real analysis - Does $f_n(z)neq 0$ for all $ngeq 1$ imply $f(z)neq 0$?

Let $z\in\mathbb{C}$, and $\left(f_n(z)\right)_{n=1}^\infty$ be a sequence of non-zero functions of $z$, i.e. $f_n(z)\neq 0$ for all $n\geq 1$. Suppose that $f_n(z)\to f(z)$ uniformly with $f$ not identically zero, and $f$ and $f_n$ holomorphic. Does this imply that $f(z)\neq 0$ ?

I have been looking at Hurwitz' theorem from complex analysis, but I get the impression under Hurwitz' theorem the condition $f_n(z)\neq 0$ has to hold on whole sets of points, and does not work for single points.

Answer

You probably want to also add the hypotheses that $f_n$ and $f$ are holomorphic. But as you pointed out, Hurwitz's theroem doesn't work at a single point. It only tells you that if $f$ has a zero at $z_0$, then the $f_n$ must have zeros near $z_0$ for sufficiently large $n$. For example, if you define $f_n(z) = z + \frac{1}{n}$, then $f_n \to f(z) = z$ uniformly on $\mathbb{C}$. Now $f(0) = 0$, but $f_n(0) \neq 0$ for all $n \geq 1$. However, each $f_n$ has a zero at $z = -\frac{1}{n}$, thus if you take any $\epsilon > 0$, then $f_n$ has a zero inside the disk $\vert z \vert < \epsilon$ for sufficiently large $n$.