Do roots of a monic polynomial contain an open cube

Question

Let us consider monic polynomials of $n^{\text{th}}$ degree $p(t) = t^n + a_{n-1} t^{n-1} + \dots + a_1 t + a_0$ with real coefficients. We know the roots are continuous functions of the coefficients. Let us define a map $F: \mathbb R^n \to \mathbb R_+^n$ by
\begin{align*}
F: (a_{n-1}, \dots, a_0) \mapsto (r_1, \dots, r_n) \mapsto (|r_1|, \dots, |r_n|),
\end{align*}
where $r_j = r_j(a_{n-1}, \dots, a_0)$ are roots of the polynomial and $|\cdot|$ denotes the modulus of a complex number. Clearly $F$ is a continuous function. Let us take the Euclidean norm on $\mathbb R^n$. Suppose for some fixed coefficients $b = (b_{n-1}, \dots, b_0)$, the roots are all nonzero and let $(c_1, \dots, c_n)$ be the roots. if we take an open ball $B_b(\varepsilon)$ around $b$, will the image $F[ B_b(\varepsilon) ]$ contain a cube, i.e., $(|c_1|-\varepsilon', |c_1|+\varepsilon') \times (|c_2|-\varepsilon', |c_2|+\varepsilon') \times \dots \times (|c_n|-\varepsilon', |c_n|+\varepsilon')$ for some possible smaller $\varepsilon'$?

Answer 1

As I commented, your function $F$ is not well-defined. However, your question still makes sense without it. You can ask, given a multiset of nonzero complex numbers $c_1,\dots,c_n$ invariant under conjugation and $\epsilon>0$, does there exist $\epsilon'>0$ such that for any $r_1,\dots,r_n\in\mathbb{R}$ with $|r_k-|c_k||<\epsilon'$ for each $k$, there exist $d_1,\dots,d_n\in\mathbb{C}$ also invariant under conjugation such that $|d_k|=r_k$ for each $i$ and coefficients of the monic polynomial of with the $d_k$ as roots are within $\epsilon$ of the coefficients of the monic polynomial with the $c_k$ as roots. (The invariance under conjugation condition here corresponds to the requirement that the polynomial has real coefficients.)

The answer to this question is no. For instance, let $n=2$, and $(c_1,c_2)=(i,-i)$ and consider $r_1=1+\epsilon'/2$ and $r_2=1$. Notice then that if $|d_k|=r_k$ for $k=1,2$, then $d_1$ and $d_2$ cannot be conjugates of each other, and so they must both be real in order to be invariant under conjugation. It follows that there are no such $d_1$ and $d_2$ that give rise to a polynomial close to the polynomial $t^2+1$ with roots $c_1$ and $c_2$.

Or, in more concrete terms, there is no polynomial with real coefficients close to $t^2+1$ whose roots have distinct absolute values, since any nearby polynomial has no real roots and hence the roots must be a complex conjugate pair.