Find closed form of composition. Consider $f(z) = \frac{1-a}{1-az}$, where $a \in(0,1)$
. Now we want to find $f(f\dots(f(z)\dots)$. It will begood if there is a closed form of it.
I've consider first step : $f(f(z)) = \frac{1 - a}{1 -a\frac{1-a}{1-az}} = \frac{(1-a)(1-az)}{1-az-a+a^2}$ , but it looks non-simplified.
I guess I miss something , any hints? 
 A: If we identify linear fractional transformation with the matrix (up to multiplicative constant), i.e.
$$
f(z) = \frac{az+b}{cz+d}\ \ \ \Longleftrightarrow \ \ \ \left[\begin{array}{cc}a&b\\c&d\end{array}\right],
$$ then composition corresponds to matrix multiplication. Thus the problem boils down to calculating
$$
A^n =\left[\begin{array}{cc}0&1-a\\-a&1\end{array}\right]^n.
$$ Fortunately, $A$ is diagonalizable with
$$
A= \left[\begin{array}{cc}1&1-a\\1&a\end{array}\right]\left[\begin{array}{cc}1-a&0\\0&a\end{array}\right]\cdot\frac{1}{2a-1}\left[\begin{array}{cc}a&a-1\\-1&1\end{array}\right],
$$ hence
$$
A^n =\left[\begin{array}{cc}1&1-a\\1&a\end{array}\right]\left[\begin{array}{cc}(1-a)^n&0\\0&a^n\end{array}\right]\cdot\frac{1}{2a-1}\left[\begin{array}{cc}a&a-1\\-1&1\end{array}\right].
$$ Up to multiplicative constant, we have
$$
A^n \sim \left[\begin{array}{cc}a(1-a)^n-(1-a)a^n&-(1-a)^{n+1}+(1-a)a^n\\a(1-a)^n-a^{n+1}&-(1-a)^{n+1}+a^{n+1}\end{array}\right],
$$ hence
$$
f^n(z) = \frac{[a(1-a)^n-(1-a)a^n]z-(1-a)^{n+1}+(1-a)a^n}{[a(1-a)^n-a^{n+1}]z-(1-a)^{n+1}+a^{n+1}}.
$$
