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Math Help - [SOLVED] Complex

  1. #1
    Member ronaldo_07's Avatar
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    [SOLVED] Complex



    Could someone help me solve these? Thanks
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  2. #2
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    Expand \frac{7}{(z- 2)(z+5)^2} as \frac{\frac{1}{7}}{z- 2}+ \frac{\frac{5}{14}}{z+ 5}- \frac{1}{(z+5)^2} using partial fractions.

    Now expand \frac{\frac{1}{7}}{z-2} as a power series in z+ 5. You can do that, for example, by writing it as \frac{\frac{1}{7}}{z+ 5- 7}= -\frac{\frac{1}{7}}{7- (z+5)} = -\frac{\frac{1}{49}}{1- \frac{z+ 5}{7}} and recognize that as the sum of a geometric series: \sum_{n=0}^\infty-\frac{1}{49}\left(\frac{z+5}{7}\right)^n. The Laurent series is -(z+5)^{-2}+ \frac{5}{14}(z+5)^{-1}+ \sum_{n=0}^\infty-\frac{1}{49}\left(\frac{z+5}{7}\right)^n
    Last edited by mr fantastic; March 4th 2009 at 05:35 PM. Reason: Fixed some latex
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