The Fundamental Theorem of Algebra (with Liouville)

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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

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