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Fundamental Theorem of Algebra
By Brad Rodgers (P1930) on Wednesday, November 22, 2000 - 12:05 am:

How do I go about proving the fundamental theorem of algebra. I would like to do this myself, but I really don't know haw to get started. Any suggestions would be very helpful.

Thanks,

Brad

By Dan Goodman (Dfmg2) on Wednesday, November 22, 2000 - 01:21 am:

Here's the geometric method (I'll only outline the proof, because the full machinery needed would take too long to go into).

You define the "winding number about 0" n(f) of a path f in the complex plane given by f:[0,1]->C where f(0)=f(1) and f is continuous. The way we define the winding number is intuitively, "the number of times the curve f winds around 0". So, for instance, n(e2pit)=1, because this is a circle going around 0 once. n(e4pit)=2 because this is a circle going around 0 twice. Anyway, you can define n(f) for any path f, although it is more tricky to do so for more complicated paths, the important fact about n(f) is that it is integer valued.

Now, let p(z) be a non constant polynomial in z, and assume p(z) is never 0 (setting up for proof by contradiction). Let f(t)=p(Re2pit), i.e. the path formed by taking the image of the circle of radius R around 0. If R=0, this path is f(t)=p(0)=x say, with x not zero.

Now, whatever R we take, f(t) is never zero as p(z) is never 0. This means if we look at the family of paths we get by varying R, they all lie in C\{0}, i.e. the complex plane without 0.

Unfortunately at this point I have to bring in another intuitive idea as the machinery required would be too involved to go into here. As we continously vary paths in C\{0}, the winding number can't change at any point, we can see this geometrically because to increase the number of times it loops round 0, you'd have to pass through 0, which you can't do because 0 is not in the set. Maybe that's not quite as intuitively satisfying as it should be, but you can take it as true that in C\{0}, what is called "homotopic transformations" (which basically means continuously changing one path to another) leave the winding number unchanged.

Given that we're done, because the winding number of f if R=0 is just the winding number of the constant path x, which is 0. Therefore, for any R the winding number of f must be 0, because we're continously transforming the paths in C\{0}. However, for very large R, p(z) is approximately kzn for some complex number k not equal to zero and some integer n>=1 [because zn is much bigger than zn-1 if |z| is very large]. So, f(z) is approximately ke2npit, which has winding number n, which is >=1.

This is a contradiction, so the assumption that p(z) is never 0 must be false. Tada! The fundamental theorem of algebra.

I think this is how Gauss proved it originally (although he lacked the machinery of winding numbers and so forth which is why he recognized his proof was lacking, and proved it again, many times). I don't know any elementary proof of FTA, but there might well be some, let's see what the others have...

By Dan Goodman (Dfmg2) on Wednesday, November 22, 2000 - 01:22 am:

Oops! You said you wanted to do it yourself and I gave you the proof. Sorry! Hopefully I was sufficiently unclear that there will be plenty for you to do :)

By Brad Rodgers (P1930) on Wednesday, November 22, 2000 - 09:44 pm:

How does that relate to proving that there can only be complex solutions to polynomial equations?

Thanks,

Brad

By Dan Goodman (Dfmg2) on Wednesday, November 22, 2000 - 09:50 pm:

It proves that every polynomial f(z) has a complex root f(z)=0, which is what the FTA says! I'm not sure I understand your question.

By Brad Rodgers (P1930) on Wednesday, November 22, 2000 - 10:49 pm:

I have a book that says that the FTA proves that given any n degree egn.

eg: axn+bxn-1+...+c=0

there can only be a solution consisting of imaginary or real numbers.

eg:x=a+bi

where b and a are real

Or the book at least says that this theorem stopped mathematicians from looking for new types of numbers.

Brad

By Dan Goodman (Dfmg2) on Wednesday, November 22, 2000 - 11:20 pm:

OK, got you. What it means is that every equation of that form has a solution a+ib (i.e. a complex number), and therefore you don't need to introduce any new sorts of numbers to solve equations. The idea in the book is probably this: the rational numbers are OK, but you can't solve X2-2=0 in rationals, so you can introduce irrationals like sqrt(2). However, there are still equations you can't solve, like X2+1=0, so you introduce the complex numbers a+ib. Once you've done this, the process has finished because all (polynomial) equations can now be solved with numbers of the form a+ib. There are other types of numbers (quaternions for instance), but these aren't needed to solve polynomials.

By Brad Rodgers (P1930) on Thursday, November 23, 2000 - 04:13 am:

But how is the above proof related to that?

By Kerwin Hui (Kwkh2) on Thursday, November 23, 2000 - 12:59 pm:

If you have a zero for f(z), say z=z0 then (z-z0) is a factor of f(z). So we can factorise:
f(z) = (z-z0)g(z), where deg(g)=deg(f)-1
So by induction, we have the polynomial f(z) has n (not necessarily distinct) complex root.(as polynomial of degree zero has zero root apart from the trivial polynomial 0)

Kerwin