Fundamental Theorem of Algebra/Proof 1

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

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

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

Then:
 * $\dfrac {\map {p_t} z} {z^m} = 1 + t \paren {a_1 \dfrac 1 z + \cdots + a_m \dfrac 1 {z^m} }$

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

Therefore, there is an $r \in \R_{>0}$ such that:
 * $\forall z \in \C: \size z = r: \forall t \in \closedint 0 1: \map {p_t} z \ne 0$

Hence the homotopy:
 * $\dfrac {p_t} {\size {p_t} }: S \to \Bbb S^1$

is well-defined for all $t$.

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

Hence $\dfrac {\map p z} {\size {\map p z} }$ must have the same degree of $\paren {z / r}^m$, which is $m$.

When $m > 0$, that is $p$ is non-constant, this result and the Extendability Theorem for Intersection Numbers imply $\dfrac {\map p z} {\size {\map p z} }$ does not extend to the disk $\Int S$, implying $\map p z = 0$ for some $z \in \Int S$.