Find $f\colon \mathbb{C} \rightarrow \mathbb{C} $ such that $f(z)=z+f(z^2)$ in a neighbourhood of $0$

Question

Find $f\colon \mathbb{C} \rightarrow \mathbb{C} $ that respects

$f(z)=z+f(z^2)$

in a neighbourhood of $0$.

I've never seen a problem like this, so i haven't any idea for solving it.

Answer 1

If, around $0$,$$f(z)=a_0+a_1z+a_2z^2+a_3z^3+\cdots,$$then$$a_0+a_1z+a_2z^2+a_3z^3+\cdots=a_0+z+a_1z^2+a_2z^4+a_3z^6+\cdots$$From this, you deduce nothing concerning $a_0(=f(0))$, but now you know that $a_1=1$. And you also know that $a_2=a_1=1$. And that $a_3=0$. And that $a_4=a_2=1$. And so on. The general rule is: $a_n=1$ if $n$ is a power of $2$ and $0$ otherwise. That is$$f(z)=a_0+z+z^2+z^4+z^8+\cdots$$and this series converges if and only if $|z|<1$.