4

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.

Znerol
  • 91
  • Looks pretty good to me. And yes, you can conclude that $f^{(n)}(0)=\lambda^n f^{(n)}(0)$ by taking derivatives. – WoolierThanThou Nov 25 '19 at 13:43
  • The equality beetwen derivatives comes from the unicity of the Taylor serie (around the same point). – nicomezi Nov 25 '19 at 13:44
  • @José: Is it really a duplicate if the question is “I just wonder if [...] my thoughts were correct?” – Martin R Nov 25 '19 at 14:10
  • 1
    @MartinR It is not really clear to me. But I think that if we see that as the real question, then this answer will not be answer (unlike yours). – José Carlos Santos Nov 25 '19 at 14:16
  • I cannot really complain because I tend to close-vote as duplicate “aggressively” :) But this looks more like a [tag:proof-verification] question to me, therefore I'll vote to reopen, and wait how the community reacts. – Martin R Nov 25 '19 at 14:21

2 Answers2

2

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.

Hagen von Eitzen
  • 374,180
1

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 $g$.

Martin R
  • 113,040