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 $z$? 

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
$$
p(z,w) = w^n -z.
$$
