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Branch of z^{1/2}

  1. Oct 9, 2011 #1
    1. The problem statement, all variables and given/known data
    Find a branch of [itex]w=z^{1/2}[/itex] which is continuous along the negative x-axis.

    3. The attempt at a solution

    The book proves that the principle square root function [itex]|z|^{1/2} \big( \cos(\theta/2) + i \sin(\theta/2) \big)[/itex], where [itex]-\pi < \theta \leq \pi[/itex] is discontinuous along the negative x-axis.

    I've defined a new branch of the square root function [itex]f_\alpha (z) = |z|^{1/2} \big( \cos(\theta/2) + i\sin(\theta/2) \big)[/itex], where [itex]\alpha < \theta \leq \alpha + 2\pi[/itex]. I know the principle square root function is discontinuous along the negative x-axis because the limit as the principle square root function is path dependent as [itex](r,\theta)[/itex] approaches an arbitrary negative number [itex]r_0 e^{i \theta_0}[/itex]. What's the best way to proceed from here? Should I choose an [itex]\alpha[/itex] such that [itex]f_\alpha[/itex] can only approach the negative x-axis one way?
     
  2. jcsd
  3. Oct 9, 2011 #2

    Dick

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    There's a lot of choices for alpha. Why not pick alpha=0? Where is the discontinuity now?
     
  4. Oct 9, 2011 #3
    Along the positive x-axis. I understand how I should approach the problem now. Nonetheless, thank you for your help! My book hid the fact that branches are discontinuous along their branch cuts at the end of an example.
     
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