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Complex integration over a square contour (part b)

  1. Nov 26, 2011 #1
    1. The problem statement, all variables and given/known data

    Let [itex]\Gamma[/itex] be the square whose sides have length 5, are parallel to the real and imaginary axis, and the center of the square is i. Compute the integral of the following function over [itex]\Gamma[/itex] in the counter-clockwise direction using parametrization. Show all work.


    2. Relevant equations

    u substitution

    3. The attempt at a solution

    Starting from the lower right point and going counter clockwise:

    1) [itex]^{3.5}_{-1.5} \int \frac{i(2.5+iy-1)}{2.5+iy+1} dy[/itex] = -0.5245 + 2.6194i

    2) [itex]^{2.5}_{-2.5} \int \frac{x+i3.5-1}{x+i3.5+1} dx[/itex] = -4.4755-2.3806i

    4) [itex]^{-2.5}_{2.5} \int \frac{x-i3.5-1}{x-i3.5+1} dx[/itex] = 3.8299-3.9026i

    All of these have been confirmed by the quad function in MATLAB. There is a problem when computing the third integral from 3.5i to -1.5i, crossing from quadrant II into quadrant III. My method seems to be the same as before but my answer is [itex]-4\pi i[/itex] more than what it should be.

    3) [itex]^{-1.5}_{3.5} \int \frac{z-1}{z+1} dz[/itex] where [itex]\stackrel{z(y)=-2.5+iy}{z'(y)=i}[/itex]

    [itex]\int \frac{i(-2.5+iy-1)}{-2.5+iy+1} dy[/itex] where [itex]\stackrel{u=-2.5+iy+1}{du=idy}[/itex]

    = [itex]\int \frac{u-2}{u} du[/itex]

    = [itex]\int 1- \frac{2}{u} du[/itex]

    = [itex]u-2*ln(u)

    = (-2.5-iy + 1)-2ln(-2.5+iy+1)^{-1.5}_{3.5}[/itex] = 1.1701+3.6638*i

    This should be 1.1701-8.9026*i

    Can you see what the problem is?

    Attached Files:

  2. jcsd
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