# For complex polynomial $f(z)$ show there are finitely many $w$ such that $f(v) = w \neq v$ implies $f'(v) =0$.

Let $$f(z) : \mathbb{C} \to \mathbb{C}$$ be a complex polynomial of degree $$n \ge 2$$. Consider the set

$$A = \{w \in \mathbb{C} : \text{there exists } v \neq w \text{ with } f(v) = w \text{ and } f'(v) \neq 0\}.$$

I would like to show that $$A$$ contains all but finitely many points of $$\mathbb{C}$$.

This fact is asserted on page 221 of Falconer's Fractal Geometry (2nd ed., within the proof of Theorem 14.10).

Here is my attempt: if $$w \in \mathbb{C} \setminus A$$, then either the $$z =w$$ is the only solution of $$f(z) - w =0$$ (in which case $$w$$ is a root of $$f(z) -w$$, of which there are only finitely many), or for all $$v \neq w$$ with $$f(v) = w$$, it holds that $$f'(v)=0$$. I cannot show why this second condition should hold for only finitely many $$w$$.

If it's needed, we can replace $$A$$ by $$A \cap J$$, where $$J$$ is the Julia set of the polynomial $$f$$, and instead try to show $$\mathbb{C} \setminus A$$ is finite, but I don't think that is needed here.

Hints or solutions are greatly appreciated.

• $f’(v)$ is another polynomial, so it must have finitely many roots. – Alex R. May 22 at 3:25
• While your study of fractal geometry is motivation for this question, the question itself actually has nothing to do with fractals. I have removed that tag. – Xander Henderson May 22 at 3:45

$$f'(v)$$ is a polynomial. It can only have finitely many roots. So there are only finitely many values of $$f(v)$$ for which $$f'(v)=0$$. $$w=f(v)$$, so there are only finitely many $$w$$.