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I don't get branch cuts in the complex plane at ALL

  1. Sep 3, 2009 #1
    Suppose you're trying to provide a branch cut in [itex]\mathbb{C}[/itex] that will define a single-value branch of [itex]f(z) = \log(z - z_0)[/itex]. I don't know where to begin. Can someone help explain this concept to me?
  2. jcsd
  3. Sep 4, 2009 #2


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    You can first shift to the origin by letting u= z- z0 so your question refers to f(u)= ln(u). Now any straight line, from 0 to infinity can be taken as a branch cut! The "standard" choice is to take the negative x-axis as branch cut so that [itex]ln(z)= ln(re^{i\theta})= ln(r)+ i\theta[/itex] and [itex]-\pi< \theta< \pi[/itex]. The point is that that choice of branch cut "separates" [itex]-\pi< \theta< \pi[/itex] from [itex]\pi< \theta< 3\pi[/itex] from [itex]3\pi< \theta< 5\pi[/itex], etc.

    In terms of your original z, a branch cut is a straight line from z0 to infinity and the "standard" choice is the line z= t+ iy0 where y0 is the imaginary part of z0 and t can be negative real number.
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