Conformal map from $D=\{ z \in \mathbb{C} : |z-2|<2 \: \: \text{and} \:\: |z-2i|<2\}$ to the unit disc $\mathbb{D}$

Question

How does one find a conformal map from the set \begin{equation} D=\{ z \in \mathbb{C} : |z-2|<2 \: \: \text{and} \:\: |z-2i|<2\} \end{equation} to the open unit disc $\mathbb{D}$ such that $1+i$ is mapped to $0$?

I don't know how to start. I drew the picture, it's just the intersection of two circles. But that's it. Can someone help me how to solve such problems? Thanks in advance.

Answer 1

Hints Recall that a linear fractional transformation maps circles (including lines, which are circles through $\infty$) to circles. So, if you find a l.f.t. $f$ that maps one of the intersection points to $0$ and the other to $\infty$, then $f$ maps the two boundary arcs to rays from the origin and hence $D$ to a sector.

Answer 2

The answer of this type of exercise is already discussed in this post

Conformal map from a lune to the unit disc in $\mathbb{C}$