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Math Help - Question about series

  1. #1
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    Question about series

    Hey folks! I was trying to find the Laurent series of f=1/(z^2+z) in the region 1<|z-1|<2.

    f can be expressed as
    1/(w+1) - 1/(w+2), where w=z -1

    f=1/(1-(-w)) - (1/2)(1/(1-(-w/2))

    Anyway, at first I tried making 1/(1-(-w))= the sum from 0 to infinity of (-w)
    ^n. However, that's only valid when |w|<1, which is not the case here, so I got a wrong answer.

    Then I looked for some examples on the internet and found the following modified geometric series

    1/(1-z)= -(1/z
    ^0 + 1/z^1 + 1/z^2+...), for |z|>1.

    Using this formula I got the right answer. My question now is, how is that formula derived?
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  2. #2
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    Re: Question about series

    Suppose |z|>1. Then \dfrac{1}{1-z} = \dfrac{\tfrac{1}{z}}{\tfrac{1}{z}-1}} = -\dfrac{1}{z}\dfrac{1}{1 - \tfrac{1}{z}}. Since |z|>1, \left|\dfrac{1}{z}\right|<1.
    Thanks from Cesc1
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  3. #3
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    Re: Question about series

    Quote Originally Posted by SlipEternal View Post
    Suppose |z|>1. Then \dfrac{1}{1-z} = \dfrac{\tfrac{1}{z}}{\tfrac{1}{z}-1}} = -\dfrac{1}{z}\dfrac{1}{1 - \tfrac{1}{z}}. Since |z|>1, \left|\dfrac{1}{z}\right|<1.
    Of course. Thank you!
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