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Math Help - find Laurent representation of f

  1. #1
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    find Laurent representation of f

    I have to find Laurent representation of a function f(x)= \frac{1}{(z-1)(z-2)} , |z|<1

    So

    f(x)= \frac{1}{(z-1)(z-2)} =  \frac{1}{z-2}- \frac{1}{z-1}

    \frac{1}{z-2} =  \frac{1}{z}  \cdot  \frac{1}{1-  \frac{2}{z} } =   \sum_{0}^{n}  \frac{2^{n}}{z^{n+1}}

    \frac{1}{z-1}= - \frac{1}{1-z} =  \sum_{0}^{n} - z^{n}

    Finally

    \frac{1}{(z-1)(z-2)} = \sum_{0}^{n}  \frac{2^{n}}{z^{n+1}} +  \sum_{0}^{n} z^{n}

    Good?
    Last edited by fqqs; May 26th 2012 at 09:10 AM.
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  2. #2
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    Re: find Laurent representation of f

    I don't think this is quite correct. Your second expansion looks good - the 1/(z-1).

    But the second one should be done in a similar way as the first I think. The way you have done it, your series converges iff |2/z|<1 iff |z|>2. Which is not what you need, you need
    |z|<1. Try taking out 1/2 in the 1/(z-2) expansion instead of 1/z. I get the following answer doing that:

    1/(z-1)(z-2) = sum_n=0^inf [1-2^(-1-n)]z^n, |z|<1

    Wolfram also gets that:

    1&#47;&#40;&#40;z-1&#41;&#40;z-2&#41;&#41; - Wolfram|Alpha

    Hope this helps.
    PS what is the quick way of using latex on this forum? lol
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  3. #3
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    Re: find Laurent representation of f

    I don't think this is quite correct. Your second expansion looks good - the 1/(z-1).

    But the second one should be done in a similar way as the first I think. The way you have done it, your series converges iff |2/z|<1 iff |z|>2. Which is not what you need, you need
    |z|<1. Try taking out 1/2 in the 1/(z-2) expansion instead of 1/z. I get the following answer doing that:

    1/(z-1)(z-2) = sum_n=0^inf [1-2^(-1-n)]z^n, |z|<1

    Wolfram also gets that:

    1/((z-1)(z-2)) - Wolfram|Alpha

    Hope this helps.
    PS what is the quick way of using latex on this forum? lol
    Thanks from fqqs
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  4. #4
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    Re: find Laurent representation of f

    Thanks, obviously you're right

    so there is no principal part in this series?


    and if our contour is defined by |z|<1 integral equals to 0 , because there are no poles.
    Last edited by fqqs; May 26th 2012 at 11:51 PM.
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  5. #5
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    Re: find Laurent representation of f

    Yes, sounds right to me
    Thanks from fqqs
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