Conformal mapping of the unit circle minus a smaller circle

Question

Map the region inside the circle $|z| = 1$ and outside the circle $|z-1/2| = 1/2$ conformally onto the unit disk.

I was thinking of using some scaling and shifting to get from the unit circle to the circle $|z-1/2| = 1/2$. but I'm not sure how that might be useful since it doesn't help with getting the region we want. The other idea I had was to work backwards starting with the unit circle to get to the region we want.

Source: Spring 1992

Answer 1

Hint: What happens to that region under the map $z\to (1+z)/(1-z)?$

Answer 2

To finish it off, @¡YOA, after the transformation $\frac{1+z}{1-z}$ suggested by @zhw. there are two more steps to take. The first one maps the vertical stripe onto $\{z: \Im z \geq0\}$ by means of $e^{i\pi z}$. (Note that $|e^{i\pi z}|= |e^{i\pi (x+iy)}|=|e^{-\pi y}| \in (-\infty, \infty)$ . And Arg$(e^{i \pi z}) = \pi x \in [0,\pi]$, as $x \in[0,1]$.)

The last transformation maps the upper half plane onto the unit disc using a usual Möbius mapping.