Let $f : \mathbb{C} \to \mathbb{C}$ be any non-constant complex polynomial and suppose $f(z_0) = z$.
If $\{w_n\}$ is a sequence of complex numbers such that $w_n \to z $, prove that there exist complex numbers $v_n$ converging to $z_0$ so that $f(v_n) = w_n$ for each $n$.
This fact is claimed on page 218 of Falconer's Fractal Geometry (2nd ed.), within the proof of Proposition 14.3.
By the fundamental theorem of algebra, it's clear that $f$ is surjective, so the $w_n$ do lie in the range of $f$. However, what's not clear to me is how to ensure we can select the $v_n$ so they converge to $z_0$.
It may also be helpful that $$f(z) = \sum_{k=0}^N \frac{f^{(k)}(z_0)}{k!}(z - z_0)^k$$ for some $N$ finite, although I haven't figured out how to use that.
Hints or solutions greatly appreciated.
Apply Rouche's Theorem to small disks $B(z_0,\frac 1 n)$. Since $f$ is non-constant $f-z$ has no zeros on the boundary of this disk for $n$ sufficiently large so $\inf \{|f(\zeta)-z|:|\zeta -z_0|=\frac 1 n \} >0$. Hence $|w_n-z| <|f(\zeta)-z|$ on the boundary for $n$ sufficiently large. By Rouche's Theorem $f-z$ and $f-w_n$ have the same number of zeros in the disk. Since $f-z$ has a zero in the disk it follows that there exists $v_n$ in the disk with $f(v_n)=w_n$. Of course, v_n \to z_0$.
Note that the argument works for analytic functions. We don't need a polynomial for this.