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

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.

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.