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Finding the Laurent series and residue of a function

  1. Apr 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the Laurent series for the given function about the specified point. Also, give the residue of the function at the point.

    $$ \frac{z^2}{z^2 - 1}, z_0 = 1 $$

    2. Relevant equations
    A Laurent expansion is comparable to a power series, except that it includes negative powers as well:
    $$ \sum_{n = - \infty}^\infty a_k (z - z_0)^n $$
    Where ##a_k## are the coefficients of the expansion.

    The residue of f at ##z_0## is the coefficient of ##(z-z_0)^{-1}## in the Laurent expansion.


    3. The attempt at a solution

    I was able to find the residue with moderate success by simply expanding the fraction:
    $$ \frac{z^2}{z^2 - 1} = \frac{z^2 - 1 + 1}{z^2 -1} = \frac{(z+1)(z-1) + 1}{(z+1)(z-1)}$$

    Afterwards I simplified this fraction by splitting it and then using partial fraction decomposition to give me:
    $$ -\frac{1}{2(z+1)} + \frac{1}{2(z-1)} + 1 $$
    Since we are finding the residue at ##z_0 = 1## of this function, then we look at the term that has ##(z - 1)## in its denominator, which would be ##\frac{1}{2}##.

    The next part would be to find the Laurent expansion of this equation, to which neither I nor my classmates had much luck. One example in the textbook suggested that I expand the numerator in powers of (z-1), which would also be something I don't understand how to do.
     
  2. jcsd
  3. Apr 8, 2013 #2

    Dick

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    Science Advisor
    Homework Helper

    If you want to expand in powers of (z-1) write u=(z-1). Then z=u+1. Substitute that into the function so you can write it totally in terms of u. Then try and expand around u=0.
     
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