I've been working on an elementary proof of the fundamental theorem of algebra, which doesn't even require calculus. The most advanced concepts are that of minima/maxima. I don't see this proof anywhere in the literature, so I suspect there might be a flaw or a jump in logic which requires more complicated theorems in order to justify, but I don't necessarily see what it would be. Here is what I have, I'd appreciate any comments on whether it's effective and/or what it needs.

Let $\displaystyle f(z) = a_0 + a_1z + a_2z^2 . . . + a_nz^n$ be a non-constant polynomial. The first thing we need to establish is that |f| attains an absolute minimum somewhere on the complex plane (some functions, for example |e^z|, do not).

For any positive real A, the following fact holds: $\displaystyle |f(z)| > A$ $\displaystyle \forall$ $\displaystyle |z| > R$ for some positive R. So it is possible to find a finite region D of the complex plain in which |f| attains some values that will be be lesser than any value |f| attains outside of D. Being a finite region, D contains an absolute minimum for |f|. Therefore, |f| attains an absolute minimum.

Suppose that f has no roots. Then |f| attains a non-zero absolute minimum somewhere on the complex plane. If there are multiple points at which this minimum is attained, choose 1 such point. Call it (a,b) (meaning that (f(a) = b.)

For simplicity: let $\displaystyle g(z) = \frac{f(z+a)}{b}$. g(z) is a polynomial such that its absolute value attains an absolute minimum at (0,1). Therefore, $\displaystyle |g(z)| \geq 1$ for all z.

Let $\displaystyle c_1$ be the coefficient of z in the expansion of g. Suppose that $\displaystyle c_1$ is non-zero. Then, for sufficiently small positive $\displaystyle \epsilon$, it is clear that $\displaystyle |g(-\frac{\epsilon}{c_1})| < 1$. In particular, if $\displaystyle c_n$ is the coefficient of z^n in g, this inequality will hold if $\displaystyle \epsilon$ is less than http://latex.codecogs.com/gif.latex?...1%5En%7D%7C%7D.

But if |g| is less than 1, this contradicts the fact that $\displaystyle |g| \geq 1$ for all complex numbers. Therefore, the coefficient of z in the expansion of g must in fact be 0.

A similar argument can be used to show that if $\displaystyle c_2$ is nonzero, a number x can be constructed such that |g(x)| < 1. To generalize, let $\displaystyle c_n$ be the lowest-power nonzero coefficient in g, and let n be its power. For sufficiently small $\displaystyle \epsilon$, $\displaystyle |g((\frac{-\epsilon}{c_n})^{1/n})| < 1$. (Note that ^(1/n) has n values. Any one of them can be chosen with the same consequence.)

Therefore, we must conclude that all the coefficients of g are zero except the constant term. But if g is constant, then f is constant. So the only polynomial who's absolute value can attain [an absolute minimum which is greater than zero] is a constant polynomial.

Edit - I just realized that my notation for f and g inconsistently switch between x and z, because I didn't write it all at once in this forum. It is fixed now, sorry for the confusion.