I want to find a conformal map from $W=\{Im(z) > 0, |z|>1\}$ (which is the upper half plane excluding the unit circle), to the unit circle itself. The problem was unclear as to whether we need to include the boundary of the circle or not. Let's assume it is included.
The obvious first choice for me was to use $1/z^2$ but I don't think this works since the positive real axis is not part of the image.
The map $\frac{1+iz}{1-iz}$ works for the entire upper half plane, so I'm not sure if I can use this to my advantage.
Can anyone suggest a method to find such a map?
Best Answer
It's often helpful to construct conformal maps in stages, and a good place to start when the boundary of your domain has two obvious special points is to send those special points to zero and infinity. In this case, I would start with the map $\frac{z+1}{z-1}$. This sends $-1$ to $0$ and $1$ to infinity, so the real line and the unit circle go to lines. Real numbers get sent to real numbers under this map, and that it is angle preserving tells you that the unit circle goes to the imaginary axis.
Figuring out whether you get the positive or negative parts of each axis will let you determine where the boundary of your domain goes, and then checking one interior point, for example zero, will completely determine the image. This will give you a nice sector, then you can use the map $z^a$ (for appropriate choice of $a$) and a map similar to the one you mentioned in your question. Make sure you draw lots of pictures!