# proof of the fundamental theorem of algebra (Liouville’s theorem)

Let $f:\u2102\to \u2102$ be a polynomial^{}, and suppose $f$ has no root in $\u2102$. We will show $f$ is constant.

Let $g=\frac{1}{f}$. Since $f$ is never zero, $g$ is defined and holomorphic on $\u2102$ (ie. it is entire). Moreover, since $f$ is a polynomial, $|f(z)|\to \mathrm{\infty}$ as $|z|\to \mathrm{\infty}$, and so $|g(z)|\to 0$ as $|z|\to \mathrm{\infty}$. Then there is some $M>0$ such that $$ whenever $|z|>M$, and $g$ is continuous^{} and so bounded^{} on the compact set $\{z\in \u2102:|z|\le M\}$.

So $g$ is bounded and entire, and therefore by Liouville’s theorem $g$ is constant. So $f$ is constant as required.$\mathrm{\square}$

Title | proof of the fundamental theorem of algebra (Liouville’s theorem) |
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Canonical name | ProofOfTheFundamentalTheoremOfAlgebraLiouvillesTheorem |

Date of creation | 2013-03-22 12:18:59 |

Last modified on | 2013-03-22 12:18:59 |

Owner | Evandar (27) |

Last modified by | Evandar (27) |

Numerical id | 6 |

Author | Evandar (27) |

Entry type | Proof |

Classification | msc 12D99 |

Classification | msc 30A99 |