Inspired by Felix’ post I want to show you how to proof the fundamental theorem of algebra using only real analysis. Some steps are quite similar to Felix’ proof, but alas, we must not use the power of holomorphic functions. I’ve seen this proof more than four years ago in my undergraduate Analysis I course at ETH, taught by Dietmar Salamon.
Theorem: Every non-constant polynomial has a complex zero.
Proof: Without loss of generality, let
be a normalized complex polynomial and define
and
and set

such that
holds. Then we get for 

and hence

This shows that, in general, zeroes of polynomials cannot be too large in terms of their coefficients. But what happens inside the circle? The set
is bounded and closed, hence compact by Heine-Borel. Polynomials are continuous, hence
has a minimum at some point
. This minimum is even a global minimum, since
for all
.
If
there is nothing to do. Hence assume that the global minimum is not zero. Then we can define the polynomial

which has the property that it has a global minimum of
at
. Hence there exist a non-zero complex number
and a positive integer
such that

Let
be any complex
-th root of
. Then
Hence there exists a polynomal
such that

Since polynomials are continuous,
has to be bounded on
, say by
. Using this and restricting
to the unit interval we get

It is now not hard to see that one can choose
small enough to make
positive, in turn getting
, in contradiction to the global mininmum of value
, QED.