latter cannot happen in since so



i.e.



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


Therefore from and , is a conformal mapping.

Finding the domain DEdit

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 6Edit

Suppose that for a sequence and any , the series



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


Solution 6Edit

Summation a_n ConvergentEdit

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 analyticEdit

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