Vrftsjtryk

$newcommand{a}{alpha} newcommand{bb}{mathbb} newcommand{b}{beta}$
The question originates from a comment by reuns of this question Geometry of the set of coefficients such that monic polynomials have roots within unit disk.

Let $a = (a_0, dots, a_{n-1}) in bb R^n$ be fixed and $a neq 0$. We consider the parametrized family of monic polynomials with degree $n$
$$ f(r, z) = z^n + ra_{n-1} z^{n-1} + dots + r a_1 z + r a_0,$$
where $r in bb R$.

For each $r in bb R$, let $L(r)$ denote the largest modulus of roots of $f(r, z)$. $L(r)$ is not a polynomial in $r$ (Is the largest root of a parametrized family of polynomials again polynomial?). But $L(r)$ should be a continuous function over $r$. I am wondering whether the number of zeros of $L(r) - 1$ is finite.

Following was some wrong attempt. See the question and comments here If coefficients are polynomial functions over $mathbb R$ of a monic polynomial, can we find $n$ continuous functions that constitute the roots?.
I think the zeros should be finite. Informally, we can consider
$$ f(r, z) = z^n + a_{n-1}(r) z^{n-1} + dots + a_1(r) + a_0(r).$$
This just rewrites $r a_i$ as a function over $r$. Since we only consider $r in bb R$, there exists $n$ continuous functions $b_0(r), dots, b_{n-1}(r)$, that for each $r in bb R$, constitutes the roots of $f(r, z)$. By Vieta's formula, all $b_j(r)$ should be polynomial functions in $r$ (This part I feel is problematic). Then for each $j$, $b_j(r) - 1$ would have at most $n$ zeros. The worst scenario would be these zeros happening in separate manner in the sense: the zeros of $b_j(r) -1 $ are those $r$'s that $b_j(r)$ also is the largest root of modulus. This gives us $n^2$ zeros of $L(r) -1 $. Is this argument correct? Could a tighter bound be provided?

Edit: Before I started the bounty, there is one answer by Akari Gale which I am not sure of . This is one of the reasons I started a bounty. As of now, there is one downvote of that answer. Could someone leave a comment on why the answer is not OK? Thanks.

If $L(r) = 1$ for some $r$, then $z=cos(x) + isin(x)$ is a root of $f(r, z)$ for some $x in mathbb{R}$. We can write
$$0 = z^n + ralpha_{n-1}z^{n-1} + ... + ralpha_1z + ralpha_0 Rightarrow r =-frac{-1}{alpha_{n-1}z^{-1} + ... + alpha_1z^{-n+1} + alpha_0z^{-n}}$$
so
$$alpha_{n-1}z^{-1} + ... + alpha_1z^{-n+1} + alpha_0z^{-n} in mathbb{R} Rightarrow $$
$$alpha_{n-1}sin(-x) + ... + alpha_1 sin((-n+1)x) + alpha_0 sin(-nx) = 0 Rightarrow$$
$$alpha_{n-1} sin(x) + ... + alpha_1 sin((n-1)x) + alpha_0 sin(nx) = 0$$
It can be written in the form $P(sin(x), cos(x)) = 0$ for $P in mathbb{R}[X, Y]$ using multiple-angle formulas. Or $P(frac{1-t^2}{1+t^2}, frac{2t}{1+t^2}) = 0$

If this polynomial after multiplication by $(1 + t^2)^n$ is not zero, then for r there is only a finite number of possible values.