Iteration of automorphisms of disks Let $L(z) = \phi_{\alpha}(z):= \frac{z-\alpha}{1-\bar \alpha z}$, $|\alpha| < 1$. Let $L_1 = L$, and $L_{n+1}(z)=L \circ L_{n}(z)$. I need to show that $L_n$ converges uniformly on compact subset of $D$. 
Clearly, $L_n$ is still an automorphism of the disk, then I want to find $a_n$ such that $\phi_{a_n} = L_n$. I compute several $L_n$ to see whether there is a pattern or not, but I cannot spot any.
 A: The transformation $\phi_{\alpha}$ corresponds naturally to the matrix 
$$
A = \begin{pmatrix}
1 & -\alpha \\
-\bar\alpha & 1
\end{pmatrix}
$$
And it is indeed a monomorphism into $\mbox{PGL}(2,\mathbb{C})$. (easy to check)
Then we convert this problem into a Linear Algebra one. This matrix has eigenvalues $1\pm |\alpha|$ and eigenvectors $( \alpha, \mp|\alpha|)$ respectively. Then let
$$
S = \begin{pmatrix}
\alpha & \alpha \\
|\alpha| & -|\alpha|
\end{pmatrix}.
$$
It follows that
$$
B=S^{-1}AS = \begin{pmatrix}
1-|\alpha| & 0 \\
0 & 1+|\alpha|
\end{pmatrix}.
$$
Therefore, the matrix corresponding to $L_n$ is 
$$
SB^nS^{-1} = \frac{1}{-2\alpha |\alpha|}\begin{pmatrix}
\alpha & \alpha \\
|\alpha| & -|\alpha|
\end{pmatrix} 
\begin{pmatrix}
(1-|\alpha|)^n & 0 \\
0 & (1+|\alpha|)^n
\end{pmatrix}
\begin{pmatrix}
-|\alpha| & -\alpha \\
-|\alpha| & \alpha
\end{pmatrix} = 
\begin{pmatrix}
\frac{1}{2}(1-|\alpha|)^n + \frac{1}{2}(1+|\alpha|)^n & \frac{\alpha((1-|\alpha|)^n -(1+|\alpha|)^n )}{2|\alpha|} \\
  \frac{|\alpha|((1-|\alpha|)^n -(1+|\alpha|)^n )}{2\alpha} & \frac{1}{2}(1-|\alpha|)^n + \frac{1}{2}(1+|\alpha|)^n
\end{pmatrix}
$$
Going back we have a transformation $L_n = \phi_\beta$ with
$$
\beta = -\frac{ \alpha((1-|\alpha|)^n -(1+|\alpha|)^n ) }{|\alpha|((1-|\alpha|)^n + (1+|\alpha|)^n)}
$$
Please let me know if any step is not clear.
