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Math Help - Laurent series

  1. #1
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    Expand in Laurent series about Zo= 0 :
     f(z) = \frac{z }{{(z+1)^2}(z-2)} i know there are three cases but i don't know how i starting ??
    this is true or not ??:
     \frac{z }{{(z+1)^2}(z-2)} = \frac {2}{9(z-2)} - \frac{2}{9(z+1)} + \frac {1}{3{(z+1)^2}}


    And how i can starting if  f(z) =Log (1+ \frac{1}{z-1} ) near Zo=1


    please help me < my exam tomorrow >
    Last edited by mr fantastic; August 16th 2009 at 07:28 PM. Reason: Merged posts and other editing
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Can you find the Laurent series of \frac{1}{1+z} about z=0? Then you can find the Laurent series of \frac{1}{(1+z)^2} by differentiating.

    To find the Laurent series of \frac{1}{z-2} about z=0 you can use the identity \frac{1}{z-2}=\frac{-1}{2}\frac{1}{1-\frac{z}{2}}.

    Then you can just multiply the series (which will work since you will not have any negative powers of z in any of your answers).
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  3. #3
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    Quote Originally Posted by Bruno J. View Post
    To find the Laurent series of \frac{1}{z-2} about z=0 you can use the identity \frac{1}{z-2}=\frac{-1}{2}\frac{1}{1-\frac{z}{2}}.
    That will give you the Laurent series that converges when |z|<2. In the region |z|>2 you need to write it as

    \frac{1}{z-2}=\frac{1}{z}\frac{1}{1-\frac{2}{z}} = \frac1z\Bigl(1+\tfrac2z + \bigl(\tfrac2z)^2+\ldots\Bigr).

    In general, the Laurent series for \frac1{z-a} will consist of powers of \tfrac za in the region |z|<a, and powers of \tfrac az in the region |z|>a.

    So for the function  \frac{z }{{(z+1)^2}(z-2)} = \frac {2}{9(z-2)} - \frac{2}{9(z+1)}  + \frac {1}{3{(z+1)^2}} there are three separate Laurent expansions, obtained by adding the power series expansions of the partial fractions. In the region |z|<1 you need to use positive powers of z/2 for the first fraction, and positive powers of z for the other two fractions. In the region 1<|z|<2 you need to use positive powers of z/2 for the first fraction, and negative powers of z for the other two fractions. In the region |z|>2 you need to use negative powers of z/2 for the first fraction, and negative powers of z for the other two fractions.
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  4. #4
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    Quote Originally Posted by Opalg View Post
    That will give you the Laurent series that converges when |z|<2. In the region |z|>2 you need to write it as

    \frac{1}{z-2}=\frac{1}{z}\frac{1}{1-\frac{2}{z}} = \frac1z\Bigl(1+\tfrac2z + \bigl(\tfrac2z)^2+\ldots\Bigr).

    In general, the Laurent series for \frac1{z-a} will consist of powers of \tfrac za in the region |z|<a, and powers of \tfrac az in the region |z|>a.

    So for the function  \frac{z }{{(z+1)^2}(z-2)} = \frac {2}{9(z-2)} - \frac{2}{9(z+1)}  + \frac {1}{3{(z+1)^2}} there are three separate Laurent expansions, obtained by adding the power series expansions of the partial fractions. In the region |z|<1 you need to use positive powers of z/2 for the first fraction, and positive powers of z for the other two fractions. In the region 1<|z|<2 you need to use positive powers of z/2 for the first fraction, and negative powers of z for the other two fractions. In the region |z|>2 you need to use negative powers of z/2 for the first fraction, and negative powers of z for the other two fractions.
    Opalg>>>thanks !
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