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Homework Help: Complex Contour Integral

  1. Feb 6, 2010 #1


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

    Calculate the following line integrals from point z'=(0,-1) to z"=(0,1) along three different contours, [itex]C_j=(0,1,2)[/itex].


    where [itex]C_0[/itex] is the straight line along the y-axis, [itex]C_1[/itex] is the right semi-circular contour of radius 1, and [itex]C_2[/itex] is the left semi-circular contour of radius 1.

    3. The attempt at a solution

    (i) Along [itex]C_0[/itex], [tex]z=iy \implies dz = idy[/tex] and the integral is

    [tex]\int_{C_0}|z|dz=i^2 \int_{-1}^1ydy=-\frac{y^2}{2}|_{-1}^1=-\frac{1}{2}+\frac{1}{2}=0[/tex]

    (ii) Along [itex]C_1[/itex], [itex]z=re^{i \theta} \implies dz = ire^{i \theta}d \theta[/tex] with [itex]\theta:\frac{3 \pi}{2} \rightarrow \frac{\pi}{2}[/itex]. Note that r=1.

    So, [tex]\int_{C_1}|z|dz = ir^2\int_{\frac{3 \pi}{2}}^{\frac{\pi}{2}}e^{2i \theta}d \theta=\frac{1}{2}e^{2 i \theta}|_{\frac{3 \pi}{2}}^{\frac{\pi}{2}}=\frac{1}{2} ( e^{i \pi}-e^{3i \pi})=0[/tex]

    (iii) Along [itex]C_2[/itex], [tex]z=re^{i \theta} \implies dz = ire^{i \theta}d \theta[/tex] with [itex]\theta:-\frac{\pi}{2} \rightarrow \frac{\pi}{2}[/itex].

    The integral is similar to (ii), and one obtains:

    [tex]\frac{1}{2} ( e^{i \pi}-e^{-i \pi})=0[/tex]

    Did I do these integrals correctly (correct limits in ii and iii)? If so then geometrically, why are these integrals equal to zero?

    Thanks for your comments.
  2. jcsd
  3. Feb 6, 2010 #2
    For (ii) and (iii), don't forget that you are integrating |z|, not z.

    The point of the problem is to show you that the path integral does depend on the path you choose. You will see later that the value of a path integral is independent of the path if a function is holomorphic. That is because holomorphic functions have antiderivatives. It's the same theorem as in multivariable calculus, when you learned that the value of a path integral over a vector field depends only on the start and end points if the vector field is the gradient of a function.

    In this example, the value of the integrals does depend on the path because |z| is not holomorphic.
    Last edited: Feb 6, 2010
  4. Feb 6, 2010 #3


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    oops forgot i was integrating |z|..i fixed (ii) quickly and got an answer of 2i, is that correct?

    \int_{C_1}|z|dz = ir^2\int_{\frac{3 \pi}{2}}^{\frac{\pi}{2}}e^{i \theta}d \theta=i [sin(\theta)-icos(\theta)]|_{\frac{3 \pi}{2}}^{\frac{\pi}{2}}=2i
    Last edited: Feb 6, 2010
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