Modulus of function is function of modulus

Question

Is there a way to characterize all non-constant entire functions such that $|f(z)| = f(|z|)$ ?

The monomials work, but I can't think of any other functions. So far, I have managed to show that $f$ is real on the real line, $f(z) \not= 0 $ if $z \not=0$, $|f'(z)| \geq f'(|z|)$, and $\forall \ \theta \in \mathbb R \ \exists \alpha \in \mathbb R $ s.t $ f(e^{i\theta}z)=e^{i\alpha}f(z) $. It would then suffice to show $\lim_{x \to \infty} f(x)= +\infty$, because $f$ would be a polynomial and have all roots at 0.

Edit: What about functions $f: U \to \mathbb C$ ?

Answer 1

An alternative proof, under the weaker assumption that $f$ is only defined and holomorphic in some disk $B_R(0)$:

If $f$ is not constant and $|f(z)| = f(|z|)$ then $f$ has no zeros except at the origin (because of the identity principle). It follows that $f(z) = z^n e^{g(z)}$ where $n$ is a positive integer and $g$ is holomorphic in $B_R(0)$. Then $$ |f(z)| = r^n e^{\operatorname{Re}(g(z))} $$ is constant on each circle $|z|=r$, so that $\operatorname{Re}(g(z))$ is constant on $|z|=r$. But the real part of $g$ is a harmonic function: If it is constant on a circle then it is constant inside the circle.

This implies that $g$ is constant, so that $f(z) = cz^n$ with some constant $c$ which is necessarily a positive real number.

Answer 2

For $r > 0$ is $$ f(r) = \max \{ |f(z)| : |z| = r \} = \max \{ |f(z)| : |z| \le r \} $$ by the maximum modulus principle, so that $f(r)$ is an increasing function of $r$.

$f(r)$ can not be bounded because then $f$ would be bounded in $\Bbb C$ and therefore constant by Liouville's theorem.

Therefore $\lim_{r \to \infty} f(r)= +\infty$. It follows (as you said) that $f$ is a polynomial whose only root is at the origin, so that $f(z) = c z^n$ for some constant $c > 0$ and some positive integer $n$.

Answer 3

Let $f(z)= \sum_{j=0}^\infty a_j z^j$ be analytic on $B\big(0,\delta\big)$ and $f(0)=0$ with multiplicity $k\in \mathbb N$.

Then $g(z):=\sum_{j=k}^\infty a_j z^{j-k}=\frac{f(z)}{z^k}$ satisfies the key property $ |g(z)| = g(|z|)$ for $z\neq 0$ and is non-zero on $\overline B\big(0,r\big)$ with $r=\frac{\delta}{m}$ for some $m\geq 2$. Compactness combined with Maximum Modulus Theorem tells us $g$ attains a max modulus with $\vert z\vert = r$, but it also attains a minimum modulus with $\vert z\vert = r$ by Minimum Modulus Theorem (equivalently: Max Modulus Theorem applied to $\frac{1}{g}$). The key property $ |g(z)| = g(|z|)$ tells us the min modulus is equal to the max modulus $\implies g$ has constant modulus on $\overline B\big(0,r\big)$ hence is a constant function.

$\implies f(z) = a_k z^k$ where $a_k \gt 0$