Find all holomorphic functions $f$ such that $f(0) = 0$ and $f(f(z)) = z$ near $0$ Using power series, I have to find all holomorphic functions $f$ such that $f(0) = 0$ and $f(f(z)) = z$ near $0$. If I'm not mistaken, $f(0)=0$ restricts the power series to a form $\sum_n a_n z^n$ but now I have no idea how to proceed. If I just plug in, I get $$\sum_n a_n \left( \sum_k a_k z^k \right)^n = z$$
for $z$ near $0$, but what next? Thank you very much for any hints.
 A: If your assumption is just that $f$ is holomorphic in a neighborhood of $0$, i.e., that it has a convergent power series about $0$ (and that seems to be the only thing you can assume if you have to "use power series"), then there are lots of solutions. In particular, for every convergent power series $g(z) = z + b_1 z + b_2 z^2  + \ldots$, the function $f(z) = g^{-1}(-g(z))$ (i.e., a local conjugate of $z \mapsto -z$ satisfies $f(f(z)) = z$. So you get infinitely many solutions, and I don't know of an easy way to classify all of them via power series coefficients.
Just as an example, it is easy to check that $f(z) = \frac{z}{z-1} = -z - z^2 - z^3 - \ldots$ is a convergent power series in $|z|<1$ with $f(f(z))=z$.
A: Important Note This answer is wrong, in specific it is so the argument I use to compare power series as Lukas kindly pointed out. I'm keeping it here only as a reference for the amount of comments made on it.
Answer So we are looking for $f(z)$ holomorphic such that there exists an open set $0\in U \subset \mathbb{C}$ with $f(f(z))=z \: \:\forall \;z\in U$. 
Let $f(z) = \sum_{n=0}^\infty a_nz^n$ be the power series of $f(z)$ around zero. From the condition $f(0)=0$ we deduce $a_0=0$.
Then $z=f(f(z))= \sum_{k=1}^\infty a_k \left(\sum_{n=1}^\infty a_n z^n\right)^k = a_1^2z +O(z^2)$. In specific $a_1=\pm 1$ and $a_i=0$ for $i>1$. This implies $f(z)= \pm z \:\: \forall z\in U$.
By the paragraph "Analytic function" of this Wikipedia page. We deduce that $f(z)=\pm z
\:\: \forall z\in U$ implies that $f(z)=\pm z \:\: \forall \; z\in \mathbb{C}$, because $f(z)$ and $\pm z$ are holomorphic functions. This implies $f(z)= \pm z$. 
