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  "body": "\n## Proof\n\n1. Let $G$, $(f_k)$, $f$, $m$ and $z_0$ be as in the statement. By [[uniform limits of holomorphic functions are holomorphic|0431RN0R1N]], $f$ is [[holomorphic|0DG0Z72CQD]] on $G$.\n2. The function $f$ is not identically zero on any disc around $z_0$. Otherwise the [[identity theorem|0YBVKMYX0W]] on the [[domain|097KNJFJ7V]] $G$ would give $f = 0$ on $G$. So the [[order|064N002TX4]] $m$ of $f$ at $z_0$ is finite.\n3. By step 2 and the [[isolated zeros theorem|0K83KEJ43P]], case 2 holds at $z_0$. Choose $\\rho_0 \u003e 0$ with $\\set{z : \\abs{z - z_0} \\le \\rho_0} \\subseteq G$ and $f(z) \\ne 0$ for $0 \u003c \\abs{z - z_0} \\le \\rho_0$.\n4. Let $0 \u003c \\rho \\le \\rho_0$. This is the meaning of \"sufficiently small\" below. Let $C$ be the circle $\\abs{z - z_0} = \\rho$.\n5. The circle $C$ is compact by the [[Heine–Borel theorem|07PFYV65MB]], and $f \\ne 0$ on $C$. By [[attainment of bounds|0V4DET59NW]], $\\abs{f}$ has a minimum $\\delta \u003e 0$ on $C$.\n6. Since $f_k \\to f$ uniformly on the compact set $C$, there is $N$ with $\\abs{f_k(z) - f(z)} \u003c \\delta$ for each $k \\ge N$ and $z \\in C$.\n7. Fix $k \\ge N$ and let $g = f_k - f$. It is holomorphic on $G$, an open set containing the closed disc $\\abs{z - z_0} \\le \\rho$.\n8. On $C$, $\\abs{g} \u003c \\delta \\le \\abs{f}$. By [[Rouché's theorem|0D430QR02W]], $f$ and $f + g = f_k$ have the same number of zeros in $\\abs{z - z_0} \u003c \\rho$, counted with order.\n9. In this disc, $f$ has only the zero $z_0$, by step 3, and its order is $m$. So $f_k$ has exactly $m$ zeros in $\\abs{z - z_0} \u003c \\rho$, counted with order, for each $k \\ge N$.\n\n## Idea\n\nChoose a small circle around $z_0$ on which $f$ has no zeros, so $\\abs{f}$ is at least some $\\delta \u003e 0$ there. For large $k$, $f_k$ differs from $f$ by less than $\\delta$ on the circle. Rouché's theorem then says that $f_k$ and $f$ have the same number of zeros inside, which is $m$.\n",
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