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

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