Find $\lambda\in \mathbb{C}$ s.t. $\exists$ entire $f$, s.t. $f(z)=f(\lambda z),~\forall z\in\mathbb{C}$ I want to find all $\lambda \in \mathbb{C}$ such that there exists an entire non-constant function $f$ with $f(z)=f(\lambda z)$ $\forall z \in \mathbb{C}$.
My idea was the following: Since $f$ must be entire on the whole complex plane it can be represented as taylor series around $0$. So I get then in total
\begin{align*}
\sum\limits_{n=0}^\infty \frac{f^{(n)}(0)}{n!}z^n=f(z)=f(\lambda z)=\sum\limits_{n=0}^\infty \frac{f^{(n)}(0)}{n!}\lambda^{n}z^n.
\end{align*}
Now I (hope I can) conclude that 
\begin{align*}
f^{(n)}(0)=f^{(n)}(0)\lambda^{n}.
\end{align*}
Therefore, either $f^{(n)}(0)=0$ or $\lambda^{n}=1$ must hold.
First I conclude that $|\lambda|=1$ must hold, therefore $\lambda=e^{i\theta}$ for some $\theta \in [0,2\pi].$ If now there exists $n\in \mathbb{N}$ such that $\theta=\frac{2\pi}{n}$ then I have $\lambda=e^{i\frac{2\pi}{n}}$ and therefore $\lambda^n=1$. For such functions I might be able to define a entire function $f$, such that all derivates in $0$ vanish except all $kn$ derivatives in $0$.
I just wonder if those are all and my thoughts were correct?
Thanks in advance.
 A: Let $M=\sup_{|z|\le 1}|f(z)|$.
Assume $|\lambda|<1$. Then for all $z$ we find $n\in\Bbb N$ with $|\lambda^n z|<1$ and hence $|f(z)|=|f(\lambda^nz)|\le M$, i.e., $f$ is bounded and hence constant.
The same argument works when $|\lambda|>1$. We conclude that $|\lambda|=1$.
Then unless $\lambda$ is a root of unity, the numbers $\lambda^n$ are dense in $S^1$. It follows that $f$ is constant on $S^1$ and by identity theorem is constant throughout.
We conclude that $\lambda^n=1$ for some $n\in\Bbb N$. But then $f(z)=z^n$ is an example of such a function.
A: You correctly argued that $f^{(n)}(0) = 0$ or $\lambda^n = 1$ must hold for all $n$, because the power series of an entire function is unique.
In order to determine all corresponding functions $f$, define $n$ as the minimum
$$
 n = \min \{ k \ge 1 | f^{(k)}(0) \ne 0 \} \, .
$$ 
Then  $\lambda=e^{i\frac{2\pi}{n}}$ and $f^{(k)}(0) = 0$ for all  $k$ which are not a multiple of $n$. It follows that
$$
 f(z) = \sum_{k=0}^\infty a_n z^{kn} 
$$
so that $f(z) = g(z^n)$ for some entire function $g$.
