Map $\{x+iy \mid x^2+y^2<1 \text{ and } x^2 + (y-1)^2<2\}$ conformally to UHP From an old qualifying exam:

Let $D$ be the domain $$D :=\{x+iy \mid x^2+y^2<1 \text{ and } x^2 +
(y-1)^2<2\}.$$ 
  
  
*
  
*Map the domain onto the upper half-plane.
  
*Obtain a function $f(z)$ analytic in the domain $D' := D \cap \{x+iy    \mid x>0\}$ and which takes on the boundary values
  $\text{Re}f(z) =    -1$ on the segment of the imaginary axis $-1<y<1$,
  and $2$ on the bounding circular arcs in $D'$, excluding the points
  $z=i$ and    $z=i-\sqrt{2}i$. Further $\text{Im}f(0)=0$. Is $f(z)$
  unique?
  

(For the first, I have to assume they want a conformal mapping, but they did not indicate it...go figure. For the second, one of the arcs is not "circular", but I guess I know what they mean.)
Let's focus on the first part for now; maybe if I get that I can get the second part. I have struggled to come up with some ideas for the first one. I don't know of any results about conformal mapping and parabolae. Maybe I should look at the polar form of a parabola?

First from the definition it seems $D := \{|x^2+y^2|<1\}$ or the unit disc itself. If that is not a mistake, then the map $z \rightarrow i \dfrac{1+z}{1-z}$ sends the unit disc to the upper half plane conformally.
However, a more interesting problem would be to map $D:= \{|x^2+y^2| >1 \text{ and } x^2 + (y-1)^2 < 2\}$ to the UHP. Here D would be the simply-connected region bounded between the two circles that are tangent to each other at $z = -i$. Since this is a special point, it would help to map this point to infinity using $z \rightarrow \dfrac{1}{z+i}$. Then the unit circle gets mapped to the line $z = - i/2$ and the larger circle gets mapped to the line $z = -i/4$. So we get an infinite strip parallel to the real axis. This is because due to conformality the circles which were tangent to each other at $-i$ get mapped to generalized circles (in this case straight lines) which are tangent to each other at infinity, which means two parallel straight lines. From here the steps are easy. 


*

*Translate the strip vertically up using $z \rightarrow z+ i/2$.

*Dilate the strip using $z \rightarrow 4\pi z$.

*Finally use $z \rightarrow e^z$ to send the above domain to the UHP.

