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

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

Let $g=\frac{1}{f}$. Since $f$ is never zero, $g$ is defined and holomorphic on $ℂ$ (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 $\{z\in ℂ:|z|\le M\}$.

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)
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