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Homework Help: Complex Integral

  1. Feb 28, 2010 #1
    1. Integrate z2/(z4-1) counterclockwise around x2 + 16y2=4

    2. Cauchy's Integral Forumula

    3. Solution
    I found the points z=1,-1,i,-i where the function is not defined. Using partial fractions to split them up, and integral them separately.

    Only points z=1,-1 lies in the contour, so...
    [tex]\oint0.25/(z-1) + 0.25/(z+1) + 1/(z^2+1) dz[/tex]
    = 0.25(2Pi I + 2Pi I) + 0 = Pi I

    Ans is 0. can anyone find my mistake?

    1. Integrate sinh2z/z4 counterclockwise around the unit circle.

    2. Cauchy's Integral Forumula

    3. Solution

    [tex]\oint sinh2z/z^4 = \oint sinh2z/(z-0)^4[/tex]
    = 2*PI*i/3! * (sinh2z)'''

    Differentiating sinh2z thrice gives 8cosh2z

    Hence, integral at z=0 = (8/3)*PI*i

    Ans is (8/3)*PI. Again, can anyone spot my mistake.
  2. jcsd
  3. Feb 28, 2010 #2
    for the first question:

    the residue at z=1 is the limit as z goes to 1 of:

    [itex](z-1) \left( 0.25 / (z-1) + 0.25 / (z+1) + 1 / (z^2+1) \right)[/itex]
    put that in and you get 0.25 + (-0.25) = 0

    i imagine this will also happen at the z=-1 pole

    then just use Cauchy's residue theorem that

    [itex] \int_\gamma f(z) dz = \displaystyle \sum_i Res(f, c_i)[/itex] where [itex]c_i[/itex] are the poles of [itex]f(z)[/itex] and you'll get the whole thing to integrate to 0+0=0
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