Fundamental Theorem of Algebra

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

Proof using Algebraic Topology
Suppose $p(z) = z^m + a_1 z^{m-1} + \ldots + a_m$.

Define a homotopy $p_t(z)=tp(z)+(1-t)z^m$.

Then $\displaystyle \frac{p_t(z)}{z^m} = 1 + t \left({a_1 \frac{1}{z} + \ldots +a_m \frac{1}{z^m}}\right)$.

The terms in the parenthesis go to $0$ as $z \to \infty$.

Therefore, there is an $r \in \R_+$ such that $\forall z \in \C$ such that $|z|=r$, $\forall t \in [0,1], p_t(z) \ne 0$.

Hence the homotopy $\displaystyle \frac{p_t}{|p_t|}:S \to \Bbb S^1$ is defined for all $t$.

This shows that for any complex polynomial $p(z)$ of order $m$, there is a circle $S$ of sufficiently large radius in $\C$ such that both $\displaystyle \frac{p(z)}{|p(z)|}$ and $\displaystyle \frac{z^m}{|z^m|}$ are homotopic maps $S \to \Bbb S^1$.

Hence $\displaystyle \frac{p(z)}{|p(z)|}$ must have the same degree of $(z/r)^m$, which is $m$.

When $m>0$, ie, $p$ is non-constant, this result and the Extendability Theorem for Intersection Numbers imply $p/|p|$ does not extend to the disk $\text{int}(S)$, implying $p(z)=0$ for some $z \in \text{int}(S)$.

Proof using Liouville's theorem from complex analysis
Let $p:\C\to\C$ be a complex polynomial with $p(z) \ne 0$ for all $z \in \C$.

Then $p$ extends to a continuous transformation of the Riemann sphere $\hat{\C} = \C \cup \{\infty\}$ (and this extension also has no zeros).

Since the Riemann sphere is compact, there is some $\varepsilon>0$ such that $|p(z)|\ge \varepsilon$ for all $z\in\C$.

Now consider the holomorphic function $g: \C \to \C$ defined by $g(z) := 1/p(z)$.

We have $|g(z)|\le 1/\varepsilon$ for all $z\in\C$.

By Liouville's Theorem, $g$ is constant. Hence $p$ is also constant, as claimed.