Let $T$ be a Möbius transformation such that $T(T(z)) = z$ for all $z \in \mathbb{\hat{C}}$. That is, $T = T^{-1}$. I want to show that $T$ must have two distinct fixed points.
I have tried using a general $$ T(z) = \frac{az + b}{cz + d} \ , $$ attempting to solve for $z$, with no luck.
I also know that $T(z)$ may only have two distinct fixed points if and only if $c \neq 0$ or $c = 0$ and $a \neq d$.
Are there any points that are guaranteed to be fixed under these conditions? How could I manage to find these points?
Best Answer
Every non-identity Möbius transformation has either one or two fixed points. A Möbius transformation with only one fixed point (the so-called “parabolic case”) is conjugate to a translation, i.e. $$ T(z) = z + a $$ if the fixed point is $\infty$, or $$ \frac{1}{T(z) - z_0} = \frac{1}{z-z_0} + a $$ if the fixed point is $z_0 \in \Bbb C$, for some $a \ne 0$. In either case $T\circ T$ is not the identity.