Problem in complex analysis regarding pairwise distinct roots of a polynomial Problem: Let $D \subseteq \mathbb{C}$ be an open set and $a_0, \dots, a_n : D \to \mathbb{C}$ continuous complex functions. For $w \in D$ we defne a polynomial
$$p_w (z) = a_n (w) z^n + \dots + a_1 (w)z + a_0(w).$$
Suppose that there exists $w_0 \in D$ such that $p_{w_0}$ has pairwise distinct zeroes and $a_n(w_0) \neq 0$.
Prove that there exists a neighborhood $U$ of $w_0$ such that $p_w$ has pairwise distinct zeroes for every $w$ in that neighborhood.
I realize that the setup of this problem hints at Rouche's theorem but I still have no idea how to solve it.
 A: The $n$ complex roots $r_1(w),...,r_n(w)$ of $p_w(z)$ depend continuously on the coefficients $a_1(w),...,a_n(w)$ (which in turn are assumed to depend continuously on $w$).
If we assume that there is a $w_0\in \mathbb{C}$ so that all roots are distinct, we can calculate something like $$g(w) = \min \{|r_i(w)-r_j(w)| : 1\leq i\neq j\leq n\},$$
from which we know $g(w_0)>0$. As a composition of continuous functions, $g$ is again continuous, and so we find a neighbourhood of $w_0$ for which $g(w)>0$ is still fulfilled.
EDIT About the continuous dependency of the roots on the coefficients:
Let's just view the coefficients $a_0,...,a_n\in\mathbb{C}$ of $p(z) = \sum\limits_{k=0}^n a_kz^k$ as independent variables, i.e. we look at the function
$$\Phi : \mathbb{C}^{n+1}\times\mathbb{C}\rightarrow \mathbb{C} , \quad (a_0,...,a_n,z)\mapsto \sum\limits_{k=0}^n a_kz^k.$$
Then $\Phi(\cdot,\cdot)$ is continuously differentiable and due to the initial assumption we know that there exists a point $(a',z') := (a_0',...,a_n',z')\in \mathbb{C}^{n+1}\times\mathbb{C}$ so that $$\Phi(a',z') = 0. $$
Also, due to the additional assumption that all roots are distinct, we have  $$\frac{\partial \Phi}{\partial z}(a',z') = \frac{d p}{d z}(z')\neq 0.$$
So due to the implicit function theorem, we find open neighbourhoods $U_{a'}\subset\mathbb{C}^n$ and $U_{z'}\subset\mathbb{C}$ with $(a',z')\in U_{a'}\times U_{z'}$ and also a continuous(ly differentiable) function $\varphi : U_{a'}\rightarrow U_{z'}$, so that $$\Phi(a,z) = 0 \Leftrightarrow z = \varphi(a) \quad \forall (a,z)\in U_{a'}\times U_{z'},$$
i.e. the root $z$ depends continuously on the coefficients $a_0,...,a_n\in\mathbb{C}$.
