the zeros of a polynomial depend in a continuous way on its coefficients.

Question

We use $P_n$ to denote the vector space of complex polynomials of degree $\leq n$, and we write $$ \mathcal{P}^{*}_n = \mathcal{P}_n \setminus \mathcal{P}_{n-1} $$ for the set of polynomials of exact degree $n$. The distance between two polynomials $P(z) = \sum_{j=0}^{n} a_jz^j$ and $Q(z) = \sum_{j=0}^{n} b_jz^j$ in $P^*_n$ is $$ d(P, Q) = \max_{j=0,\ldots,n} |a_j - b_j|. $$ a) Let $P \in \mathcal{P}^*_n$ be a polynomial of degree $n$ with zeros $z_1, \ldots, z_n$ in the complex plane, where zeros are listed according to their multiplicity. Prove that for every $\varepsilon > 0$ there exists $\delta > 0$ such that every $Q \in \mathcal{P}^*_n$ with $d(P, Q) < \delta$ has zeros $w_1, \ldots, w_n$ satisfying $$ \max_{j=1,\ldots,n} |z_j - w_{\sigma(j)}| < \varepsilon $$ for some permutation $\sigma$ of the set $\{1, \ldots, n\}$.

Hint: It is possible to use Rouché’s theorem.

b) Is the following statement also true?

For every $n$, there is a continuous mapping $$ Z_n : \mathcal{P}^*_n \rightarrow \mathbb{C}^n $$ with the property that if $P \in \mathcal{P}^*_n$ and $\mathcal{Z}_n(P) = (z_1, \ldots, z_n)$, then $z_1, \ldots, z_n$ are the zeros of $P$, listed according to their multiplicity.

Prove or disprove this statement.

I was going through some Complex Analysis books and came across this question. I have no idea how to solve it, so maybe someone can help me?

Thanks in advance!

Answer 1

For b, take a look at the family of polynomials $\frac{z^2}{t}-1$, where $t\in\mathbb{C}_0$. The roots of these polynomials are $\pm \sqrt{t}$. Recall that the square root needs a branch cut $B$ to be well-defined on $\mathbb{C}\setminus B$, hence, there is no continuous mapping for this family of functions, so this is a counterexample, with $n=2$.

EDIT: technically the square root needs 2 branch cuts, but for this argument I only used the absence of branch cuts, so the same reasoning applies.

Answer 2

For a, you can use the following:

Take $\varepsilon$ small enough such that the circles $C_j=\{z:|z-z_j|=\varepsilon\}, \hspace{1mm} j=1,...,n$ are disjoint. You can now define something like \begin{equation*} \rho_j= \min_{z\in C_j} |P(z)| \end{equation*} At the same time, you can look for a $\delta$ such that \begin{equation*} \delta \max_{z \in C_j} \left\{\sum_{j=0}^n |z^j|\right\}<\rho_j \end{equation*} If $d(P,Q)<\delta$, you can estimate to the point where you have that \begin{equation*} |Q(z)-P(z)|< |P(z)| \end{equation*} Where now using Rouché's Theorem, you get the answer.