Möbius transformations on $D$ such that $f(D)=D$

I need to find all Möbius transformations on unit disk such that $f(D)=D$, please help!

Solution 1:

Mapping from $D$ to $D$ is onto, so exist $z_0 \in D$ such that $B(z_0)=0$. Using correspondence principle of border (I don't know English version of this theorem) we find that $B(T)=T$, where $T=\{\omega \in \mathbb{C}: |\omega|=1\}$, and $B(z_0^*)=\infty$ (which is same as $B\left(\dfrac{1}{\overline{z_0}}\right)$, where $z_0^*$ is inverse point of $z_0$ with respect to the circle $T$).

Now, $$\omega=B(z)=K \dfrac{z-z_0}{z-\frac{1}{\overline{z_0}}}=-\overline{z_0}K\dfrac{z-z_0}{1-\overline{z_0}z}=K_1\dfrac{z-z_0}{1-\overline{z_0}z},\qquad K_1=-\overline{z_0}K.$$

Using that $|B(e^{it})|=1, \, t \in [0,2\pi)$ we find $$1=|B(e^{it})|=|K_1|\frac{|e^{it}-z_0|}{|1-\overline{z_0}e^{it}|}=\frac{|K_1|}{|e^{it}|}\frac{|e^{it}-z_0|}{|e^{-it}-\overline{z_0}|}=|K_1|,$$ so $K_1=e^{i\alpha}$ for some $\alpha \in [0, 2\pi)$. So, $$\omega=B(z)=e^{i\alpha} \frac{z-z_0}{1-\overline{z_0}z},\, \alpha \in [0,2\pi),\, |z_0|<1$$

Solution 2:

Outline of solution:
1) First, show that $\varphi$ is a Moebius transformation such that $\varphi(\mathbb{R})=\mathbb{R}$ if and only if $\varphi(z)=\frac{az+b}{cz+d}$ where $a,b,c,d\in\mathbb{R}$. You can do this by looking at $\varphi(0),\varphi(\infty)$ and $\varphi(1)$.
2) Take any Moebius transformation $\psi(z)$ such that $\psi(D)=\mathbb{R}$. (e.g. $\psi(z)=\frac{iz+1}{z+i}$). Show that a Moebius transformation $f(z)$ satisfies $f(D)=D$ if and only if $\varphi=\psi\circ f\circ\psi^{-1}$ satisfies $\varphi(\mathbb{R})=\mathbb{R}$.
3) Use the first two steps and some algebraic manipulations to show that $f(z)=\frac{w_1z+w_2}{\overline{w_2}z+\overline{w_1}}$ for some $w_1,w_2\in\mathbb{C}$ with $|w_1|\neq|w_2|$.