{
  "aliases": [
    "dalembert-gauss-theorem",
    "dalemberts-theorem"
  ],
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  "body": "\n## Statement\n\nLet $n \\ge 1$ be an integer and let $a_0, a_1, \\dots, a_n$ be [[complex numbers|09FTAET5Z4]] with $a_n \\neq 0$.\nLet $p(z) = a_n z^n + a_{n-1} z^{n-1} + \\cdots + a_1 z + a_0$.\nThen there is a complex number $z_0$ with $p(z_0) = 0$.\n\n## Remarks\n\nProof with [[Liouville's theorem|0EHKVRF4J6]]: if $p$ has no zero, then $1/p$ is entire. Since $|p(z)| \\to \\infty$ as $|z| \\to \\infty$, $1/p$ is bounded, so it is constant, which is false for $n \\ge 1$.\nBy induction on $n$, it follows that $p(z) = a_n (z - z_1) \\cdots (z - z_n)$ for some $z_1, \\dots, z_n \\in \\mathbb{C}$. So $p$ has exactly $n$ roots, counted with multiplicity.\nThe theorem holds for polynomials with real coefficients too, but the roots can be non-real: $z^2 + 1$ has the roots $i$ and $-i$.\n\n## Explanations\n\n### L1 human\n\nEach polynomial equation of degree at least $1$ has a solution in the complex numbers.\nSo we never need to add new numbers to $\\mathbb{C}$ to solve polynomial equations, unlike the step from $\\mathbb{R}$ to $\\mathbb{C}$ for $x^2 = -1$.\n",
  "commit": "f591989ed15704e9d8755a7a5a6426d73de7a3ae",
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