Finding a Möbius transformation sending a set onto the open unit disk. 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.
 A: 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 $w^* = -w+ r^2/(2\bar w)$.
