1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
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?

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

    StatusX

    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Complex Integral
  1. Complex integral (Replies: 7)

  2. Complex integrals (Replies: 5)

  3. Complex integration (Replies: 1)

  4. Complex integration (Replies: 3)

  5. Complex integration (Replies: 6)

Loading...