Prove that holomorphic functions preserve their norms under rotations.

Question

I want to prove (or disprove) that any non-constant holomorphic function preserves its absolute value under rotations of its domain. The precise formulation of the problem is as follows:

Let $f:\mathbb{C}\rightarrow \mathbb{C}$ be holomorphic on an open subset $U\subseteq\mathbb{C}$, and let $s\in \mathbb{C}$ be a complex number such that $|s|=1$ (this condition ensures that multiplying any complex number by $s$ will only rotate that number by some angle $\theta$) and $su\in U$ for every $u\in U$. Prove (or disprove) that, for every $z\in U$, the equation: $$|f(z)|=|f(sz)|$$ holds.

progress so far:

I managed to prove that there exists a function $g$ that serves as an upper bound for $f$ and $f’=f(sz)$. Details of the proof write as follows:

Let $f’(z)=f(sz)$. If $f$ is holomorphic on $U$, then it is analytic. If it is analytic, we can define $f$ as a series expansion, and get: $$f(z)=a_0+a_1z+a_2z^2+\cdots$$ for some complex constants $a_i$. Likewise, we get: $$f’(z)=a_0+a_1sz+a_2(sz)^2+\cdots$$ Taking the modulus, and appliying the triangle inequality: $$|f|=\left|\sum_{i\in\mathbb{N}}a_iz^i\right|\leq \sum_{i\in\mathbb{N}}|a_iz^i|$$ $$|f’|=\left|\sum_{i\in\mathbb{N}}a_i(sz)^i\right|\leq \sum_{i\in\mathbb{N}}|a_i(sz)^i|$$ Applying the properties of the complex absolute value, we get: $$\sum_{i\in\mathbb{N}}|a_i(sz)^i|=\sum_{i\in\mathbb{N}}|s||a_iz^i|$$ By hypothesis, $|s|=1$, so we can write: $$|f’|\leq \sum_{i\in\mathbb{N}}|a_iz^i|$$ Finally, we can see that the function $g(z)=\sum_{i\in\mathbb{N}}|a_iz^i|$ is an upper bound for both functions, $|f|$ and $|f’|$, thus we have: $$\{|f|;|f’|\}\leq g$$ As stated.

Then, i tried to apply Louville’s theorem to derive some other facts about $g$, but $g$ isn’t even holomorphic! Other topological theorems dealing with open maps prove that $g$ is continuous… but from here on I’m genuinely stuck.

Answer 1

Another counterexample is $f(z)=e^z$ and $s=-1$. We have for any $z\in \mathbb{C}\setminus i \mathbb{R}$

$$ \vert f(z) \vert = e^{\text{Re}(z)} \neq e^{-\text{Re}(z)} = \vert f(-z)\vert. $$

Answer 2

Nevermind, i found a counterexample:

Let $U\subset\mathbb{C}$ be the open ball centered at the origin with radius 1: $$U=B_1(0)$$ and let $f$ be the factorial function $$f(z)=\Gamma(z+1)=\int_0^{\infty}x^ze^{-x}\mathrm{d}x$$ Note that $\Gamma(z+1)$ is holomorphic in $B_1$. Finally, let $s=-1$ and notice $|s|=1$. Since the point $p=1/2$ is in $B_1$, by hypotheses, or simple inspection, we know that $sp\in B_1$, with $sp=-1/2$.

If the theorem holds, then the equality: $$|\Gamma(p+1)|=|\Gamma(sp+1)|$$ must be true, and since $p=1/2$ $$|\Gamma(3/2)|=|\Gamma(1/2)|$$ must hold. This equation is clearly false, since $\Gamma(3/2)=\frac{\sqrt{\pi}}{2}$ and $\Gamma(1/2)=\sqrt{\pi}$, and thus $$|\frac{\sqrt{\pi}}{2}|\neq|\sqrt{\pi}|$$