How do I show that each Möbius transformation that preservers the open unit circle (maps it to itself) must be of the form: $c \frac{z-z_0}{\bar{z_0}z-1}, |c|=1, |z_0|<1$ ? I've seen previous answers saying that the fact that a Möbius transformation is determined by its values on the points $0, 1, \infty$ can be used to easily show this, although I didn't manage to use this in order to solve the problem.
Help would be appreciated!
Let $Tz=\frac{az+b}{cz+d}$ map the unit disc onto itself. Let $z_0$ be the point which is mapped by $T$ to $0$. By the symmetry principle, $\frac{1}{\overline{z_0}}$ is mapped to $\infty$. Thus $T$ is of the form $Tz = c \frac{z-z_0}{\overline{z_0}z-1}$. Since $1$ is mapped onto the unit circle, we have $1 = |T1| = |c||\frac{1-z_0}{\overline{z_0}-1}|=|c|$ (where $|\frac{1-z_0}{\overline{z_0}-1}|=1$ by an isosceles triangle argument.