Show that any complex polynomial of degree $n \ge 2$ has finitely many neutral or attracting orbits

Question

Let $f: \mathbb{C} \to \mathbb{C}$ be a complex polynomial of degree $n \ge 2$. A point $w \in \mathbb{C}$ is a periodic point of $f$ of minimal period $p$ if $f^{\circ p}(w) =w$ and $p$ is the smallest positive integer such that this equality holds. The orbit of $w$ is the set $\{w, f(w) , \dots, f^{\circ (p-1) }(w) \}$.

A periodic point $w$ is called attracting if $|(f^{\circ p})'(w)| < 1$, neutral if $|(f^{\circ p})'(w)| = 1$, and repelling if $|(f^{\circ p})'(w)| > 1$.

I would like to show that $f$ can only finitely many neutral or attracting orbits.

If I have this fact, then I can complete an argument (along the lines of the proof of Theorem 14.10 in Falconer's Fractal geometry, 2nd ed.) which shows the Julia set of $f$ is exactly the closure of the set of repelling periodic points of $f$.

For any fixed $p$, $f^{\circ p}$ can of course have only finitely many fixed points (of any type), but when one allows $p$ of any size it's not clear to me how the number of attractive or neutral orbits remains bounded. The size of the derivative will need to be used but I don't yet see how.

Hints, solutions, or reference suggestions are greatly appreciated.

Answer 1

I'll just give some hints (using $d$ for the degree of $f$, rather than $n$, which I reserve for the iterates):

1. Note that $(f^{\circ n})'(z)$ is just a product of $f'$ evaluated at $z,f(z),f^{\circ 2}(z),\dots,f^{\circ(n-1)}(z)$, so the number of attracting periodic orbit is at most $d-1$, since its basin of attraction must contain a critical point (excluding the point at infinity).
2. For the number of neutrals, the idea is to perturb $f$ in some way so that the neutral become either attracting or repelling, with at least half of them becoming attracting (since we know how to count them). Note that $z^d$ has no neutral periodic orbits. Conjugating $f$ by $z\mapsto cz$, we may assume $f(z)=z^d+\dots$. Then we use a carefully chosen $t\in\mathbb{C}$ with the map $f_t(z)=(1-t)f(z)+tz^d$. Try this with the condition $z_0(t)$ being an order $m$ neutral periodic orbit and you eventually get half of the directions $t$ (interpreting $t$ as a curve parametrised by $\tau\in(-\epsilon,\epsilon)$) result in $z_0$ becoming attracting. Double count the other way, since there are at most $d$ attracting orbits, this gives at most $2d$ neutral perioidic orbits.