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# supplement - Some Applications of the Residue Theorem...

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Some Applications of the Residue Theorem * Supplementary Lecture Notes MATH 322, Complex Analysis Winter 2005 Pawe± l Hitczenko Department of Mathematics Drexel University Philadelphia, PA 19104, U.S.A. email: [email protected] * I would like to thank Frederick Akalin for pointing out a couple of typos. 1

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1 Introduction These notes supplement a freely downloadable book Complex Analysis by George Cain (henceforth referred to as Cain’s notes), that I served as a primary text for an undergraduate level course in complex analysis. Throughout these notes I will make occasional references to results stated in these notes. The aim of my notes is to provide a few examples of applications of the residue theorem. The main goal is to illustrate how this theorem can be used to evaluate various types of integrals of real valued functions of real variable. Following Sec. 10.1 of Cain’s notes, let us recall that if C is a simple, closed contour and f is analytic within the region bounded by C except for ﬁnitely many points z 0 ,z 1 ,...,z k then Z C f ( z ) dz = 2 πi k X j =0 Res z = z j f ( z ) , where Res z = a f ( z ) is the residue of f at a . 2 Evaluation of Real-Valued Integrals. 2.1 Deﬁnite integrals involving trigonometric functions We begin by brieﬂy discussing integrals of the form Z 2 π 0 F (sin at, cos bt ) dt. (1) Our method is easily adaptable for integrals over a diﬀerent range, for example between 0 and π or between ± π . Given the form of an integrand in (1) one can reasonably hope that the integral results from the usual parameterization of the unit circle z = e it , 0 t 2 π . So, let’s try z = e it . Then (see Sec. 3.3 of Cain’s notes), cos bt = e ibt + e - ibt 2 = z b + 1 /z b 2 , sin at = e iat - e - iat 2 i = z a - 1 /z a 2 i . Moreover, dz = ie it dt , so that dt = dz iz . Putting all of this into (1) yields Z 2 π 0 F (sin at, cos bt ) dt = Z C F ± z a - 1 /z a 2 i , z b + 1 /z b 2 ² dz iz , where C is the unit circle. This integral is well within what contour integrals are about and we might be able to evaluate it with the aid of the residue theorem. 2
It is a good moment to look at an example. We will show that Z 2 π 0 cos 3 t 5 - 4 cos t dt = π 12 . (2) Following our program, upon making all these substitutions, the integral in (1) becomes Z C ( z 3 + 1 /z 3 ) / 2 5 - 4( z + 1 /z ) / 2 dz iz = 1 i Z C z 6 + 1 z 3 (10 z - 4 z 2 - 4) dz = - 1 2 i Z C z 6 + 1 z 3 (2 z 2 - 5 z + 2) dz = - 1 2 i Z C z 6 + 1 z 3 (2 z - 1)( z - 2) dz. The integrand has singularities at z 0 = 0, z 1 = 1 / 2, and z 2 = 2, but since the last one is outside the unit circle we only need to worry about the ﬁrst two. Furthermore, it is clear that z 0 = 0 is a pole of order 3 and that z 1 = 1 / 2 is a simple pole. One way of seeing it, is to notice that within a small circle around z 0 = 0 (say with radius 1 / 4) the function z 6 + 1 (2 z - 1)( z - 2) is analytic and so its Laurent series will have all coeﬃcients corresponding to the negative powers of z zero. Moreover, since its value at

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## This note was uploaded on 07/18/2011 for the course AEB 6933 taught by Professor Carriker during the Fall '09 term at University of Florida.

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supplement - Some Applications of the Residue Theorem...

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