$f$ analytic and $|f|$ a function of $|z|$

Question

Suppose $f$ is analytic inside the unit disc and that $|f(z)|$ depends only on $|z|$. Prove that we can write $f(z) = Cz^N$, for all $z$ in the disc.

In the suggested proof, it is stated like it's obvious that $|f|$ only depends on $|z|$ must mean that $|f|$ is strictly increasing. From that point on, the proof is easy, but I'm wondering why that is obvious? I understand that each circle in the z-plane is mapped to a new circle in the f-plane, but why can't the radius of this new circle be some weird oscillating function of $|z|$?

Answer 1

To turn John's comment on the maximum modulus principle into an answer:

If $|f(z)|$ is a function that depends only on $|z|$, then, as you pointed out, a circle in the $z$-plane is mapped to a circle in the $f$-plane. Now, on any closed disk where $|f(z)|$ has a local maximum, $|f(z)|$ takes its maximum on the boundary. Therefore the $|f(z)|$ takes its local maximum on a closed disk at every point of the boundary of that disk. So if $U$ is a circle in the $z$-plane whose interior contains a circle $V$, $|f(U)| > |f(V)|$.

Answer 2

If $f$ is not constant, then $|f(z)|=\mathrm{e}^{i h(z)}g(|z|)$, where $g: [0,1)\to[0,\infty)$ strictly increasing and $h :D\to\mathbb R$, due to the maximum principle.

Case I. $f(0)\ne 0$. Then $F(z)=1/f(z)$ is also analytic in $D$, and $G(r)=\max_{|z|\le r}|F(z)|$ is decreasing, which violates the maximum principle. Thus if $f(0)\ne 0$, then $f$ has to be constant.

Case II. $f(0)=0$. Then, as $f$ is non-constant, $z=0$ is an isolated root, and let $m$ be its order. Then $f(z)=z^m w(z)$, where $w(0)\ne 0$. But $w$ also has the same property: it is analytic in $D$ and $|w(z)|$ is a function of $|z|$, and due to the conclusion of Case I, $w$ is constant, which implies that $f(z)=cz^m$.