Power Series of Iterations of a Rational Function I would like to ask you for help in the following problem.
Let $z$ be a nonzero complex number in the unit ball and let $f(x)$ be a function defined by
$$
f(x)=\frac{1}{x^2+z}.
$$
So, $f(0)=1/z$ and
$$
f(f(0))=\frac{1}{\left(\frac{1}{z}\right)^2+z}=\frac{z^2}{z^3+1}=z^2(1-z^3+z^6-\cdots)=z^2-z^5+z^8-\cdots.
$$
By using Mathematica software, I saw that the iterations $f^3(0), f^4(0), \ldots$ when considered as a power series in $z$, always have the form
$$
f^{2n+1}(0)=\frac{1}{z}+a_1z^2+a_2z^5+a_3z^8+\cdots
$$
or
$$
f^{2n}(0)=a_1z^2+a_2z^5+a_3z^8+\cdots.
$$
So the only powers in the series are from exponents of form $3k+2$, with $k\geq -1$.
Unfortunately, I was not able to prove this and so I would like to ask you guys for some suggestion. I tried to use higher order derivatives at $z=0$ together with a Fáa di Bruno generalization of the Chain Rule. But without sucess
 A: The $n$-th iterate of $f$ at $x=0$ is a function of $z$, so let us write that as $F_n(z)$. The $F_n$ satisfy the recursion
$$
 F_0(z) = 0 \, , \, F_{n+1}(z) = \frac{1}{F_n(z)^2+z} \, .
$$
The desired power series representation suggests to define
$g_n(z) = z F_n(z)$. The $g_n$ are rational functions in $z$ satisfying the recursion
$$
 g_0(z) = 0 \, , \, 
g_{n+1}(z) = \frac{z}{(g_n(z)/z)^2 + z} = \frac{z^3}{g_n(z)^2+z^3} \, .
$$
This shows that $g_n(z) = h_n(z^3)$ where the $h_n$ are rational functions in $z$ satisfying the recursion
$$
 h_0(z) = 0 \, , \, 
h_{n+1}(z) = \frac{z}{h_n(z)^2+z} \, .
$$
The first functions are
$$
\begin{align}
  h_0(z) &= 0 \, ,\\
  h_1(z) &= 1 \, ,\\
  h_2(z) &= \frac{z}{z+1} \, ,\\
  h_3(z) &= \frac{z^2+2z+1}{z^2+3z+1} \, .\\
\end{align}
$$
Using the recursive relation for $h_n$ one can now show that
$$
 h_n(z) = \begin{cases}
z + O(z^2) & \text{ if $n$ is even,}\\ 
1 + O(z) & \text{ if $n$ is odd.}\\ 
\end{cases}
$$
The desired representation for $F_n(z) = \frac 1z h_n(z^3)$ follows.
The power series for $h_n$ (and the Laurent series for $F_n$) converge in a neighborhood of $z=0$, but not necessarily in the entire unit disk. For example, the distance of the origin to the nearest zero of the denominator of $h_3(z)$ is $(3-\sqrt 5)/2 \approx 0.38$.
