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Unformatted text preview: P ROBLEM SET 8 Partial solution 1. Note that f ( z ) = 1 z 1 1 z +1 . Since 1 z +1 is analytic at z = 1 it has a Taylor series expansion near z = 1 . The Taylor series can be determined in a standard way by computing derivatives, but also more efficiently by using that when  a  < 1 then 1 a + a 2 a 3 + a 2 = 1 / (1 + a ) . In particular 1 1 + z = 1 2 + ( z 1) = 1 2 1 1 + z 1 2 = 1 2 (1 z 1 2 + ( z 1) 2 2 2 ) when  z 1 2  < 1 that is when  z 1  < 2 . So the Taylor series converges when  z 1  < 2 . This can also be verified by using the root or ratio tests for power series. Therefore the Laurent series for f ( z ) is f ( z ) = 1 z 1 X k =0 ( 1) k ( z 1) k 2 k +1 = X l = 1 ( 1) l +1 ( z 1) l 2 l +2 and converges when  z 1  < 2 . The residue at 1 is the coefficient with the ( z 1) 1 term, that is 1 2 . Sec. 2.5 p14. Since f has a zero of order m at z there exists an analytic function g ( z ) near z such that g ( z ) 6 = 0 and f ( z ) = (...
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This document was uploaded on 03/27/2010.
 Spring '09

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