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!

Homework Help: [Complex Analysis] Branch cuts of the logarithm

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data

    Consider a branch of [itex]\log{z}[/itex] analytic in the domain created with the branch cut [itex]x=−y, x≥0.[/itex] If, for this branch, [itex]\log{1}=-2\pi i[/itex], find the following.

    2. Relevant equations

    [tex]\log{z} = \ln{r} + i(\theta + 2k\pi)[/tex]
    3. The attempt at a solution

    This one is actually given in the textbook (odd numbered problem), but I'm having trouble understanding how the answer was arrived at.

    The answer given: [itex]0.693 - i\frac{11\pi}{6}[/itex]

    I can see easily that [itex]\log{\sqrt{(1)^2 + (\sqrt{3})^2}}=\ln{2} = 0.693...[/itex] The real part here makes sense since it's the (real) log of the modulus of the given complex number [itex]\sqrt{3}+i[/itex].

    I can also understand that the branch cut is made along [itex]x=-y[/itex]. Where I'm getting confused is how the cut actually affects this log. So [itex]r = 2, \theta=\frac{\pi}{6}[/itex]. Winding around counterclockwise from 0, we reach [itex]\frac{\pi}{6}[/itex] easily, since it does not cross the branch cut at all.

    Does the restriction [itex]\log{1}=-2\pi i[/itex] actually restrict this to moving around the circle clockwise from [itex]-\frac{\pi}{4}[/itex] such that [itex]-\frac{9\pi}{4} < \theta \le -\frac{\pi}{4}[/itex]? When using this log with principal values and restricted to a domain of analyticity of [itex]-\pi < \theta \le \pi[/itex] we traditionally wind around counterclockwise toward [itex]\pi[/itex] and clockwise toward [itex]-\pi[/itex]. This one, if I understand it correctly, winds around [itex]-2\pi[/itex] from the cut so that it's restricted to one set of values for an otherwise multi-valued log.

    Why do both of these ways of figuring a single-valued log (the traditional principal valued log cut on the negative x axis and the one used in this problem) seem to involve winding around the axis different ways? What should I be understanding here that I'm not?
  2. jcsd
  3. Oct 2, 2011 #2
    The issue is that the angle between the real axis and your point is π/6 , so the total angle from the ray is the angle from the ray to 1--which you're told is 2π , minus the angle from the ray to 31/2+i , which is π/6.
  4. Oct 2, 2011 #3
    It's actually [itex]-2\pi[/itex], not [itex]2\pi[/itex] where the log of 1 is defined. That is what is throwing me. I'm trying to get a clear picture in my head instead of just a plug and chug with the single-valued (analytic) definition of the log in complex, which works but doesn't lead me to using or understanding the nature of the branch cut involved.
  5. Oct 2, 2011 #4
    I think then it has to see with which is the positive direction, i.e., the direction in which angles increase; I don't know if there are standard rules for this.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook