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Homework Help: Application of Cauchy's residue theorem

  1. Aug 21, 2013 #1
    Really need help for this one. Cheers!

    1. The problem statement, all variables and given/known data

    Question: calculate function z/(1-cos z) integrated in ac ounterclockwise circular contour given by |z-2pi|= 1

    2. Relevant equations

    3. The attempt at a solution

    Clearly the pole in the given contour is 2pi. But the problem is: if it's a simple pole, then apply formula

    we have Residue=lim (z-2pi) * z/(1-cos z) where z->2pi. This limit does not exist.

    So I reckon 2pi might be a higher order pole but this actually makes no sense and even if it's true,

    there is a really nasty differentiation.

    Any thoughts.?
    Last edited: Aug 21, 2013
  2. jcsd
  3. Aug 21, 2013 #2
    Please use the standard format of PF!
  4. Aug 21, 2013 #3


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    Yes, please use LaTeX.

    What you need is a Laurent expansion around the pole at [itex]z=2 \pi[/itex]. For your function
    [tex]f(z)=\frac{z}{1-\cos z}[/tex]
    you have
    [tex]1-\cos z=\frac{(z-2 \pi)^2}{2}+\mathcal{O}[(z-2 \pi)^4].[/tex]
    You need the single-order pole contribution. You can get this by either using the standard formula
    [tex]\text{res}_{z \rightarrow 2 \pi} f(z)=\lim)_{z \rightarrow 2 \pi} \frac{\mathrm{d}}{\mathrm{d} z} (z-2 \pi)^2 f(z),[/tex]
    which is a bit cumbersome here, or you use the series expansion of [itex]1-\cos z[/itex] as given above and use [itex]z=(z-2 \pi)+2 \pi[/itex] in the numerator.
  5. Aug 21, 2013 #4
    sorry for not using LaTex, havent used this forum in ages.

    And thanks so much for the help! very smart move!!

    Correct me if im wrong, the answer is 4i*pi?
    Last edited: Aug 21, 2013
  6. Aug 22, 2013 #5


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    Looks good!
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