A Möbius map from the unit disc onto itself I know that this has been asked a few times, but I found no thread that fully derived the results.
I want to show that for any Möbius transformation from the unit disc onto itself it has the form
$e^{i\alpha} \frac{z-z_0}{\overline{z_0}z-1}\quad,\quad \alpha \in \mathbb{R},\ \vert z_0 \vert < 1$
I found that for $\frac{a z + b}{ c z + d}$ it's necessary that:
$\forall z\ \vert z \vert = 1\ :\  \vert az+b \vert^2 = \vert cz+d \vert^2$
which implies (by using $\vert\ k \vert^2 = k\overline{k}$ and assigning $z=1,-1,i$)
$\vert a\vert^2 + \vert b \vert^2 = \vert c \vert^2 + \vert d \vert^2$
$a\overline{b} = c\overline{d}$
From here didn't get much further. The closest I got is:
$\frac{az+b}{cz+d}=\frac{\overline{c}}{\overline{a}}\frac{\vert a \vert^2z+\overline{a}b}{\vert c \vert^2 z+\underbrace{\overline{c}d}_{z_0}} = \frac{\overline{c}}{\overline{a}}\frac{\vert a \vert^2z+z_0}{\vert c \vert^2 z+z_0}$
It can go further, but I didn't get something similar to $z-z_0$ nor $z_0 z - 1$.
 A: Note that such a Mobius transform must have a root (otherwise it won't map onto the unit ball), which we might as well call $z_0$. So, our Mobius transform must have the form
$$a\frac{z - z_0}{cz + d}$$
for some $z_0$ in the disk. Using continuity of the inverse, we can expect this root not to occur on the boundary, hence $|z_0| < 1$.
Clearly $d \neq 0$, since taking $z \to 0$ should yield a finite result. So, we may divide the denominator and $a$ by $-d$. Absorbing $-d$ into the expression yields
$$a \frac{z - z_0}{cz - 1}.$$
Using your steps,
$$|a|^2(1 + |z_0|^2) = |c|^2 + 1$$
$$|a|^2 \overline{z_0} = c$$
From substituting the second equation into the first, we get
$$|a|^2(1 + |z_0|^2) = |a|^4|z_0|^2 + 1 \implies |a|^2 = \frac{1 + |z_0|^2 \pm \sqrt{(1 + |z_0|^2)^2 - 4|z_0|^2}}{2|z_0|^2},$$
which, when simplified, yields $|a|^2 = 1$ or $|a|^2 = \frac{1}{|z_0|^2}$. In the first case, we get $c = \overline{z_0}$, and $|a| = 1$, and we are done.
In the second case, we obtain $c = \frac{1}{z_0}$. If this is the case, our general form becomes
$$a \frac{z - z_0}{\frac{z}{z_0} - 1} = az_0,$$
which is clearly a false solution unless $|a| = 1$. But, this solution can be expressed in the previous form, so we are done indeed.
