"If $U$ is an open subset of the complex plane, then matrices $X\in\textrm{M}(n,\mathbb C)$ all of whose eigenvalues belong to $U$ make up an open subset of $\textrm{M}(n,\mathbb C)$." Trying to prove this by using the argument principle, I was led to the following question: do approximately the same polynomials have approximately the same roots? More precisely, fix a complex polynomial $p(z)=a\_0+a\_1 z+\ldots+a\_n z^n$ with $a\_n\neq 0$. Let us say that a polynomial $q(z)=b\_0+b\_1 z+\ldots+b\_n z^n$ is in the $\varepsilon$-neighborhood of $p(z)$ if $|b\_i-a\_i|<\varepsilon$ for all $0\leq i\leq n$. Now suppose all roots of $p(z)$ lie in, say, the open unit disk. Does $p(z)$ have an $\varepsilon$-neighborhood consisting of polynomials with the same property?



Do approximately the same polynomials have approximately the same roots?
I think so, yes. Suppose all the roots lie *inside* the unit disc. By compactness $|p(z)|$ takes a positive minimum $\delta$ on the contour (the unit circle). For $q(z)$ in the epsilon-neighbourhood, $q(z)-p(z)$ is a polynomial with epsilon coefficients, so we can take epsilon small enough so that on the contour $|q(z)-p(z)|<\delta<|p(z)|$. Then appeal to Rouché's theorem to learn that $p(z)$ and $q(z)$ have the same number of zeros in the region.


