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 $z$? 

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.
