Möbius transformation from a unit disc to its complement

Question

I am trying to find all möbius transformations that map $D_{1}(0)$ to its complement.

My Idea: I am trying to find 3 mappings say $T_1, T_2,T_3$ where

1. $T_1$ maps $D_1(0)$ to $K(0,1)$ "specific"
2. $T_2$ maps $K(0,1)$ to itself $K(0,1)$ "arbitrary"
3. $T_3$ maps $K(0,1)$ to $[D_1(0)]^c$ "again specific"

And then $f=T_3 \circ T_2 \circ T_1$ will be that arbitrary map which I want to find...

Now I know how to map $D_1(0)$ to $K(0,1)$, so I'll find $T_1$, I also have $T_2$ the arbitrary one.. I just need to find $T_3$...

My question is " is there any such Möbius transformation exist? If yes then what will be that ?"

Note: A möbius transformation is a map of the form $T(z) = \frac {az+b}{cz+d}$, where $a,b,c,d \in \Bbb C$ such that $ad-bc \neq 0$. And here by $K(0,1)$ my means unit circle and by $D_1(0)$ my means unit disc.

Answer 1

given such a map $\phi$, $\frac{1}{\phi(z)}:\mathbb{D}\to \mathbb{D}$ is an automorphism of the disk , and thus is of the form $e^{i\vartheta}\frac{z-\alpha}{1-\overline{\alpha}z}$ (with $\vartheta\in [0,2\pi], \alpha\in \mathbb{D}$).

This gives us $$\phi(z)=e^{i\theta}\frac{1-\overline{\alpha}z}{z-\alpha}$$

Conversely, every map of this form maps $\mathbb{D}$ to its complement