1. Limited time only! Sign up for a free 30min personal 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!

Branch Cuts in the Complex Plane

  1. Feb 2, 2010 #1
    1. The problem statement, all variables and given/known data

    The function

    f(z) = (1-z2)1/2

    of the complex variable z is defined to be real and positive on the real axis in the range -1 < x < 1. Using cuts running along the real axis for 1 < x < infinity and -infinity < x < -1, show how f(z) is made single-valued and evaluate it on the upper and lower sides of both cuts.


    3. The attempt at a solution

    I started by expressing f(z) in terms of complex exponentials:

    f(z) = (1-z2)1/2
    = (e-i2pi - r1r2ei2(theta1 + theta2))1/2
    = (r1r2)1/2ei(theta1 + theta2 - pi)

    But I don't understand how to evaluate this on both sides of the cuts. I've searched the internet quite a bit trying to find something to help explain it to me but couldn't find anything so I've come to you guys!

    Thanks for any help!
    Andrew
     
  2. jcsd
  3. Feb 3, 2010 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Remind yourself of how sqrt(z) is defined around 0 with a branch cut along the negative x axis. You define choose the range of argument theta=(-pi,pi) for z not on the negative real axis and then if z=r*e^(i*theta), define sqrt(z)=r^(1/2)*exp(i*theta/2), right? For your function you want to define sqrt(1+z) around z=(-1) and sqrt(1-z) around z=1 with their suggested cuts in an analogous way. Then multiply them together to get f(z).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Branch Cuts in the Complex Plane
  1. Branch cut (Replies: 8)

Loading...