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!

Branches of the complex logarithm

  1. Jan 28, 2013 #1
    1. The problem statement, all variables and given/known data
    Find a branch of log(z − 1) that is analytic inside the unit circle. What is the value of this branch at z = 0?

    2. The attempt at a solution

    Alright so clearly the log(z) function is analytic at all points accept for the negative real axis.

    So log(z-1) will be analytic at all points x ≤ 1. My problem is choosing a branch. I'm really bad at doing this. I understand if the function was something like log(z+1) it would already be analytic inside the unit circle using the principle branch. I just don't see how I can choose a branch that is analytic inside the unit circle. I think if I understand this problem I can be prepared for my test tomorrow. Thanks for any help in advance.
     
  2. jcsd
  3. Jan 28, 2013 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    As just the real axis is problematic, you can begin in the upper [or lower] part of the plane, choose a branch there, and try to extend this to the interior of the unit circle afterwards.
     
  4. Jan 28, 2013 #3
    So if I understand right from what I've got in my notes I can just start my branch at some arbitrary point say something like.

    [itex]\tau[/itex]<arg(w)≤[itex]\tau[/itex]+2[itex]\pi[/itex]

    Where [itex]\tau[/itex] = [itex]\pi[/itex]/4 maybe?

    However, then the function would not be analytic at the angle [itex]\tau[/itex] on the unit circle am I right?
     
  5. Jan 28, 2013 #4
    Tell me if this works:

    log(z-1) = log(-1) + log(1-z) = Log|-1| + i[itex]\cdot[/itex]arg(-1) + log(1-z)

    = 0 + i[itex]\cdot[/itex][itex]\pi[/itex] + log(1-z)

    So I end up with f(z) = log(1-z) + i[itex]\cdot[/itex][itex]\pi[/itex]

    Then I could use the principle argument because now the cut is at real values of x ≥ 1. Which is outside the inside of the unit circle. I have no idea if my logic here is right though.
     
  6. Jan 29, 2013 #5

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    That should work, right.
     
  7. Jan 29, 2013 #6
    Awesome! I think I finally get this. Hope I do well on the test today. Thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Branches of the complex logarithm
  1. Complex Logarithms (Replies: 1)

  2. Complex Logarithm (Replies: 1)

Loading...