Finding a Möbius transformation sending a set onto the open unit disk.

Question

Let $g : G \longrightarrow \mathbb C$ be an analytic function on a region $G.$ Let $a \in G.$ Then for any $r \gt 0$ there exists a Möbius transformation $T_r$ such that $T_r \left (\mathbb C \setminus \overline {B(-g(a),r)} \right ) = D,$ where $D$ is the open unit disk. What will happen if we further require $T(g(a)) = 0\ $?

I am thinking about the following map $$z \longmapsto \frac {r} {z + g(a)}.$$ Then clearly it's a Möbius transformation taking the desired domain into $D$ misses only the origin. Is there any way to fix this? Any help in this regard would be warmly appreciated.

Thanks for investing your valuable time in reading my question.

Answer 1

Some preliminary remarks:

• The domain $G$ and the analytic function $g$ are irrelevant. With $w = g(a)$ you are looking for a Möbius transformation $T$ which maps the exterior of the disk $B(-w, r)$ onto the unit disk, with $T(w) = 0$.
• Möbius transformations map disk on the extended complex plane $\hat{\Bbb C} = \Bbb C \cup \{ \infty \}$ to disks or lines on the extended complex plane. $\mathbb C \setminus \overline {B(-w,r)} $ cannot be mapped to the full unit disk, the point $T(\infty)$ will always be missing in the image.
• $T(-w) = 0$ is only possible if $w$ lies in the exterior of $B(-w, r)$, i.e. if $2|w| > r$.

So we can formulate the problem as follows: Given $w \in \Bbb C$ and $r > 0$ with $2|w| > r$, find a Möbius transformation $T$ such that $T \left (\hat{\Bbb C} \setminus \overline {B(-w,r)} \right ) = \Bbb D$ and $T(w) = 0$.

One can start as you did: $T_1(z) = r/(z+w)$ maps $\hat{\Bbb C} \setminus \overline {B(-w,r)} $ onto the unit disk, with $T_1(w) = r/(2w) =: \alpha$. The choose $T_2$ as an automorphism of the unit disk with $T_2(\alpha) = 0$. These automorphism are well-known: $$ T_2(z) = c\frac{z-\alpha}{1-\overline{\alpha} z} $$ with an arbitrary factor $c$ of modulus one. Then the composition $T = T_2 \circ T_1$ solves the given problem.

Another option to get the same result is to determine the reflection point $w^*$ of $w$ with respect to the disk $B(-w, r)$. Since Möbius transformations preserve symmetry with respect to a circle or line, $T(w) = 0$ implies $T(w^*) = \infty$. The transformation is therefore of the form $$ T(z) = d \frac{z-w}{z-w^*} $$ for some constant $d$ of modulus one. The reflection $w^*$ point of $w$ with respect to a circle $B(z_0, r)$ is determined by the formula $$ (w^* - z_0)\overline{(w-z_0)} = r^2 \, , $$ see for example here. In our case that gives $w^* = -w+ r^2/(2\bar w)$.