Find a conformal map from a half plane deprived of a disk onto an annulus + question on a connected harmonic function Let $D = \{z\in \mathbb{C} :$ Re $ {z} \geq 0\} \setminus \{ z:|z-2|\leq1\}$. 


*

*Find a conformal map from D onto an annulus $r < |z| <1$.

*Find a continuous bounded function on $\bar{D}$ which is harmonic in $D$, vanishes on the imaginary axis, and takes value $1$ on $|z-2|=1$.


For 1, I tried to find a linear fractional transformation but it doesn't work for all $r$.
For 2, I considered a holomorphic function $f$ such that $u = $Re $f$ is harmonic, $u(z) = 0$ on the imaginary axis and $u(z) = 1$ on $|z-2|=1$, but couldn't find such f. How can I find such f?
 A: For part 1, take $T(i z)$ from here. For part 2, take $\operatorname{Re} \ln T(i z)$ with $|a| = 1$ and divide by $\ln(2 - \sqrt 3)$ to set its value on $\{z: |z - 2| = 1\}$ to $1$.
A: Here is a treatment of Part 1.
I will address here the fully analogous problem of mapping the upper half plane $Im(z)>0$ minus the disk defined by $|z-2i| <1$ onto an annulus. This is motivated by the recent question here.
My intention here is to show that "inversion transform" gives a direct mean to find the right transformation.

Fig. 1: $\textit{The image of the blue region by transform (2) is the red region}$, $\textit{an annulus centered in}$ $(0,-\sqrt{3}/2)$.
Some recalls about inversion can be useful:
In its most basic version, inversion is the map
$$z \mapsto f(z)=Z=\frac{1}{\bar{z}}.$$
Under its general form, involving a translation of origin and a scaling, inversion is defined by the following relationship:
$$Z-c=\frac{\alpha^2}{\overline{z-c}} \ \iff \ Z=F(z)=\frac{c\bar{z}+(\alpha^2-|c|^2)}{\bar{z}-\bar{c}}\tag{1}$$
where $c \in \mathbb{C}$ is the center of inversion, and $\alpha^2$ is its so-called "power".
Geometric interpretation of this transform :
The image $M'$ of any point $M$ (see figure) is such that

*

*the center of inversion $C$, points $M$ and $M'$ are aligned;


*the product of distances $CM \times CM'=\alpha^2$ (the power of inversion).
In particular, points $M$ belonging to the "circle of inversion" with center $C$ and radius |a| (in green on the figure) are invariant by the transformation ($M'=M$).
For symmetry reasons, we are going to look for the adequate center $C$ on the imaginary axis, i.e., $C=ia$.
In order that the origin $O$ is kept unchanged, $O$ must belong to the circle of inversion, which forces $\alpha = |a|$.
Therefore, (1) becomes:
$$Z=F(z)=\frac{ia\bar{z}}{\bar{z}+ia}\tag{2}$$
Using (2), the image of point $A=i$ is $A'=\frac{ia}{a-1}$, and the image of point $B=3i$ is $B'= \frac{-3ia}{a-3}$.
Let us now constraint circle $\Gamma'$ - the image of circle $\Gamma$ - to have the same center as circle $L'$ (image of $L$), i.e., point $\frac{a}{2}i$ :
$$\frac12(A'+B')=\frac12(\frac{ia}{a-1}+\frac{-3ia}{a-3})=\frac{a}{2}i$$
resulting in
$$a^2=3 \ \implies \ a=-\sqrt{3}$$
(see, once again, the figure).
Remark 1 : It can be advantageous to take the conjugation in (2) :
$$Z=\overline{F(z)}=\frac{-iaz}{z-ia}\tag{3}$$
(which will generate another annulus symmetris of the red one with respect to $x$ axis).
The advantage is that, with (3), we have a true Möbius transformation.
Remark 2 : Once a certain annular region is obtained, it is easy to map it holomorphicaly onto any another annular region.
Remark 3 : The common tangents to circles $\Gamma$ and $\Gamma'$ (dotted lines) meet in the center of inversion $C$. Do you see why ?
Remark 4: Inversion transform is implemented in GeoGebra (the software  used for the generation of this figure) under the name "Reflect about (inversion) Circle".
