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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

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

  2. 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 $z$?

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

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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.

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