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 be a non-constant polynomial with coefficients in
. Prove
has a root in
.
Solution: First, we note that Liouville’s theorem states that any bounded entire function (has infinitely many derivatives, is analytic/holomorphic) is constant. In other words, if we can find a number
such that
for all
, then
for some constant
. The proof of this follows from Cauchy’s integral formula, and we take it for granted here.
Suppose to the contrary that is non-constant and has no roots. Then the function
is defined everywhere on
and is analytic (the denominator is never zero by assumption). Now if
Then we have
Now each term in the denominator of the rightmost term, we have as
. Therefore,
as
.
In particular, we may pick sufficiently large so that
whenever
. On the other hand, since
is continuous on the closed, bounded disk centered at
of radius
,
is also bounded whenever
. Together these two imply that
is bounded everywhere on
, and so
, so
is constant as well. This contradicts the assumption that
is non-constant, and hence
has a zero.