Complex Integrals and the Residue TheoremDate: 12/18/2000 at 03:02:18 From: Lydia Monroe Subject: Complex Analysis: Integrals To Dr. Math: I would be grateful if you could help me solve this integral, which involves complex numbers: the integral over C of (z^2/((z-1)^2*(z+1)))dz where C is the circle C = {z| |z - 2i| = 2}. I thought maybe I could use the Taylor Series to solve this, but I'm not sure. If I wanted to integrate around the problem points using the Cauchy Integral Problem, how would I set the problem up? I'm confused by the (z-1)^2 part. If it weren't squared, I think I could figure it out, but do I integrate over three curves, one for each z-1 and one for z+1? I'm confused... It's a closed curve (I think) with counter-clockwise orientation. Thank you for any help you can give. Date: 12/19/2000 at 13:12:04 From: Doctor Mitteldorf Subject: Re: Complex Analysis: Integrals Dear Lydia, The rule is to evaluate the function at each of the poles contained within the contour of integration, then multiply by 2pi*i. The situation at z = -1 is straightforward. The function behaves like: (-1)^2 -------------- (-2)^2 * (z+1) so the residue is (-1)^2/(-2)^2 = 1/4. The situation at z = 1 is a little more complicated. (z-1)^(-2) by itself would not produce any residue at all; however, there's a part of the behavior at z = 1 that actually behaves like (z-1)^(-1). To find this, we must expand the rest of the function in a Taylor series about z = 1, and find the part the term that is linear in (z-1). This, when multiplied by (z-1)^(-2), gives a term that behaves like (z-1)^(-1). So we need to evaluabe the first derivative of F(z)= z^2/(z+1) at the point z = 1. I calculate that: z^2-2z F'(z) = ------- (z+1)^2 so that F'(1)= -1/4. Hence the residue is +1/4, exactly canceling the residue from the simple pole at z = -1. My calculation is that the integral is zero. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ Date: 12/20/2000 at 17:03:32 From: Lydia Monroe Subject: Re: Complex Analysis: Integrals I don't know about poles and residues (though I'm going to look them up now) and though I got your answer, it's a little confusing for me. I tried to work it out with 'brute force' and I got a different answer. I guess I must have made an error. (I got (-pi*i - pi) as the answer.) I used the Cauchy Integral theorem and integrated over the problem point i (-1 is not inside or on C). I got that the integral of (z^2/(z+1))/(z-1)^2, which is the same as the original integral, is equal to 2pi*i (-1/(i+1)), which comes to 2pi*i (-1/2+(1/2)i) or ((-pi*i) - pi). Where did I go wrong? I also did this same integral over c: {z| |z+1| = (1/2)} and got pi, and also over c: {z| |z| = 3}, which includes both problem points, so I just added the results from my two earlier integrals and got (- pi*i). Thank you for your help. Date: 12/20/2000 at 17:32:05 From: Doctor Mitteldorf Subject: Re: Complex Analysis: Integrals Lydia - The residue theorem is really the heart of what makes these complex integrals fun and interesting. It also simplifies computation enormously, and makes many integrals computable that would otherwise be opaque. (There are even some real integrals that can most easily be calculated by thinking in terms of integrals over a contour including half a plane.) Thanks for affording me the opportunity to review the subject and remember things I haven't thought about in many years. I'd be interested to see how you set up the "brute force" version of these integrals. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ Date: 01/11/2001 at 02:06:27 From: Lydia Monroe Subject: Complex Analysis: Integrals To Dr. Mitteldorf - I recently wrote you about another complex analysis problem that I was struggling with and you suggested that I use poles and residues to solve the problem. Consequently I have been studying that area of complex analysis on my own and wondered if you could tell me whether the answer I have calculated for one particular problem is correct. The problem goes: Find the residue of (sin z)/z^4 at z = 0. Now (sin z)/z^4 is the same as (1/z^4)*(sin z), which can be written as the multiple of a series: (1/z^4)(z-((z^3)/3!)+...) and this is the same as 1/z^3 + higher terms. So I thought perhaps then the residue would be 1/z^2, but I'm very uncertain about residues and such. I took a class in complex analysis last semester but it suddenly sped up at the end of the semester and I only remember my professor mentioning residues once. There is no second semester course on complex analysis at my school. Is this unusual? (It is a top-ranked school.) Thank you for your time. This forum is of tremendous help. Date: 01/11/2001 at 06:32:54 From: Doctor Mitteldorf Subject: Re: Complex Analysis: Integrals Dear Lydia, The logic of residue calculus is based on the following chain of reasoning: 1) If a function is analytic everywhere, then the integral over any path is 0. 2) Therefore you can distort any path integral, as long as you're moving it only within a region where the function is analytic. 3) Suppose you have an integral of a function with no singularities except isolated poles: you can use the above principle to change your path integral into a sum of infinitesimal circular loops around the poles. 4) You can evaluate these infinitesimal loop integrals around poles by analyzing the function at the pole in terms of its Taylor series. Use the fact that the loop integral of z^n is identically zero for every value of n, positive or negative, except n = -1. 5) For n = 1 the value of the integral is 2pi*i. It follows that you can evaluate the contribution from any pole by representing the function at the pole as a Taylor series, ignoring all Taylor terms except n = -1, and then multiplying the n = -1 coefficient by 2pi*i. These ideas are strange and defy our intuition. Each step of the above requires some analysis in order to convince yourself it's true; in addition, it helps to go through some examples so that you're comfortable with the ideas. I encourage you to go through this process on your own if there isn't a teacher available. Perhaps you can find a text that will help. The text I was taught from years ago was Ahlfors: _Complex Analysis_. In the case of your function (1/z^4)*(sin z), you've done the analysis correctly: write it as (1/z^4)(z-((z^3)/3!)+...). Then the only term that matters is the second term, which multiplies out to be (1/6)z^-1. The residue is 1/6, so the integral is 1/3 pi*i. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/