hw10_soln - Complex Analysis Fall 2007 Homework 10:...

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Unformatted text preview: Complex Analysis Fall 2007 Homework 10: Solutions 3.3.2 We have 1 z ( z + 1) = 1 z 1 z + 1 = 1 z 2 1 1 + 1 /z = 1 z 2 X n =0 (- 1) n z n since | z | > 1 implies that | 1 /z | < 1. Multiplying the 1 /z 2 into the series and reindexing we have 1 z ( z + 1) = X n =2 (- 1) n z n for | z | > 1. Since Laurent series are unique, this must be the desired expansion. 3.3.4 We first note that we have the partial fraction expansion 1 z ( z- 1)( z- 2) = 1 z- 1 z- 1 + 1 z- 2 . (a) For 0 < | z | < 1 we have- 1 z- 1 = 1 1- z = X n =0 z n and 1 z- 2 =- 1 2 1 1- z/ 2 =- 1 2 X n =0 z n 2 n . Hence, in this region we have 1 z ( z- 1)( z- 2) = 1 z X n =0 z n- 1 2 X n =0 z n 2 n ! = 1 z X n =0 1- 1 2 n +1 z n . Multiplying the 1 /z through the sum and reindexing we have 1 z ( z- 1)( z- 2) = X n =0 1- 1 2 n +2 z n + 1 2 z for 0 < | z | < 1. Uniqueness of Laurent series guarantees this is the desired expansion....
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This note was uploaded on 02/29/2012 for the course MATH 4364 taught by Professor Ryandaileda during the Fall '11 term at Trinity University.

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hw10_soln - Complex Analysis Fall 2007 Homework 10:...

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