TITLE: Möbius transformation that preservers the unit circle
QUESTION [5 upvotes]: 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!

REPLY [2 votes]: 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.