Roots of a polynomial with holomorphic coefficients Let $f_1,f_2,\dots,f_n : \mathbb{D} \to \mathbb{C}$ be holomorphic functions and consider the polynomial 
$$ w^n + f_1(z)w^{n-1} + \dots + f_n(z). $$
Suppose, I happen to know that


*

*For each $z$, the roots of the above polynomial are all in $\mathbb{D}$.

*For each $z$, one of the roots is an $n$-th root of $z$.
Given these two conditions, is it true that for each $z$, all the roots of the above polynomial are nothing but $n$-th roots of $z$? 
 A: It is true. Here's the proof. Let $\zeta=e^{\frac{2\pi i}{n}}$ be the $n$-th root of unity. Note that by the assumption, for each $z\in \mathbb{D}$, there is $k\in \{0,1,\ldots,n-1\}$ such that
$$
p(z^n, z\zeta^k) = 0
$$ where $$
p(z,w) = w^n + f_1(z)w^{n-1} + \cdots + f_n(z).
$$ Let $D_k$ be the set of all $z\in \mathbb{D}$ such that $p(z^n, z\zeta^k) = 0$. Then each $D_k$ is closed in $\mathbb{D}$ and we have
$$
\mathbb{D} = \bigcup_{0\leq k\leq n-1} D_k.
$$ Note that by pigeonhole principle, one of $D_k$ has $0$ as its limit point. Then Identity theorem implies that for some $k$,
$$
p(z^n, z\zeta^k) \equiv 0\quad\cdots(*),
$$ for all $z\in\mathbb{D}$. Note that change of variable $z\mapsto z\zeta^j$ yields
$$
p(z^n, z\zeta^{k+j})=0,\quad\forall j,
$$ and hence
$$
p(z^n, z\zeta^{j})=0, \quad \forall j=0,1,\ldots, n-1.
$$ It is saying that $p(z,w)=0$ for $\textbf{all}$ roots of $w^n = z$, giving us the desired result that
$$
p(z,w) = w^n -z.
$$
A: Here's a proof for $n=2$, using only assumption 2. The assumption that $w^2 + f_1(z)w + f_2(z)$ always has a square root of $z$ as a root means that the polynomials $w^2 + f_1(z)w + f_2(z)$ and $w^2 - z$ (thought of as polynomials in $w$) share a root for every $z$. This implies that their resultant vanishes identically in $z$:
$$
\mathop{\rm Res}\big(w^2 + f_1(z)w + f_2(z),w^2 - z\big) = -z f_1(z)^2 + f_2(z)^2 +2 f_2(z) z + z^2 = 0.
$$
In particular, this implies that $f_2(z) + z = \pm f_1(z) \sqrt z$ for every $z$ near $0$, say. But since $f_2(z)+z$ and $f_1(z)$ are both assumed holomorphic near $z=0$, the only possibility (by considering the rate of vanishing of both sides as $|z|\to0$) is that both sides equal $0$; that is, $f_1(z) = 0$ and $f_2(z) = -z$. In other words, $w^2 + f_1(z)w + f_2(z) = w^2 - z$, and the two roots are now clearly the two square roots of $z$.
It would be nice if the same proof extended to $n\ge3$ (my hunch: why not?), but someone who knew more about resultants would need to figure out the generalization.
