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Complex integrals

  1. Jan 29, 2012 #1
    By parameterizing the curve (not by Cauchy's theorem) and using the series of sin z, nd
    the value of
    ∫z^k sin(z)dz around a closed Contour C where C is the unit circle z=e^(iθ), for 0≤θ<2π




    What do they mean by using series of sin z ? I mean if I expand it .. I get e^(iθ)- e^(3iθ)/3! ---
    and so on ..Not sure what Iam meant to do with that since its an infinite series ..Do I say for small z sinz≈z , but not sure ..

    thanks
     
  2. jcsd
  3. Jan 29, 2012 #2

    tiny-tim

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    hi rbnphlp! :smile:
    they mean sinz = z - z3/3! + z5/5! - …

    (learn it … also cosz = 1 - z2/2! + z4/4! - … :wink:)
    (i haven't tried it :redface:, but …)

    my guess is that all the integrals are going to be 0 except for one …

    if so, it won't matter that there's an infinite number of them! :biggrin:
     
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