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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


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    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.


    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


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    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!
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