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  "body": "\n## Proof\n\nWe argue by contradiction.\n\n1. Let $n \\ge 1$, let $a_0, \\dots, a_n$ be [[complex numbers|09FTAET5Z4]] with $a_n \\ne 0$, and let $p(z) = \\sum_{k=0}^{n} a_k z^k$.\n2. Assume that $p(z) \\ne 0$ for each $z \\in \\C$.\n3. The polynomial $p$ is a power series with finitely many nonzero coefficients, so its radius of convergence is $\\infty$. By [[power series are holomorphic|0YMP521ZR5]], $p$ is [[holomorphic|0DG0Z72CQD]] on $\\C$.\n4. Then $p$ is continuous: $p(z + h) - p(z) = h \\cdot \\frac{p(z + h) - p(z)}{h} \\to 0 \\cdot p'(z) = 0$ as $h \\to 0$.\n5. Let $g = 1/p$, defined on $\\C$ by step 2. For $z \\in \\C$ and small $h \\ne 0$, $$\\frac{g(z + h) - g(z)}{h} = - \\frac{p(z + h) - p(z)}{h} \\cdot \\frac{1}{p(z + h) \\, p(z)}.$$\n6. By steps 3 and 4, the right side of step 5 tends to $-p'(z)/p(z)^2$. So $g$ is [[complex differentiable|0D1C9Y1EXA]] at each point, and $g$ is [[entire|0419794R0T]].\n7. Let $S = \\sum_{k=0}^{n-1} \\abs{a_k}$ and $R_0 = \\max(1, 2S/\\abs{a_n})$. Let $\\abs{z} \\ge R_0$.\n8. By the [[triangle inequality|03HF5PMY5T]], $\\abs{p(z)} \\ge \\abs{a_n} \\abs{z}^n - \\sum_{k=0}^{n-1} \\abs{a_k} \\abs{z}^k$.\n9. Since $\\abs{z} \\ge 1$, we have $\\abs{z}^k \\le \\abs{z}^{n-1}$ for $k \\le n - 1$. So $\\abs{p(z)} \\ge \\abs{z}^{n-1} (\\abs{a_n} \\abs{z} - S)$.\n10. Since $\\abs{z} \\ge 2S/\\abs{a_n}$, we have $S \\le \\abs{a_n} \\abs{z} / 2$. So $\\abs{p(z)} \\ge \\abs{a_n} \\abs{z}^n / 2 \\ge \\abs{a_n}/2$.\n11. The closed disc $K = \\set{z : \\abs{z} \\le R_0}$ is closed and bounded, so it is compact by the [[Heine–Borel theorem|07PFYV65MB]].\n12. The function $\\abs{p}$ is continuous by step 4. It [[attains its minimum|0V4DET59NW]] $m$ on $K$, and $m \u003e 0$ by step 2.\n13. By steps 10 and 12, $\\abs{g(z)} \\le \\max(1/m, 2/\\abs{a_n})$ for each $z \\in \\C$. So $g$ is bounded.\n14. By [[Liouville's theorem|0EHKVRF4J6]], $g$ is constant. Hence $p = 1/g$ is constant, and $\\abs{p}$ is bounded on $\\C$.\n15. But step 10 gives $\\abs{p(z)} \\ge \\abs{a_n} \\abs{z}^n / 2$ for $\\abs{z} \\ge R_0$. Since $n \\ge 1$, this tends to $\\infty$ as $\\abs{z} \\to \\infty$.\n16. Steps 14 and 15 contradict each other. So assumption 2 is false, and there is $z_0 \\in \\C$ with $p(z_0) = 0$.\n\n## Idea\n\nIf $p$ had no zero, then $1/p$ would be an entire function. It is small far away, because $\\abs{p}$ grows like $\\abs{z}^n$, and it is bounded on a large disc by compactness. Liouville's theorem then makes $1/p$ constant, which a nonconstant polynomial cannot be.\n",
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