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   Author  Topic: fund. theorem of algebra  (Read 4267 times)
Maria
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fund. theorem of algebra  
« on: Feb 10th, 2006, 3:01am »
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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!!  Shocked    
 
(This problem is from the complex analysis book by Greene & Krantz)
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Icarus
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Re: fund. theorem of algebra  
« Reply #1 on: Feb 10th, 2006, 9:12pm »
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Note that if the leading term of P(z) = azn, then the leading term of P'(z) = nazn-1. 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 limR-->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.
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Re: fund. theorem of algebra  
« Reply #2 on: Feb 10th, 2006, 10:11pm »
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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.
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Michael Dagg
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Re: fund. theorem of algebra  
« Reply #3 on: Feb 11th, 2006, 8:04am »
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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 11th, 2006, 8:05am by Michael Dagg » IP Logged

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Michael Dagg
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Re: fund. theorem of algebra  
« Reply #4 on: Feb 11th, 2006, 8:30am »
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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.
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