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Unformatted text preview: 13 Definite integrals Read: Boas Ch. 14. 13.1 Laurent series: Def.: Laurent series (LS). If f ( z ) is analytic in a region R , then f ( z ) = ∞ X n =0 a n ( z z ) n + ∞ X n =1 b n ( z z ) n (1) converges in R , with a n = 1 2 πi I C f ( z ) dz ( z z ) n +1 ; b n = 1 2 πi I C f ( z ) dz ( z z ) n +1 (2) If you recall, for an ordinary power series we need x < const. ≡  a n +1 a n  , by the ratio test, for convergence. Analog for complex numbers is  z  < const 2 . This applies then to the first part of the series. But a function may have some singularities outside the region R , so we need to account for them as well; hence the second part of the series. Here there may be a region of space where 1 /  z  < const 1 is required for convergence. The figure shows that these regions can overlap, giving the annulus R where the function is analytic and both parts of the series converge. This depends of course on what the a n and b n are. How do you find them? Well, Figure 1: A Laurent series might converge only in an annulus R around a given point z , here between circles C 1 and C 2 in complex plane, because the “aseries” converges inside C 2 and the “bseries” converges outside C 1 . you can use Eq. (2), but this might involve doing rather a lot of integrals :). In practice, we can often use “partial fraction” tricks, like the following. 1 Ex.: 1 z ( z 1) = 1 z 1 1 z = 1 z (1 z ) 1 = 1 z 1 z z 3 . . . , (3) so you can just read off the coefficients. Note only one of the b n is nonzero for this case. If f ( z ) is regular in the annulus, no matter how small we make C 1 , and yet f ( z ) is not regular everywhere inside C 2 , we have a special situation of an isolated singularity . There are three subpossibilities: 1. LS may have all b n = 0 f ( z ) = ∞ X n =0 ( z z ) n , z 6 = z (4) This is called a removable singularity. You may think in fact it’s not a singu larity at all, but take the example f ( z ) = ‰ sin z z z 6 = 0 2 z = 0 (5) The function is certainly not analytic at z = 0. However we can “remove” the singularity by redefining the function to have the value 1 at z = 0 (lim z → sin z/z = 1). Not a very interesting case, anyway. 2. LS may have b n 6 = 0 for n < m ; b n = 0 for n ≥ m . This is called a pole of order m . f ( z ) = z + 3 z 2 ( z 1) 3 ( z + 1) . (6) This function has poles of order 1 at z = 1, of order 2 at z = 0, and of order 3 at z = 1. N.B. If f ( z ) has a pole of order m at z , then ( z z ) m f ( z ) is regular around z ....
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This note was uploaded on 04/17/2008 for the course PHZ 3113 taught by Professor Hirshfeld during the Fall '07 term at University of Florida.
 Fall '07
 HIRSHFELD

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