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

Solution 1:

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).

Solution 2:

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 $|a|<1$.