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Order of a pole

  1. Aug 17, 2008 #1
    I still don't quite fully understand about the order of poles and calculating residues.

    Take f(z) = 1/sin z at z=0 for example.
    When I try putting that into Laurent expansion about z=0,
    1/ sin z = 2i/ (e^z - e^(-z)) = 2ie^(-z) / ( 1 - e^(-2z))
    = 2ie^(-z) [ 1 - e^(-2z) + e^(-4z) - e^(-6z) + ...] using geometric series
    = 2i [ e^(-z) - e^(-3z) + e^(-5z) - e^(-7z) + ... ]

    so when I expand out those "e"s using e^(g(z)) = sum_{i>=0} (g(z))^i / i!,
    I get no "negative exponent" for z since they're all positive

    but then lim_{z -> 0} f(z) = infty so z=0 has to be a pole, right? So then how do we find out the order (of that pole) via Laurent expansion, and consequently how to find its residue at z=0 (which should be equal to 1...)?

    Can anyone explain what's going on?

    Also, as another example, take f(z) = 1/sin(z^2) at z=0.
    Same problem as before...
    f(z) = 2i / (e^(z^2) - e^(-z^2))
    = 2i [ e^(-z^2) - e^(-3z^2) + e^(-5z^2) - ... ]
    How to use the Laurent exansion to find the order of the pole and the residue?
  2. jcsd
  3. Aug 17, 2008 #2


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    Your first problem is that you applied the geometric series formula outside of its domain of convergence. While the 'magic' of algebra and analytic functions will often forgive such sins, any hope of that happening is dashed by the second problem: expanding out those e's doesn't give an expression that can be rearranged into a power series.
  4. Aug 18, 2008 #3
    How should we go about finding orders and residues in those cases then?
  5. Aug 18, 2008 #4


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    In this particular case, you can use the fact that [itex]\lim_{z\rightarrow 0} z/sin z= 1[/itex]. 1/sin z= (z/sin z)(1/z) so the coefficient of 1/z in a Laurent series is 1.
  6. Aug 18, 2008 #5
    Thanks mate.

    Are there any good online ebooks with more examples like these available that you know of? Most of the ones I've searched/seen have either no examples or only very straightforward ones like 'polynomial denominators'...
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