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# HW8sol - P ROBLEM SET 8 Partial solution 1 1 1 1 Note that...

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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 k =0 ( - 1) k ( z - 1) k 2 k +1 = 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 0 there exists an analytic function g ( z ) near z 0 such that g ( z 0 ) = 0 and f ( z ) = ( z - z 0 ) m g ( z ) near z 0 . Therefore Res ( f f , z 0 ) = Res m ( z - z 0 ) m - 1 g ( z ) + ( z - z 0 ) m g ( z ) ( z - z 0 ) m g ( z ) , z 0 = Res ( m z - z 0 , z 0 )+ Res ( g g , z 0 ) By Cauchy’s formula Res ( m z - z 0 , z 0 ) = m while since g /g is analytic at z 0

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HW8sol - P ROBLEM SET 8 Partial solution 1 1 1 1 Note that...

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