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

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 $$\frac{p_t}{|p_t|}:S \to \mathbb{S}^1$$ 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 $$\frac{p(z)}{|p(z)|}$$ and $$\frac{z^m}{|z^m|}$$ are homotopic maps $$S \to \mathbb{S}^1$$.

Hence $$\frac{p(z)}{|p(z)|}$$ 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 self-map 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.