Sequence of analytic functions and the limit function has same number of zeros

Question

Let $\{ f_n \}$ be a sequence of analytic functions on the closed ball $\overline{B}(0,R)$ that converges uniformly on this close ball to an analytic function $f.$ Assume that $f$ has no zeros on $|z|=R.$ Prove that for $n$ large $f_n$ has the same number of zeros in $B(0,R)$ as $f.$ My approach: Intutively this is clear. For a proof I tried, from the definition, for any $\epsilon >0,$ there exists $N > 0$ such that $|f_n(z)-f(z)| < \epsilon$ for $n > N$ and for all $z \in B(0,R).$ Since elements in the sequence and the limit function are analytic they both have Taylor expansions about $0,$ as well as two expressions can be written using the Argument principle. But I wasn't successful in showing the desired result. Any help is appreciated.

Answer 1

You may use Rouché's theorem: If $|g(z)-f(z)|<|f(z)|$ on the boundary of a compact set then $f$ and $g$ has the same number of zeros inside.

Since $f$ is continuous and has no zeros on the (compact) boundary $|z|=R$ there is $r>0$ so that $|f(z)|\geq r>0$ on the boundary. Now pick $N$ so that for $n\geq N$ we have $|f_n(z)-f(z)|\leq r/2< r \leq |f(z)|$ on the boundary and apply Rouché.