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?
 A: The region is the intersection of the unit disk (centered at the origin) and a disk of radius $\sqrt2$ centered at $(0,1)$. The points on the boundaries of both are $\pm1$, and we want to send one of these to zero, the other to infinity. If you take
$$f_1(z)=\frac{z-1}{i(z+1)}\,,$$ then the upper arc (unit circle) goes to the positive real axis, and the lower arc goes from $0$ to $\infty$ by way of $1-i$. The interior point $0$ of the domain goes to $i$, so you see that the image is the wedge of angle $3\pi/4$ at the origin. Just take the $4/3$-power now to expand it to the UHP.
A: Yes, in the context of a complex analysis exam you can assume that "map $D$ onto $D'$" means to map conformally (biholomorphically). Appealing to the fact that both domains have the same cardinality, and thus a bijection exists is not sufficient. :) 

one of the arcs is not "circular" 

Yes, and the question is correctly worded: the boundary value should be $-1$ on the vertical line segment (non-circular) and $2$ on the circular arcs. 
When you map $D$ onto the UHP following Lubin's method, the points $\pm 1$ end up at $0$ and $\infty$. Where does the vertical line segment $[i, i-i\sqrt{2}]$ go?  Since $\pm 1$ are symmetric about it, $0$ and $\infty$ will be symmetric about its image under the Möbius transformation used by Lubin. Hence, the Möbius transformation sends this segment to a circular arc centered at $0$. The subsequent power map keeps the circular shape of this arc. You'll find the radius by looking at where $0$ goes, for example. 
Basically, you will find that the image of $D'$ is a half-disk. You need a holomorphic function with real values $-1$ and $2$ on two parts of its boundary. Idea: use another  Möbius  map to transform the half-disk to a sector in the plane, in which a multiple of $i\log z$ does the job. (The real part of $i\log z$ is $-\arg z$, which takes on two different constants on the half-line bounding the sector).

Concerning uniqueness of $f$, it is tempting to argue as follows: if $f_1$ and $f_2$ are two such functions, then $h:=\operatorname{Re}(f_1-f_2)$ is a harmonic function vanishing on the boundary. By the maximum principle $h$ is identically zero. It follows that $f_1-f_2$ is a purely imaginary constant, which is zero because $\operatorname{Im}(f_1-f_2)$ vanishes at $0$.  (There's a nonzero chance that this is what the problem author  had in mind.)
Problem is, we can't apply the maximum principle because we don't know anything about the behavior of $h$ at two boundary points. On any simply connected domain there is a nonzero harmonic function that has zero boundary values at every point except one. Indeed, on the unit disk the function $\operatorname{Re}\frac{1+z}{1-z}$ has this property (the Poisson kernel, basically). By composition with a conformal map, we can transplant this example to other domains. 
Thus, there are infinitely many distinct harmonic functions which attain the values $-1$ and $2$ on the indicated arcs. Each of them can be written (uniquely) as the real part of a holomorphic function whose imaginary part vanishes at $0$. All these are different holomorphic functions that satisfy the given requirements.
One can restore uniqueness by imposing the additional condition that $\operatorname{Re} f$ is bounded on $D'$. Then a different version of the maximum principle applies: if $h$ is a harmonic function in a bounded domain $\Omega$, $h$ is bounded from above in $\Omega$, and  $\limsup_{z\to \zeta}h(z)\le M$  for all $\zeta\in\partial \Omega$ except for some finite set, then $h\le M$ in $\Omega$. In particular, a bounded harmonic function whose boundary values on $\partial \Omega$ are zero at all but finitely many points must be identically zero. 
A: 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.

