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  "body": "\n## Proof\n\n1. Let $D$, $f$, $g$, $S$ and $z_0$ be as in the statement. Let $h = f - g$. It is [[holomorphic|0DG0Z72CQD]] on $D$, and $S$ is its set of zeros.\n2. Let $A$ be the set of $z \\in D$ such that $h = 0$ on some open disc with center $z$ contained in $D$.\n3. $A$ is open. If $h = 0$ on $\\set{w : \\abs{w - z} \u003c r}$ and $\\abs{w - z} \u003c r$, then $h = 0$ on the disc with center $w$ and radius $r - \\abs{w - z}$.\n4. $D \\setminus A$ is open. Let $z \\in D \\setminus A$. By the [[isolated zeros theorem|0K83KEJ43P]] at $z$, case 1 fails, so case 2 holds.\n5. After shrinking the radius of case 2, there is $r \u003e 0$ with $\\set{w : \\abs{w - z} \u003c r} \\subseteq D$ and $h(w) \\ne 0$ for $0 \u003c \\abs{w - z} \u003c r$.\n6. Let $w$ satisfy $0 \u003c \\abs{w - z} \u003c r$. Since $h(w) \\ne 0$, $w \\notin A$. Also $z \\notin A$. So the disc of step 5 lies in $D \\setminus A$.\n7. $z_0 \\in A$. By assumption, each disc with center $z_0$ contains a point $w \\ne z_0$ with $h(w) = 0$. So case 2 of the isolated zeros theorem fails at $z_0$.\n8. Hence case 1 holds at $z_0$: $h = 0$ on a disc with center $z_0$. Shrinking it so that it lies in $D$ shows $z_0 \\in A$.\n9. By steps 3, 4 and 6, $D = A \\cup (D \\setminus A)$ with both sets open and disjoint. Since the [[domain|097KNJFJ7V]] $D$ is [[connected|0D939CMME7]], one of them is empty.\n10. By step 8, $A$ is not empty. So $A = D$, and $h = 0$ on $D$. That is, $f(z) = g(z)$ for each $z \\in D$.\n\n## Idea\n\nThe set of points near which $f - g$ vanishes identically is open by definition. By the isolated zeros theorem its complement is also open, and the accumulation point lies in it. A connected set cannot split into two disjoint nonempty open parts, so the set is all of $D$.\n",
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