Show any involutive Möbius transformation must have two distinct fixed points. 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?
 A: 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.
A: Every mobius transform can be represented as some $2 \times 2$ complex matrix $A$, in such a way that we consider $A$ and $\lambda A$ to represent the same transformation for $\lambda$ any non-zero scalar.
Let $A$ be a $2 \times 2$ matrix. We suppose that $A^2 = \lambda I$ for some $\lambda \neq 0$, so that projectively the matrix is an involution. The characteristic polynomial of the matrix is $x^2 - \lambda = (x+\sqrt \lambda)(x - \sqrt \lambda)$. Therefore the matrix has two distinct eigenvectors. When those eigenvectors are interpreted as homogeneous coordinates, they are fixed points.

An alternative answer might be to notice that, without loss of generality, one of the fixed points can always be moved to zero via conjugation. So let $f(z)$ be a Mobius transformation: The general form of $f(z)$ when $f(0)=0$ is $f(z)=\frac{a}{\frac{c}{z} + d}$. Then taking into account the fact that $f(f(z)) = z$ reduces that to $f(z) = - \frac{1}{d + \frac{1}{z}}$, at which point the remaining fixed point is at $z=-2/d$. If $d = \infty$, which is the only way for the other fixed point to be at $0$, then $f(z)$ is no longer a Mobius transformation, let alone an involution.
The idea here was inspired by Martin R.
