Author 
Topic: fund. theorem of algebra (Read 4267 times) 

Maria
Newbie
Posts: 2


fund. theorem of algebra
« on: Feb 10^{th}, 2006, 3:01am » 
Quote Modify

Here is the problem: "Give another proof of the fundamental theorem of algebra as follows: let P(z) be a nonconstant polynomial. Fix a point Q in C. Consider the integral of 1/(2*Pi*i)* P'(z)/P(Z) over the contour of D(Q,R). Argue that, as R > infinity, this expression tends to a nonzero constant." I think this is impossible without using the fund. theorem of algebra!! (This problem is from the complex analysis book by Greene & Krantz)


IP Logged 



Icarus
wu::riddles Moderator Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863


Re: fund. theorem of algebra
« Reply #1 on: Feb 10^{th}, 2006, 9:12pm » 
Quote Modify

Note that if the leading term of P(z) = az^{n}, then the leading term of P'(z) = naz^{n1}. Hence, lim_{z>oo} zP'(z)/P(z) = n. In particular, if h > 0 is arbitrary, we can find an N such that if z > N, then  zP'(z)/P(z)  n  < h. If P(z) has no zeros, then P'(z)/P(z) is analytic everywhere, so we can switch the contour from D(Q, R) to D(0, R) without changing the value. Choose R > N, and note that _{}n/z dz = 2ni. Hence,  _{} P1(z)/P(z) dz  _{ }n/z dz  <= _{ } P'(z)/P(z)  n/z  dz <= _{ } h/z  dz <= h _{ }dz/z = 2h Since h is arbitrary, it must be lim_{R>oo} _{ }P'(z)/P(z) dz = _{ }n/z dz = 2ni. Note that by Cauchy's theorem, if P(z) has no zeros, so P'(z)/P(z) is analytic everywhere, then _{ }P'(z)/P(z) dz = 0.


IP Logged 
"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? "  Anonymous



Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948


Re: fund. theorem of algebra
« Reply #2 on: Feb 10^{th}, 2006, 10:11pm » 
Quote Modify

Note that a proof by contradiction is not necessary here. For, in any case, P has at most finitely many zeros, so for R sufficiently large the center of the disk doesn't matter. Then, by the argument principle, the integral is the number of zeros of P(z) within this disk. Thus it's eventually constant, and we find that that constant is n = deg(P). So, the fundamental theorem of algebra can be made to follow from Liouville's theorem, Rouche's theorem, the argument principle, the open mapping theorem / maximum modulus principle, Jensen's formula, Hadamard's factorization theorem, . . . but I'm still not convinced.


IP Logged 



Michael Dagg
Senior Riddler
Gender:
Posts: 500


Re: fund. theorem of algebra
« Reply #3 on: Feb 11^{th}, 2006, 8:04am » 
Quote Modify

I will add, in order to argue this, you need to prove some facts about winding numbers, and in particular that the above expression is the winding number of the image of the contour D(Q,R) under P about the origin. (This is the principle of the argument, and doesn't need the fundamental theorem of algebra in its proof.) Then a separate argument can be used to show that when R is sufficiently large this winding number is equal to the degree of P (because the leading term dominates, and causes the whole thing to go round n times  this isn't too hard to prove). When R=0, the expression is 0, and it always takes integer values, so there must be discontinuities. But these can arise only if P has zeros.

« Last Edit: Feb 11^{th}, 2006, 8:05am by Michael Dagg » 
IP Logged 
Regards, Michael Dagg



Icarus
wu::riddles Moderator Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863


Re: fund. theorem of algebra
« Reply #4 on: Feb 11^{th}, 2006, 8:30am » 
Quote Modify

I was aiming a little lower with my post: Nothing past a limited formulation of Cauchy's theorem. In fact, for these purposes, the easily proven version wherein the partial derivatives are assumed continuous is sufficient.


IP Logged 
"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? "  Anonymous



