I have one of those "find the map" problems that is really giving me a lot of trouble. Let $B_1(1)$ be the ball of radius $1$ centered at $1$. We have the following domain: $\mathbb{D} \cap B_1(1) \cap \mathbb{H}$, the intersection of two discs with the upper half-plane, and we want a conformal isomorphism to $\mathbb{D}$ (or $\mathbb{H}$ of course). For convenience, the two circles intersect in $\mathbb{H}$ at $(\frac{1}{2}, \frac{\sqrt{3}}{2})$.
I know how to do this for the intersection of two discs, but not these three regions (three discs, really). I feel like the way to proceed is either to find a map sending this region to a half-strip (which is also the intersection of three "discs"), or use some LFT sending the three "corners" to judiciously chosen points, but I can't figure it out at all. If we can map it to a slitted plane, or plane missing a bounded interval, or a half-strip, or a sector, then I can proceed from there, but otherwise I am totally stumped.
Alternatively, does anyone know if the function used for the intersection of just the two discs behaves well on the "top half" of this region? For all I know it maps it to a quadrant or something. EDIT: This approach doesn't seem fruitful after looking into it a bit more.
Anyone see a way to proceed? You would be saving my bacon!
Let $a=\dfrac12+\dfrac{\sqrt3}{2}\,i$. Then $\dfrac{z-a}{z-\bar a}$ transforms the intersection of the two discs into an angle. The image of the segment $[0,1]$ is an arc of circumference perpendicular to the sides of the angle. Thus, the image of the intersection of the disks with the upper half-plane is either a circular sector or the outside of the sector. Can you continuue from here?