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Homework Help: Branch cuts in complex analysis

  1. Oct 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Given that the standard square root sqrt(anything) has a branch cut from (-inf,0), find the branch cuts of the following:

    2. Relevant equations

    3. The attempt at a solution
    I understand what branch cuts do (multivalue functions -> single value), I am just having trouble since all three of those expressions can be reduced to the same expression - I don't see why they would need different cuts? I can't really find any examples online and my book only has less than one paragraph explaining branch cuts (let alone examples).
  2. jcsd
  3. Oct 3, 2012 #2
    No they can't! That is precisely the point. Perhaps you should calculate what you get when you plug in, say, z=±2 into those expressions.
  4. Oct 3, 2012 #3
    2+sqrt(2^2-1) = 2+sqrt(3)
    -2+sqrt((-2)^2-1) = -2+sqrt(3)

    2+isqrt(1-2^2) = 2+isqrt(-3) = 2+sqrt(3)
    -2+isqrt(1-(-2)^2) = -2+isqrt(-3) = -2+sqrt(3)

    2+sqrt(2+1)sqrt(2-1) = 2+sqrt(3)sqrt(1) = 2+sqrt(3)
    -2+sqrt(-2+1)sqrt(-2-1) = 2+sqrt(-1)sqrt(-3) = -2+sqrt(3)

    am I missing something?
  5. Oct 3, 2012 #4
    It's tough. Really tough as you indicated by the lack of good information in your text about it. How about this, suppose I have a function that ISN"t a polynomial. I mean trigs, logs, other non-polynomials. Call that function G. And I tell you wherever G is zero or infinity, there is a branch-point and to make a branch-cut, I define a line from one branch-point to another. So Take:


    and I know [itex]G(z)=\sqrt{z^2-1}[/itex] is not a polynomial and that the quantity [itex]z^2-1=0[/itex] when [itex]z=\pm 1[/itex]. Therefore, from what I said above, the branch points are infinity and [itex]\pm 1[/itex]. Thus I can make branch-cuts in several ways, from -1 to infinity AND 1 to infinity, or a branch-cut from -1 to 1 or I can even make one from -1, through 1 and through to infinity. Now in this particular example, the standard way of making branch-cuts would be two ways:

    (1) make a cut from -1 to 1

    (2) make two cuts: one from -infty to -1 and another one from 1 to infty

    Now how about this: even though that's not exactly what you might want, try and use what I said to devise branch-cuts for the other ones.
  6. Oct 3, 2012 #5
    Are you using √(-1) = i or √(-1) = -i?
  7. Oct 3, 2012 #6
    OK, so if I try to map just the square roots to the range (-inf, 0) I get:

    1) [tex]z^2-1=-R[/tex] where R ranges from (0, inf).
    Solving for Z yields [tex]z = \pm \sqrt{1-R}[/tex], which for R<1 gives (-1,1), and for R>1 gives (-i inf, +i inf) so I get branch cuts from (-1,1) and the imaginary axis.

    2) solving [tex]1-z^2=-R[/tex] yields [tex]z = \pm \sqrt{1+R}[/tex], which gives me (-inf, -1) and (1, inf) for branch cuts.

    3) Solving [tex]z+1=-R, z-1=-R[/tex] gives (-inf, -1) and (-inf, 1), but the overlap cancels(?) to just give (-1,1) as the branch cut.
  8. Oct 3, 2012 #7
    Ok, lemme' tell you something that is a key to understanding branch-cuts: they're arbitrary. That is, you can extend them from the branch-points any way you like. Now, granted, we try and standardize it somewhat as in the case of sqrt(z) where we just arbitrarly set the cut from -infty to 0 but that's not set in stone. I could equally have it run up the imaginary axis from zero or even a stair-step pattern up to infinity. So in the case of

    you should look at that and immediately conclude the branch points are at -1 and 1 and the branch-cuts can basically extend anyway from those points, like I said, even stair-steps going up from 1 and stairsteps going down from -1. But to make it simple, we just extend the cut from -1 to 1 (branch-point to branch-point) or depending on the application, from -infty to -1 along the real axis is one cut and 1 to +infty on the real axis as another cut with infinity being another branch-point.

    However, I'm not really sure what your particular question is asking.
    Last edited: Oct 3, 2012
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