Show that if $f_n$ converges uniformly to $f$, and $f$ has $k$ zeroes. Then $f_n$ has $k$ zeroes when $n>N$ The question
Let $(f_n)_{n≥1} ∈ \mathcal{H}(Ω)$ such that $\ f_n \overset{Ω}{\implies} f$. Show that if $f$ has $k$ zeros (counting multiplicities) at $Ω$, then for a $N≥1$ the function $f_n$ has at least $k$ zeros at $Ω$, for all $n≥N$.
What I've tried
If $z_0$ is a zero of $f$ with $m_0$ multiplicity, then:
$$f^{(j_0)}(z_0) = 0,\quad ∀j_0 < m_0$$
Applying Weierstrass we know that $\ f_n^{(k)}(z) \overset{Ω}{\implies} f^{(k)}(z)$. So the pointwise convergence is also guaranteed at any zero of $f$. Then I can find an $N_0$ such that for every $n≥N_0$:
$$ \left|f_n^{(j)}(z_0) - f^{(j)}(z_0)\right| =   \left|f_n^{(j)}(z_0)\right| ≤ ε,\quad ∀j<m_0 $$
Then one can choose $N = \max\{N_i\}$ (where $N_i$ is the index found for each zero of $f$), so $|f_n^{(j)}|$ is small enough at every zero when $n≥N$.

I only show that the convergence states at the zeros, but I haven't shown that $f_n$ is exactly zero at those points when $n≥N$.
Any help?
 A: Well, take a loop $\gamma$ enclosing all the zeros of $f$, such that $f$ does not have a zero on that loop. Let $\delta$ be the minimum of $|f|$ on $\gamma$. This exists since $\gamma$ is compact. Let $\epsilon = \delta/2$. Then for $n \gg 0$, $|f_n (z) - f(z)| < \epsilon$. So $|f_n(z)| > |f(z)| -\epsilon \ge \delta - \delta/2  = \delta /2$ on the loop $\gamma$. Therefore, for $n \gg 0$, $f_n(z)$ does not vanish on the loop $\gamma$. WLOG assume the same for all $n \ge 1$ as the initial things do not matter.
Now by argument principle, for $n \ge 1 $ ,$$ Z_{\gamma}(f_n) = \int_{\gamma} \frac{f_n '}{f_n } , $$ where $Z_{\gamma}(f_n)$ denotes the zeroes inside $\gamma$.
But on $\gamma$, $\frac{f_n '}{f_n }$ converges uniformly to $\frac{f'}{f}$.
Therefore by uniform convergence, $$\lim _{n \to \infty} Z_{\gamma}(f_n) =   \lim_{n \to \infty}\int_{\gamma} \frac{f_n '}{f_n }  = \int_{\gamma} \frac{f '}{f } = Z_{\gamma}(f) = k $$
So for $n \gg 0$, $f_n(z)$ has $k$ zeros inside the loop $\gamma$, in particular it has at least $k$ zeros in the domain.
