Fundamental Theorem of Algebra/Proof 2

Theorem
Every non-constant polynomial with coefficients in $\C$ has a root in $\C$.

Proof
Let $p: \C \to \C$ be a complex polynomial with $p \left({z}\right) \ne 0$ for all $z \in \C$.

Then $p$ extends to a continuous transformation of the Riemann sphere:
 * $\hat \C = \C \cup \left\{{\infty}\right\}$

This extension also has no zeroes.

By Riemann Sphere is Compact, there is some $\epsilon \in \R_{>0}$ such that:
 * $\forall z \in \C: \left|{p \left({z}\right)}\right| \ge \epsilon$

Now consider the holomorphic function $g: \C \to \C$ defined by:
 * $g \left({z}\right) := \dfrac 1 {p \left({z}\right)}$

We have:
 * $\forall z \in \C: \left|{g \left({z}\right)}\right| \le \dfrac 1 \epsilon$

By Liouville's Theorem, $g$ is constant.

Hence $p$ is also constant, as claimed.