Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Complex Integral

  1. Oct 17, 2006 #1
    Here is the exercise:

    Use the indefinite integral to compute [tex] \int_{C} \sqrt{z}dz[/tex] where C is a path from z = i to z = -1 and lying in the third quadrant. Note: [tex] \sqrt{z} = e^{(1/2)lnz}[/tex] where the principal branch of lnz is defined on [tex]C \setminus [0,\infty][/tex].

    I am just a little unsure of why he gave us that note (although I do use it). Here is what I did for the exercise:

    [tex] \int_{C} \sqrt{z}dz = \left[ \frac{2}{3}z^{3/2} \right]_{i}^{-1} [/tex]

    [tex]= \frac{2}{3} \left[ e^{(3/2)ln(-1)} - e^{(3/2)ln(i)} \right] [/tex]

    [tex]= \frac{2}{3} \left[ e^{(3/2)\pi i} - e^{(3/2)(\pi/2) i} \right] [/tex]

    [tex]= \frac{2}{3}(1 - 2i)[/tex]

    Everything look fine?

    Last edited: Oct 17, 2006
  2. jcsd
  3. Oct 17, 2006 #2


    User Avatar
    Homework Helper

    I think you computed e^(3/2 pi/2 i) incorrectly, but other than that it looks ok. The reason he was explicit about the branch cut was because you'd get a different answer if the cut passed through the path.
  4. Oct 17, 2006 #3
    Woops, I forgot the whole [itex] \frac{\sqrt{2}}{2}[/itex] part in there. Thanks!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook