I am reading this link on complex dynamics and in Problem 12-1 it asks the reader to prove, using the Maximum Modulus Principle, that Herman rings cannot occur for polynomials.
I have seen this question apparently answered here, but I do not really understand it.
As far as I know, for a function $f$, a Fatou component $U$ is a Herman ring if $U$ is conformally isomorphic with an annulus $A_{r,R}$ and $f^p$ is conformally conjugate to a rotation (with irrational angle) of $A_{r,R}$. Hence $U$ contains infinitely many invariant curves $\gamma$ in its interior.
How can I prove the claim using the Maximum Modulus Principle?
Let $f: \mathbb{C} \to \mathbb{C}$ be a polynomial. If $\deg(f) \leq 1$, then by extending $f$ to an endomorphism of the Riemann sphere, you can show that the Fatou set of $f$ is either the whole sphere or it is biholomorphic to the plane. So in either case, $f$ cannot exhibit a Herman ring.
For the case that $\deg(f) \geq 2$, let $U$ be a bounded Fatou component and let $\Gamma$ be a simple loop in $U$. Let $V$ be the bounded connected component of $\mathbb{C} \setminus \Gamma$ and note that $\partial V = \Gamma$. By the maximum modulus principle, for all $n \in \mathbb{N}$, $$\max_{z \in \overline{V}} |f^n(z)| = \max_{z \in \Gamma} |f^n(z)| \leq M,$$ for some $M > 0$. The inequality follows because $\Gamma$ is a compact subset of $U$ and the sequence of iterates $(f^n)$ is normal on $U$, therefore $(f^n)$ is uniformly bounded on $\Gamma$. This shows that the orbit of any $z \in V$ under $f$ is bounded, and so $V$ is an open set contained in the filled Julia set $K(f)$. Thus, $V \subset \mathbb{C} \setminus J(f)$, where $J(f)$ is the Julia set of $f$, since $V$ cannot intersect $\partial K(f) = J(f)$. Hence, $V \subset U$, as $U$ is a connected component of $\mathbb{C} \setminus J(f)$ and $U \cap V \neq \varnothing$. Since $\Gamma$ was an arbitrary simple loop in $U$, this implies that $U$ cannot be a Herman ring.