Showing G 1:1 conformal mapping[edit | edit source]
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.
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
Suppose that for a sequence and any , the series
is convergent. Show that is analytic on and has analytic continuation to
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.
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