# Finding a conformal map of lunar domain to upper half disk

Does there exist a conformal map from the region $\Omega = \{z :|z|<1\} \cap \{z: |z- \frac{1+i}{\sqrt2}|<1\}$ onto the region $\{z: |z|<1, \operatorname{Im}z>0\}$?

I think I need to find at least three intersection point of the two circles and mapped them to real axis using the formula of fractional linear transformation. I even have difficulty finding the intersection points. I would really appreciate if someone do this rigorously. This is not a homework problem. This is from the collection of previous qual exams.

• If I haven't mistaken, there is such a map due to Riemann Mapping Theorem. Dec 30 '12 at 4:53
• Most likely you're expected to prove that the Riemann mapping theorem applies. For this note that both domains are convex and so are simply connected. Or are you asked to give an explicit map? Dec 30 '12 at 6:45
• As has been said, it's not apparent that you need an explicit map. But for what it's worth, I compute the intersection points as $\left(-\sqrt{1/2-\sqrt{12}/8},1/4(\sqrt{2}+\sqrt{6})\right)$ and $\left(\sqrt{1/2+1/8\sqrt{12}},1/4(\sqrt{2}-\sqrt{6})\right)$. They might have been nicer in polar coordinates. Note there are only two, not three. Dec 30 '12 at 6:57
• @Jose27, Yes we do need an explicit map. Dec 30 '12 at 17:41
• @KevinCarlson, I would assume they are nicer in polar coordinates, the intersection point you gave looks not very nice. But that is something anyway. Thaks Dec 30 '12 at 17:42

Find a linear fractional transformation mapping the intersection points to $0$ and $\infty$. The two boundary arcs will be mapped to rays from $0$ to $\infty$, so the image will be a sector of some opening angle $\alpha$ (which is the angle at which the two circles or circle/line segment intersect.) Then compose with a power map $z \mapsto z^\beta$ such that $\alpha \beta = \pi$, which maps the sector to a half-plane.