Iteration of fractional linear maps

Our concern here are functions of the form $f(z)=\frac{az+b}{cz+d}$ with $c\neq 0$.

Fixpoints

In the case $c\neq 0$, the fixpoints are given by

$L_{1,2}=-\frac{d-a}{2c} \pm \sqrt{\left(\frac{d-a}{2c}\right)^2 + \frac{b}{c}}$, or for later use:

$L_{1,2}=\frac{a_1}{c} \pm \frac{a_2}{c} \underbrace{\sqrt{1-\frac{ad-bc}{a_2^2}}}_{=:q}$, where $a_1=\frac{a-d}{2}$ and $a_2=\frac{a+d}{2}$.

Derivative at the fixpoints

The derivative of the fractional linear map $f$ is:

$f'(z)=\frac{ad-bz}{(cz+d)^2}$

If we plug in $z=L_1$ we get:

$f'(L_1) = \frac{ad-bz}{(a_1+a_2 q + d)^2} = \frac{ad-bz}{a_2^2}/\left(1+q\right)^2 = \frac{1-q^2}{\left(1+q\right)^2} = \frac{1-q}{1+q}$

$f'(L_2) = \frac{1+q}{1-q} = 1/f'(L_1)$.

Schröder coordinate at the fixpoints

$\sigma_k(f(z))=f'(L_k) \sigma(z)$, $\sigma_k$ is unique up to a multiplicative constant.

$\sigma_1(z) = \frac{z-L_1}{z-L_2}$.

$\sigma_1\left(\frac{az+b}{cz+d}\right) = \frac{(a-cL_2)z + b-dL_2}{(a-cL_1)z + b -dL_2} = \frac{a_2(1-q)z+}{a_2(1+q)z + }$.