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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.

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    $\begingroup$ $f’(v)$ is another polynomial, so it must have finitely many roots. $\endgroup$ – Alex R. May 22 at 3:25
  • $\begingroup$ 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. $\endgroup$ – Xander Henderson May 22 at 3:45
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$f'(v)$ 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 $w$.

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