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ch16

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269 CHAPTER 16 Laurent Series. Residue Integration This is another powerful and elegant integration method that has no analog in calculus. It uses Laurent series (roughly, series of positive and negative powers of z ), more precisely, it uses just a single term of such a series (the term in 1/( z z 0 ), whose coefficient is called the residue of the sum of the series that converges near z 0 ). SECTION 16.1. Laurent Series, page 701 Purpose. To define Laurent series, to investigate their convergence in an annulus (a ring, in contrast to Taylor series, which converge in a disk), to discuss examples. Major Content, Important Concepts Laurent series Convergence (Theorem 1) Principal part of a Laurent series Techniques of development (Examples 1–5) SOLUTIONS TO PROBLEM SET 16.1, page 707 2. z • • • , 0 z R 4. ( 1 • • • ) 2 z 2 z 4 • • • , 0 z R 6. We obtain e z m 0 n 0 z n n 0 ( n m 0 ) z n z z 2 • • • , 0 z R 1. 8. Successive differentiation and use of cos 1 _ 4 sin 1 _ 4 1/ 2 gives Y • • • Z , R . This can also be obtained from sin z sin [( z ) ] [ sin ( z ) cos ( z )] and substitution of the usual series on the right. 4 4 1 2 4 4 z 4 4! 1 3! 1 2! ( z 4 ) 1 ( z 4 ) 2 1 ( z 4 ) 3 1 2 65 24 8 3 5 2 2 z 1 z 2 1 m ! 1 z 2 z m m ! 1 z 2 1 1 z 1 z 2 4 45 2 3 1 z 2 (2 z ) 6 6! (2 z ) 4 4! (2 z ) 2 2! 1 z 2 1 24 z 3 1 2 z im16.qxd 9/21/05 1:05 PM Page 269

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270 Instructor’s Manual 10. n 0 ( z ) 2 n 4 ( z ) 2 • • • , 0 z R 12. 3 i ( z i ) 14. n 0 z • • • , 0 z R 16. We obtain ` n 0 ( z 1) n 1 , 0 z 1 2, ` n 0 , z 1 2 18. n 0 ( 1) n ( z 1) n , 0 z 1 1, n 0 , z 1 1 20. n 0 ( z 1) n 4 , a n sinh 1 ( n even), a n cosh 1 ( n odd), z 1 0 22. ` n 0 ( ) , z i 1, n 0 ( ) , z i 1 24. Team Project. (a) Let a n ( z z 0 ) n and c n ( z z 0 ) n be two Laurent series of the same function ƒ( z ) in the same annulus. We multiply both series by ( z z 0 ) k 1 and integrate along a circle with center at z 0 in the interior of the annulus. Since the series converge uniformly, we may integrate term by term. This yields 2 ia k 2 ic k . Thus, a k c k for all k 0, 1, • • • . (b) No, because tan (1/ z ) is singular at 1/ z /2, 3 /2, • • • , hence at z 2/ , 2/3 , • • • , which accumulate at 0. (c) These series are obtained by termwise integration of the integrand. The second function is Si( z )/ z 3 , where Si( z ) is the sine integral [see (40) in App. A.31]. Answer: • • • , • • • . z 2 5!5 1 3!3 1 z 2 z 2 4!4 z 3!3 1 2!2 1 z i n ( z i ) n 2 2 n 1 ( z i ) 2 ( 1 z i i ) 2 1 z 2 ( z i ) n i n 2 2 n 1 [ i ( z i )] 2 1 z 2 a n n ! ( 1) n ( z 1) n 1 ( 1) n 1 2 n ( z 1) n 2 1 ( z 1) 2 ( 1 z 2 1 ) 1 ( z 1) 2 2( z 1) 1 1 ( z 1 1) 2 1 1 z 2 ( 1) n 1 2 n 1 1 120 z 3 1 6 z 1 (2 n 1)! z 2 n 1 3 z i i ( z i ) 2 1 6! 1 4! 1 2!( z ) 2 1 ( z ) 4 ( 1) n 1 (2 n )! im16.qxd 9/21/05 1:05 PM Page 270
Instructor’s Manual 271 SECTION 16.2. Singularities and Zeros. Infinity, page 707 Purpose. Singularities just appeared in connection with the convergence of Taylor and Laurent series in the last sections, and since we now have the instrument for their classification and discussion (i.e., Laurent series), this seems the right time for doing so.

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ch16 - im16.qxd 1:05 PM Page 269 CHAPTER 16 Laurent Series...

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