# Do roots of a monic polynomial contain an open cube?

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 $\varepsilon'$?

• Your map $F$ is not well-defined: the roots are only defined as an unordered collection, not as an ordered $n$-tuple. It would be well-defined (and continuous) if you changed the codomain to $\mathbb{R}^n_+/S_n$. – Eric Wofsey Jul 8 '18 at 6:57
• You could still consider your $F$ as a "multi-valued function" and ask your question about the image $F[B_b(\epsilon)]$, though. – Eric Wofsey Jul 8 '18 at 6:58

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.
• Thanks. Will the ''conclusion'' hold: if we let the roots to be identical real numbers, i.e., $c=(\beta, \beta, \dots, \beta)$? Here by conclusion I mean: an open cube $(|\beta| - \varepsilon', |\beta| + \varepsilon' )^n$ such that all the conjugate invariant subset is contained in the image of $F[B_b(\varepsilon)]$? – user1101010 Jul 8 '18 at 16:04
• Yes, it is true if the roots are all real. This follows immediately from the fact that the function taking a collection of roots to the coefficients of the corresponding polynomial is continuous (since it's just given by the elementary symmetric polynomials). So, just pick a collection of real roots near $\beta$ with the desired absolute values, and if they are sufficiently near $\beta$, the resulting polynomial will be sufficiently near the initial polynomial. – Eric Wofsey Jul 8 '18 at 16:30