TITLE: Conformal map from disk with smaller disk removed to upper half plane
QUESTION [6 upvotes]: I'm working on a problem that was a previous complex qualifying exam at my university. I believe I have a solution, but I'm not entirely confident in it. The problem is as follows:

Find a one-to-one conformal map of the region $\Omega=\{z\in\mathbb{C}\,|\,|z|<2\text{ and }|z-1|>1\}$ onto the upper half plane.

Here's my attempt:
The region $\Omega$ can be mapped to the vertical strip $\{z\in\mathbb{C}\,|\, -1/2<\text{Re }(z)<-1/4\}$ by the map $z\mapsto\frac{1}{z-2}$. Then $z\mapsto\left(\frac{4\pi}{z-2}+2\pi\right)i$  maps $\Omega$ to the horizontal strip $\{z\in\mathbb{C}\,|\,0<\text{Im }(z)<\pi\}$. Lastly
$$
z\mapsto \exp\left(\frac{4\pi}{z-2}i+2\pi i\right)
$$
or
$$
z\mapsto\exp\left(\frac{4\pi}{z-2}i\right)
$$
should map to the upper half plane.
Is this correct? Is there any easier way to see this? Any input would be greatly appreciated, thanks in advance.

REPLY [3 votes]: This is correct, and there is no easier approach. Any time you have a domain with two mutually tangent circles in its boundary, you are probably going to send the point of tangency to $\infty$, to turn the circles into parallel lines. And the subsequent step of mapping an infinite strip onto halfplane with exponential functions is also standard.