Fundamental Theorem of Algebra

= Theorem =

Every non-constant polynomial with coefficients in $$\mathbb{C}$$ has a root in $$\mathbb{C}$$.

= Proof =

Suppose $$p(z) = z^m + a_1 z^{m-1} + ... + a_m$$. Define a homotopy $$p_t(z)=tp(z)+(1-t)z^m$$. Then $$\tfrac{p_t(z)}{z^m} = 1 + t(a_1 \tfrac{1}{z} + ... +a_m \tfrac{1}{z^m})$$. The terms in the parenthesis go to 0 as $$z \to \infty$$. Therefore, there is an $$r \in \mathbb{R}_+$$ such that $$\forall z \in \mathbb{C}$$ such that $$|z|=r$$, $$\forall t \in [0,1], p_t(z) \neq 0$$. Hence the homotopy $$\tfrac{p_t}{|p_t|}:S \to \mathbb{S}^1$$ 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 $$\mathbb{C}$$ such that both $$\tfrac{p(z)}{|p(z)|}$$ and $$\tfrac{z^m}{|z^m|}$$ are homotopic maps $$S \to \mathbb{S}^1$$.

Hence $$\tfrac{p(z)}{|p(z)|}$$ 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 $$z \in \text{int}(S)$$.