Fundamental theorem of algebra

The fundamental theorem of algebra tells us that for an $n$-th degree polynomial function $P(z)$, we have $n$ roots (whether real or complex). Alternatively we can reformulate by saying

An extension of this is that for any polynomial $P(z)$, if $z=a+bi$ is a root, then its complex conjugate $\bar{z}=a-bi$ is also a root.

Proof

It suffices to prove that any non-constant polynomial has at least one complex root. For example, $z^2-3z+2=(z-2)(z-1)=0$. This is our original remark that we must prove.

We assume that this assumption is not true, that $P(z)$ is constant, i.e., $P(z)\ne 0$. Let:

$$f(z)=\frac{1}{P(z)}$$

then $f(z)$ is analytic everywhere. It’s easy to prove $f(z)$ is bounded:

$$\underset{P\to \infty }{\lim }=\frac{1}{P(z)}=0$$ $$|f(z_0)|\le 1;|z_1|\ge R_0$$

which acts like an upper bound.

Then by Liouville’s theorem, $f(z)$ must be constant, so $P(z)=\frac{1}{C}$ must be constant. However, in the beginning of the proof we assumed that $P(z)$ was not constant. By contradiction, the theorem is proved.¹

Footnotes

1. ”Like thunder and lightning, the theorem is proved.” - Prof. Nachman ↩