# Fundamental theorem of algebra

I really think this is a misnomer. The theorem has very little to do with algebra. The theorem, is an assertion about the complex numbers which are very much an analytic object. Anyway, the proof is pretty short, so here goes.

Theorem 1 Every polynomial with coefficients in ${{\mathbb C}}$ has at least one root in ${{\mathbb C}}$.

Proof: Let ${P:{\mathbb C}\rightarrow{\mathbb C}}$ be a non-constant polynomial. Assume for the sake of contradiction that ${P}$ has no roots in ${{\mathbb C}}$. Then indeed, there is some small open set in ${{\mathbb C}}$ that must not be in the image of ${P}$. Why? If ${P(z)=a_n\cdot z^n+\cdots a_1\cdot z+a_0}$ assuming ${a_n\ne0}$, then

$\displaystyle \lim_{z\rightarrow\infty} \frac{P(z)}{z^n}=a_n,$

so for ${z}$ far enough away from zero, ${P(z)/z^n}$ must be close to ${a_n}$. In particular, for large enough ${|z|}$ we can make ${|P(z)|/|z|^n}$ at least ${|a_n|/2}$. Hence, ${|P(z)|}$ must be, (for large ${|z|}$), at least ${|a_n|\cdot|z|^n}$, and hence, far away from zero. Let ${R}$ be any such value which is large enough to guarantee that for ${|z|>R}$, we have the result just described.

But what about the small complex numbers? Couldn’t those get close to zero? Alas, no. the set ${\{z\mid |z|\le R\}}$ is compact, and so if ${P(z)}$ gets arbitrarily close to zero on this set, it in fact hits zero. By assumption, ${P(z)}$ is never zero, and so it must be bounded away from zero. So let’s say that if ${|w|<\varepsilon}$, then ${w}$ is not in the image of ${P}$. In other words, for each ${z\in{\mathbb C}}$, ${|P(z)|\ge\varepsilon}$.

Now if ${P}$ is never zero, then ${f=1/P}$ is defined on all of ${{\mathbb C}}$. But remember that ${|P(z)|\ge\varepsilon}$. This means that ${|f(z)|\le 1/\varepsilon}$. But now ${f}$ is bounded and entire (holomorphic on all of ${{\mathbb C}}$), so ${f}$ must be a constant, implying that ${P}$ must be a constant polynomial. This contradicts our assumption that ${P}$ was non-constant, so it must be that every polynomial has a root. $\Box$

This is just one of many proofs. You can read a bunch of them on Adam Azzam’s blog (which you should already be reading).