$f$ is an analytic function in $\mathbb C$. Prove or disprove that if $f(z)=f(3z)$ then $f(z)=C$, while $C$ is constant complex number.

Question

$f$ is an analytic function in $\mathbb C$. Prove or disprove that if $f(z)=f(3z)$ then $f(z)=C$, while $C$ is constant complex number. I tried to disprove by trying to make $f(z)=e^{2\pi iz}$ this function does not equal constant because $f(0)=e^{2\pi i0}=e^0=1$ while $f(\frac{1}{2})=e^{2\pi i(\frac{1}{2})}=e^{\pi i}=-1$ , $f(0)\neq f(\frac{1}{2})$. but then I was skeptic about the condition $f(z)=f(3z)$. Is my function $f(z)=e^{2\pi iz}$ fulfills the condition $f(z)=f(3z)$? Same question as: "Is this equation correct:$$e^{2\pi iz}=e^{6\pi iz} $$??".

Answer 1

From the comments, $f$ is analytic in $\mathbb{C}$.

This implies $f$ is continuous, hence bounded in the unit disk by a positive number, say $B$. For all $z\in\mathbb{C}$, we may choose an integer $n$ such that $3^n>|z|$, hence $3^{-n}z$ lies in the unit disk. So from the assumption $f(z)=f(3z)$, $|f(z)|= |f(3^{-n}z)|\le B$. Since $f$ analytic and bounded in the whole complex plane, applying Liouville’s Theorem yields $f$ is constant.

By the way, $e^{2\pi iz}=e^{6\pi iz}$ is not correct for all $z$. For example, let $z=1/3$, then $e^{2\pi iz}= e^{2/3\cdot\pi i}=-\frac12+\frac{\sqrt 3}{2}i$, but $e^{6\pi iz} = e^{2\pi i} = 1$.

Answer 2

My first answer was totally false. I thought that $f$ is complex periodic (as it is on the real line).

However, I found another way.

I propose this method because, as the properties of the function $f$ are really simple, there should be a simple way to solve the problem without the use of theorems.

Since $f$ is analytical, there exists the coefficients $\{a_n\}$ such that :

$$f(z) = \sum^\infty_0 a_n z^n $$

Now, $f(z) = f(3z)$ :

$$\sum^\infty_0 a_n z^n = \sum^\infty_0 3^n a_n z^n := \sum^\infty_0 b_n z^n $$

by unicity of the coefficients, $$\forall n \in \mathbb N, a_n = b_n$$ then : $$\forall n > 0, a_n = 0 $$

Answer 3

A different but perhaps more visible or direct way to see that $f(z)$ must be constant is to note that $f(z) = f(3z)$ implies $|f’(z)| = 3|f’(3z)|$. Thus if $f’(z_0)\not=0$ for some $z_0\not=0$ then the derivative $f’$ blows up along the geometric sequence $z_0, z_0/3, z_0/9,\ldots$ which can’t happen for $f$ analytic. Thus $f’(z) = 0$ for all $z\not=0$ and indeed also for $z=0$, so $f$ is constant.