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:

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

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:

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

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

Schröder coordinate at the fixpoints

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

Then $\sigma_1(z) = \frac{z-L_1}{z-L_2}$, $\sigma_2(z)=\frac{z-L_2}{z-L_1}$.

Proof

$\sigma_1\left(\frac{az+b}{cz+d}\right) = \frac{(a-cL_1)z + b-dL_1}{(a-cL_2)z + b -dL_2} = \frac{a_2(1-q)z+b- dL_1}{a_2(1+q)z + b -dL_2 }=\frac{1-q}{1+q}\frac{z-\frac{dL_1-b}{cL_2+d}}{z-\frac{dL_2-b}{cL_1+d}}$.

$c^2 L_1 L_2 = (-d+a_2+a_2 q)(-d+a_2 - a_2 q) = (-d+a_2)^2 - a_2^2 q^2 = \frac{(a-d)^2}{4} - \frac{(a+d)^2}{4} + ad - bc = -bc$

$(cL_2+d)L_1 = -b + dL_1$

$\sigma_1\left(\frac{az+b}{cz+d}\right) = \frac{1-q}{1+q} \frac{z-L_1}{z-L_2} = f'(L_1) \sigma_1(z)$

Iterates

The iterates $f^t$ of $f$ can then be given by:

$f^t(z) = \sigma_k^{-1}(f'(L_k)^t \sigma_k(z))$

$\sigma_1^{-1}(w) = \frac{L_2w-L_1}{w-1}$

$f^t(z) = \frac{L_2K^t\frac{z-L_1}{z-L_2} - L_1}{K^t\frac{z-L_1}{z-L_2}-1} = \frac{(L_2K^t-L_1)z -L_2K^tL_1 + L_2L_1}{(K^t-1)z-K^tL_1+L_2} = $