Find a conformal map from a half plane deprived of a disk onto an annulus + question on a connected harmonic function

Question

Let $D = \{z\in \mathbb{C} :$ Re $ {z} \geq 0\} \setminus \{ z:|z-2|\leq1\}$.

1. Find a conformal map from D onto an annulus $r < |z| <1$.
2. 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?

Answer 1

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 $1$.

Answer 2

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".