Fundamental Theorem of Algebra/Proof 1

Proof
Let $p \left({z}\right)$ be a polynomial in $\C$:
 * $p \left({z}\right) = z^m + a_1 z^{m-1} + \cdots + a_m$

where not all of $a_1, \ldots, a_m$ are zero.

Define a homotopy:
 * $p_t \left({z}\right) = t p \left({z}\right) + \left({1-t}\right) z^m$

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

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

Therefore, there is an $r \in \R_{>0}$ such that:
 * $\forall z \in \C: \left|{z}\right| = r: \forall t \in \left[{0 \,.\,.\, 1}\right]: p_t \left({z}\right) \ne 0$

Hence the homotopy:
 * $\displaystyle \frac{p_t}{ \left|{p_t}\right|}:S \to \Bbb S^1$

is well-defined for all $t$.

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

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

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