# Fundamental theorem of algebra.
The fundamental theorem of algebra (FTA) states
> Every nonconstant polynomial $p(z) \in \mathbf{C}[z]$ has a root in $\mathbf{C}$.
There are many proofs of it, in topology, in complex analysis, and in algebra. But in fact there is a simple proof of it in basic analysis, relying on just the extreme value theorem.
First we establish some basics. Any polynomial $p(z)$ is continuous, and so is its absolute value $|p(z)|$.
The general strategy goes as follows.
(0) Take a nonconstant complex polynomial $p(z) \in \mathbf{C}[z]$, and suppose to the contrary that it has no complex root whatsoever.
(1) Establish that $|p(z)|$ has a global minimum at some $z_{0}$ using continuity arguments.
(2) But show that one can produce some small perturbation $\epsilon$ such that $|p(z_{0}+\epsilon)| < |p(z_{0})|$, giving a contradiction!
Recall extreme value theorem,
> If $f:K \to \mathbf{R}$ is a continuous real-valued function on a compact set $K$, then $f$ attains a maximum and minimum value on $K$.
So consider first a closed disk $D_{1}$ of say radius $1$, then $|p(z)|$ has some minimum value $u \ge 0$ on $D_{1}$. Then as $|p(z)| \to \infty$ as $|z| \to \infty$, there exists some radius $R$ such that $|z| > R > 1$ implies $|p(z)| > u$. Since on the closed disk $D_{R}$ of radius $R$, $|p(z)|$ attains some possibly new minimum value $w \le u$ at some $z_{0} \in D_{R}$, and that $|p(z)| > u \ge w$ outside the disk $D_{R}$, we see that $|p(z)|$ attains a global minimum at $z_{0}$ with some value $w \ge 0$.
If this minimum value $w = 0$, then we are done, $p(z_{0}) = 0$, where $z_{0}$ is a complex root of $p$.
So suppose to the contrary that $|p(z_{0})| = w > 0$.