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hw10_soln

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

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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 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) = 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 = n =0 z n and 1 z - 2 = - 1 2 1 1 - z/ 2 = - 1 2 n =0 z n 2 n . Hence, in this region we have 1 z ( z - 1)( z - 2) = 1 z n =0 z n - 1 2 n =0 z n 2 n = 1 z 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) = 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. (b) When 1 < | z | < 2 we have - 1 z - 1 = - 1 z 1 1 - 1 /z = - 1 z n =0 1 z n = - n =1 1 z n 1

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and, as above, 1 z - 2 = - 1 2 1 1 - z/ 2 = - 1 2 n =0 z n 2 n . Thus 1 z ( z - 1)( z - 2) = 1 z - n =1 1 z n - 1 2 n =0 z n 2 n = - n =0 z n - 1 2 n +1 - n =1 1 z n +1 .
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hw10_soln - Complex Analysis Fall 2007 Homework 10...

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