For which $a$ does the equation $f(z) = f(az) $ has a non constant solution $f$ For which $a \in \mathbb{C} -\ \{0,1\}$ does the equation $f(z) = f(az) $ has a non constant solution $f$ with $f$ being analytical in a neighborhood of $0$.
My attempt:
First, we can see that any such solution must satisfy:
$f(z)=f(a^kz)$ for all $k \in \mathbb{N} $.
If $|a|<1$:
The series $z_{k} = a^k$ converges to 0 which is an accumulation point, and $f(z_i)=f(z_j)$ for all $i, j\in \mathbb{N} $. Thus $f$ must be constant.
If $|a|=1$:
For all $a \neq 1$ , $f$ must be constant on any circle around $0$, so again $f$ must be constant.
My quesions are: 
Am I correct with my consclusions? 
Also, I'm stuck in the case where $|a|>1$. Any ideas?
Thanks

You have treated the case $0<|a|<1$.
The case $|a|>1$ reduces to the case $|a|<1$ because $f(z)=f(az)$ for all $z$ implies $f(a^{-1}z)=f(z)$ for all $z$.
The case $|a|=1, a\ne1$ splits into two subcases:


*

*$a^k=1$ for some $k\in\mathbb N$. Then $f(z)=z^k$ is a nonconstant function satisfying the equation

*otherwise, the sequence $a^k$ is dense on $S^1$, which implies that $f$ is constant on $S^1$ and ultimately constant on $\mathbb C$


(Of course the exceptional cases $a=0, a=1$ directly imply $f$ constant and $f$ arbitrary, respectively).
