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Complex square roots

  1. Sep 5, 2004 #1
    I've got a homework problem from my function theory course that I never fully understood. It's getting hot again since the exams are nearing...please help :smile:

    It goes simply like this: Compute the Integral
    [tex]\int_{\vert z \vert=1}\frac{dz} {\sqrt{6 z^2-5 z +1}} [/tex]
    The square root is chosen such that [itex] \sqrt{2}>0 [/itex]

    One has to use that the integrand is holomorphic in C without the interval from 1/3 to 1/2 on the real line. But why is that? Why must one cut out the whole interval and not only the two "critical" points?
    We didn't do much on square roots in class, the best I could find in my textbook was that a function that is holomorphic in a simply connected region + no zeroes does have a n-th root...I don't believe that's useful here.

    Edit: I know how to actually solve the problem, that's why I didn't post this in the homework section. I just don't really understand the square root thing.
     
    Last edited: Sep 5, 2004
  2. jcsd
  3. Sep 5, 2004 #2

    Hurkyl

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    Do you remember anything about "branch cuts"?

    If not, let's just consider [itex]f(x) := \sqrt{x}[/itex]. As you know, in its full generality, this is a multivalued function. We can give a rule for selecting which of the values to use at any point, such as the principal value. The problem is that we have to make a "branch cut" someplace where we jump discontinuously from one selection to the other.

    For instance, take the value on the unit circle parametrized by [itex]e^{it}[/itex]. On the circle, near t = 0, we have [itex]f = e^{it/2}[/itex]. If we extend this continuously all the way around the circle, we still have that [itex]f = e^{it/2}[/itex] near t = [itex]2\pi[/itex]. However, this is a problem because while t = 0 and t = [itex]2\pi[/itex] correspond to the same point, we have f(0) = 1 but f([itex]2\pi[/itex]) = -1.


    With the square root in which you're interested, it turns out that the interval [1/3,1/2] can be used as the branch cut.
     
  4. Sep 5, 2004 #3
    Thank you, Hurkyl.

    I'm still not 100% clear...
    I understand now why to cut the interval out, but then: let's say we have the square root with sqrt(2)>0 on the complex plane without the negative real axis, and we have f(z)=6z^2-5z+1. If we want to take the square root of f, we have to take out all the points where f(z) lies on the neg. real axis, right?

    So, if this logic is reasonable, we have to cut out [1/3,1/2], but also the line z=5/12+i*R, is that correct? I'm unsecure about this because the solution paper doesn't mention this (it may be wrong).
     
  5. Sep 5, 2004 #4
    Ah, it's all right, I've just figured out how that works. Good. Very good. The exams may come :biggrin:
     
  6. Sep 5, 2004 #5

    Hurkyl

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    Well, there are always lots of ways to select which branch you want to use at a point. I was wondering why it was at all possible for the cut to be selected as it is (aside from the fact it almost looks like [itex]\sqrt{x^2}[/itex], and your logic almost explains why.

    The domain of your function, g(z), is split into two disconnected parts: the part left of z = 5/12 + iR, and the part to the right. (minus, of course, the interval [1/3, 1/2]) Along the boundary, the function jumps to the other branch.

    In this case, I can make a new function, h(z), that selects the same branch as g(z) on the right side, but the opposite branch on the left side. Then, the line 5/12 + iR (except for R=0) will be a removable singularity -- I can extend h(z) analytically to this line as well, so its domain is everything but [1/3,1/2].

    But I see you've already got it. Good! :smile:
     
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