There is a proof in an ALGEBRA BOOK published by the Russian Mathematician A. Kurosh. There is a translated version in English try to find it...
Higher Algebra ? Kurosh | Mir Books
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 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: 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 . g(z) is a polynomial such that its absolute value attains an absolute minimum at (0,1). Therefore, for all z.
Let be the coefficient of z in the expansion of g. Suppose that is non-zero. Then, for sufficiently small positive , it is clear that . In particular, if is the coefficient of z^n in g, this inequality will hold if is less than .
But if |g| is less than 1, this contradicts the fact that 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 is nonzero, a number x can be constructed such that |g(x)| < 1. To generalize, let be the lowest-power nonzero coefficient in g, and let n be its power. For sufficiently small , . (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.
There is a proof in an ALGEBRA BOOK published by the Russian Mathematician A. Kurosh. There is a translated version in English try to find it...
Higher Algebra ? Kurosh | Mir Books