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        general >> complex analysis >> fund. theorem of algebra
        
(Message started by: Maria on Feb 10^th, 2006, 3:01am)

Title: fund. theorem of algebra
Post by Maria on Feb 10^th, 2006, 3:01am

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!! :o

(This problem is from the complex analysis book by Greene & Krantz)

Title: Re: fund. theorem of algebra
Post by Icarus on Feb 10^th, 2006, 9:12pm

Note that if the leading term of P(z) = azⁿ, then the leading term of P'(z) = naz^n-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 _{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/oint.gif}n/z dz = 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifni.

Hence,
| _{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/oint.gif} P1(z)/P(z) dz - _{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/oint.gif}n/z dz | <= _{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/oint.gif}| P'(z)/P(z) - n/z | |dz| <= _{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/oint.gif}| h/z | |dz| <= h _{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/oint.gif}|dz/z| = 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifh

Since h is arbitrary, it must be lim_R-->oo _{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/oint.gif}P'(z)/P(z) dz = _{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/oint.gif}n/z dz = 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifni.

Note that by Cauchy's theorem, if P(z) has no zeros, so P'(z)/P(z) is analytic everywhere, then _{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/oint.gif}P'(z)/P(z) dz = 0.

Title: Re: fund. theorem of algebra
Post by Eigenray on Feb 10^th, 2006, 10:11pm

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.

Title: Re: fund. theorem of algebra
Post by Michael_Dagg on Feb 11^th, 2006, 8:04am

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.

Title: Re: fund. theorem of algebra
Post by Icarus on Feb 11^th, 2006, 8:30am

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.