Roots of a polynomial depend continuously on its coefficients

Idea. If $|p(z)|\le\delta$ then the product of the $n$ distances $|z-r_i|$ is at most $\delta$, so one of them is at most $\delta^{1/n}$. Smallness of $p(z)$ at a root of $q$ comes from the closeness of the coefficients.

Step 1: Cauchy bound. Every root $z$ of a monic $z^n+a_{n-1}z^{n-1}+\dots+a_0$ satisfies $|z|\le1+\max_k|a_k|$. (If $|z|>1$ then $|z|^n=|\sum a_kz^k|\le\max|a_k|\frac{|z|^n-1}{|z|-1}<\max|a_k|\frac{|z|^n}{|z|-1}$, so $|z|-1<\max|a_k|$.) Put $m=\max_k|c_k|$ and $R=2+m$. Since $\eta\le1$, $\max|d_k|\le m+1$, so all roots of $p$ and of $q$ lie in $\{|z|\le R\}$, and $R\ge2\ge1$.

Step 2: $p(z)$ is small at a root of $q$. Let $q(z_0)=0$, $|z_0|\le R$. Then
$$|p(z_0)|=|p(z_0)-q(z_0)|=\Big|\sum_{k<n}(c_k-d_k)z_0^k\Big|\le\eta\sum_{k=0}^{n-1}R^k\le\eta\,nR^{n-1}=:\delta.$$

Step 3: some root is close. $|p(z_0)|=\prod_{i=1}^n|z_0-r_i|\le\delta$, so $\min_i|z_0-r_i|^n\le\prod_i|z_0-r_i|\le\delta$? Careful: this needs *every* factor to be $\ge\min$, so $\min_i|z_0-r_i|^n\le\prod_i|z_0-r_i|$. Hence
$$\min_i|z_0-r_i|\le\delta^{1/n}=(nR^{n-1})^{1/n}\eta^{1/n}=:C\eta^{1/n}.$$

Step 4: converse. Let $r_i$ be a root of $p$; $|r_i|\le R$. Then $|q(r_i)|=|q(r_i)-p(r_i)|\le\delta$ by the same computation, and $q(r_i)=\prod_j(r_i-s_j)$ with $s_j$ the roots of $q$, so some $s_j$ satisfies $|r_i-s_j|\le\delta^{1/n}$.

Sharpness of the exponent $\frac1n$. For $p=z^n$ and $q=z^n-\eta$ the roots of $q$ have modulus $\eta^{1/n}$: with $n=3$, $\eta=10^{-6}$, the roots have modulus $0.01$ (checked numerically). So one can't replace $\eta^{1/n}$ by $\eta$ in general; with *simple* roots one can (implicit function theorem).

Remark. This is a "quantitative" form of the theorem that roots are continuous functions of coefficients. It applies to characteristic polynomials, since the coefficients of $\chi_A$ are polynomial in the entries of $A$.