# The Fundamental Theorem of Algebra (with Liouville)

This proof assumes knowledge of complex analysis, specifically the notions of analytic functions and Liouville’s Theorem (which we will state below). The fundamental theorem of algebra has quite a few number of proofs (enough to fill a book!). In fact, it seems a new tool in mathematics can prove its worth by being able to prove the fundamental theorem in a different way. This series of proofs of the fundamental theorem also highlights how in mathematics there are many many ways to prove a single theorem, and in re-proving an established theorem we introduce new concepts and strategies. And perhaps most of all, we accentuate the unifying beauty of mathematics.

Problem: Let $p(x)$ be a non-constant polynomial with coefficients in $\mathbb{C}$. Prove $p(x)$ has a root in $\mathbb{C}$.

Solution: First, we note that Liouville’s theorem states that any bounded entire function (has infinitely many derivatives, is analytic/holomorphic) $\mathbb{C} \to \mathbb{C}$ is constant. In other words, if we can find a number $M$ such that $|f(z)| < M$ for all $z \in \mathbb{C}$, then $f(z) = c$ for some constant $c$. The proof of this follows from Cauchy’s integral formula, and we take it for granted here.

Suppose to the contrary that $p(x)$ is non-constant and has no roots. Then the function $f(z) = \frac{1}{p(z)}$ is defined everywhere on $\mathbb{C}$ and is analytic (the denominator is never zero by assumption). Now if $\displaystyle p(z) = a_nz^n + a_{n-1}z^{n-1} + \dots + a_0$

Then we have $\displaystyle |f(z)| = \left | \frac{1}{p(z)} \right | = \frac{1}{|z|^n} \frac{1}{\left | a_n + \frac{a_{n-1}}{z} + \dots + \frac{a_0}{z^n} \right |}$

Now each term in the denominator of the rightmost term, we have $\frac{a_i}{z^{n-i}} \to 0$ as $z \to \infty$. Therefore, $|f(z)| \to \frac{1}{|a_n||z|^n} \to 0$ as $z \to \infty$.

In particular, we may pick $R$ sufficiently large so that $|f(x)| < 1$ whenever $|z| > R$. On the other hand, since $f(z)$ is continuous on the closed, bounded disk centered at $0$ of radius $R$, $f(z)$ is also bounded whenever $|z| \leq R$. Together these two imply that $f(z)$ is bounded everywhere on $\mathbb{C}$, and so $f = 1/p = c$, so $p = 1/c$ is constant as well. This contradicts the assumption that $p$ is non-constant, and hence $p$ has a zero. $\square$