Rouche's Theorem applied to a sequence of analytic functions converging to a non-constant function with zero in a domain

Question

Suppose a sequence of analytic functions $\{f_n\}$ in domain $D$ converge uniformly on every compact subset of $D$ to a non-constant function $f(z)$. Let $f(a) = 0$ for some $a \in D$. Using Rouche's Theorem, show that there is a sequence $a_n \in D$ such that $\lim_{n \to \infty} a_n = a$ and $f_n(a_n) = 0$ for sufficiently large $n$.

Is the sequence $\{a_n\}$ suppose to be a sequence of zeros? I thought Rouche's Theorem is used to show the number of zeros and poles of some function is the same as another function if this inequality is satisfied but I am not sure how the theorem applies in this context. Any tips would be appreicated.

Answer 1

An outline for which you can fill in the details:

1. Show/note that $f$ is analytic in $D$.

2. Thus there is a ball $B$ with $a \in B$ so that the only zero $f$ has in $B$ is at $a$.

3. Let $\gamma$ denote the boundary of $B$ oriented counterclockwise and set $c = \min_{z \in \gamma} |f(z)|$.

4. Since $f_n \to f$ uniformly on compact sets, for $n$ large enough we have $ \max_{z \in \gamma} |f_n(z) - f(z)| = o(1).$

5. What happens if you apply Rouche's theorem to $f$ and $f_n - f$ on the contour $\gamma$?

Answer 2

Choose $g=f-f_n$. Also choose circles $C_n$ centered at $a$ whose radius converges to $0$ as $n\to\infty$. Then show that for sufficiently large $n$, Rouché's theorem can be applied. It will tell you that $f$ and $f_n$ have the same number of roots within that circle. And since the circles get arbitrarily small, the roots will converge to $a$.