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

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Let $f \colon \mathbb{C}\rightarrow\mathbb{C}$ be a polynomial, and suppose $f$ has no root in $\mathbb{C}$.  We will show $f$ is constant.

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

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