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

Thank you all very much. For completeness, I will write here a sctach of the soultion:
For $|a|=1$ we can take $f(z)=z^k$ with $k = 2\pi/Arg(a) $.
For $|a|>1$, we can notice that actually $f(z)=f(a^kz)$ for all $k \in \mathbb{Z}$, so the solution is similar to the case $|a|<1$.
