I was reading this article which claimed the following:
$$|z-\alpha|\leq n\left|\frac{p(z)}{p'(z)}\right|$$
Where $p$ is a polynomial with degree $n$, and $\alpha$ is the zero of $p$ closest to $z$.
Why is this the case?
I know that if we replace $z$ with $\delta=z-\alpha$, then we get an equivalent polynomial $q(\delta)$ with degree $n$, which would mean $\left|\frac{p(z)}{p'(z)}\right|=\left|\frac{q(\delta)}{q'(\delta)}\right|$. However, I can't blindly apply the triangle inequality, so I'm struggling to figure out how to prove this.
If $a_1, \ldots, a_n$ are the zeros of $p$, counted with multiplicity, then $$ p(z) = c(z - a_1) \cdots (z-a_n) $$ implies $$ \frac{p'(z)}{p(z)} = \frac{1}{z-a_1} + \cdots +\frac{1}{z-a_n} $$ and therefore $$ \left| \frac{p'(z)}{p(z)} \right| \le \sum_{k=1}^n \frac{1}{|z-a_k|} $$ for all $z \in \Bbb C \setminus \{ a_1, \ldots, a_n \}$.
Now assume that $p'(z) \ne 0$ and let $\alpha \in \{ a_1, \ldots, a_n \}$ be the zero closest to $z$. If $p(z) \ne 0$ then $$ \left| \frac{p'(z)}{p(z)} \right| \le \frac{n}{|z-\alpha|} \implies |z-\alpha| \le n \left| \frac{p(z)}{p'(z)} \right| \, . $$ If $p(z)= 0$ then $\alpha = z$ and the last inequality holds trivially.