# 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

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 $$z$$?

## 2 Answers

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

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.