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Residues of reciprocal polynomials and functions involving reciprocal polynomials

  1. Jan 10, 2010 #1
    I have need to calculate the residues of some functions of the form [itex]\frac{f(x)}{p(x)}[/itex] where p(x) is a polynomial. To be more specific I have already calculated the 2 residues of [itex]\frac{1}{x^2+a^2}[/itex]. That one was quite easy. Now I'm asked to calculate the residues of
    [itex]\left(\frac{1}{x^2+a^2}\right)^2[/itex] and [itex]\frac{z^2}{x^2+a^2}[/itex]
    How would I do that? I have trouble splitting the fraction up into a series. Is there any general tips for calculating residues of functions with polynomials in the denominator?
  2. jcsd
  3. Jan 11, 2010 #2
    You just try to apply the Cauchy theorem. It's not so hard. For example, the function
    [itex]f(z) = \left( \frac{1}{z^2 + a^2}\right) ^2 [/itex]​
    has two singular point [itex]z_{1,2} = \pm \, a i [/itex], where [itex] i^2 = -1[/itex]. As the definition of residue, we have
    [itex] Res \left[ f , z = ai \right] = \frac{1}{ 2 \pi i } \oint _{|z-ai| = \epsilon} \frac{1}{(z^2 + a^2)^2} dz [/itex]​
    with [itex] \epsilon [/itex] small enough. We write
    [itex] \frac{1}{ 2 \pi i } \oint _{|z-ai| = \epsilon} \frac{1}{(z^2 + a^2)^2} dz = \frac{1}{ 2 \pi i } \oint _{|z-ai| = \epsilon} \frac{1/(z+ai)^2}{(z - ai)^2} dz [/itex]​
    and , by applying the Cauchy theorem, we have
    [itex] Res \left[ f , z = ai \right] = \frac{1}{ 2 \pi i } \oint _{|z-ai| = \epsilon} \frac{1/(z+ai)^2}{(z - ai)^2} dz = \left. \frac{1}{1!} \frac{d}{dz} \left( \frac{1}{(z+ai)^2}\right) \right|_{z=ai} = \frac{1}{4a^3 i} [/itex]​
    the first one
    [itex]Res [ f , z = - ai] = \frac{1}{ 2 \pi i } \oint _{|z + ai| = \epsilon} \frac{z^2/(z-ai)}{(z+ai)} dz = \left. \left( \frac{z^2}{(z-ai)^2}\right) \right|_{z=-ai} = \frac{a}{2i} [/itex]​
    In fact, there're so many way to defeat these problems in textbook of complex analysis.
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