Showing G 1:1 conformal mappingEdit
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 DEdit
To find , we only need to consider the image of the boundaries.
Consider the right hand boundary,
Now, consider the left hand boundary .
Now consider the bottom boundary .
Hence, the boundary of maps to the real line. Using the test point , we find
We then conclude
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