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Math Help - Laurent series and residue

  1. #1
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    Laurent series and residue

    Find the Laurent series for the given function about the indicated point. Also, give the residue of the function at the point.

    (1) \frac{z^2}{z^2-1}; z_0=1

    Thanks
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  2. #2
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    This is what I got so far:

    \frac{z^2}{z^2-1} = \frac{1}{z^2-1} + 1

    = 1 + \frac{A}{z-1}+\frac{B}{z+1}

    A=1/2, B=-1/2

    = 1+\frac{1}{2}(\frac{1}{z-1})-\frac{1}{2}(\frac{1}{z+1})

     1 +\frac{1}{2}\frac{1}{z-1} = \frac{1}{2}\frac{1}{z+1}

    Any idea what I do from there? Thanks!
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  3. #3
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    You have 1+1/2\frac{1}{z-1}-1/2\frac{1}{z+1} so need to express the last one in terms of z-1. That's not hard, just subtract and add one in the denominator:

    \frac{1}{z+1}=\frac{1}{z+1-1+1}=\frac{1}{2+(z-1)}=\frac{1}{2-(-(z-1))}=1/2\frac{1}{1-\frac{-(z-1)}{2}}

    Now, just finish it by expressing the last expression as a sum.
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  4. #4
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    Quote Originally Posted by shadow_2145 View Post
    Find the Laurent series for the given function about the indicated point. Also, give the residue of the function at the point.

    (1) \frac{z^2}{z^2-1}; z_0=1

    Thanks
    The function has form \frac{f(z)}{g(z)} and g'(z_0) \not = 0.
    Therefore, the residue is given by \frac{f(z_0)}{g'(z_0)} = \frac{1}{2}.
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