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How to interpret the phase along a contour of sqrt(z^2+1)

  1. Jun 7, 2009 #1
    This is what I did:


    Take the function sqrt(z^2-1). I try to think of it as rotating the values of the contour on this function. I already know the resulting integral on top of the segment is -i*integral from -1 to 1. Substitution of z = i*x we get the contour for sqrt(z^2+1) witht the branch points. the problem is that sqrt (z^2-1) becomes isqrt(1-z^2) in absolute value thats where the i comes from in the original integral. Thus, we have the integral is really sqrt(1-z^2)now that its rotated we take the phases from each side and get negative 2I where we have subsituted in the original integral z = xe^ipi/2 to get the desired limits. Is this the right way to interpret the integral? because I dont know any other way to yield the right answer, but I dont know if this is the right thing to do in general. Please help, any help will be appreciated!
     
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  3. Jun 7, 2009 #2

    Mute

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    First, note that

    [tex]\sqrt{z^2 + 1} = \sqrt{(z+i)(z-i)}.[/tex]

    You see that there are two branch points, i and -i. These are the points about which if you were to try to follow the value of the function around these points on some closed contour you would find you've acquired a phase of -1. So, you need to introduce a branch cut(s) to make the expression single valued. There are two (main) choices.

    The first is that you can have a branch cut from [itex]+i[/itex] out to [itex]+i\infty[/itex], and similarly from [itex]-i[/itex] out to [itex]-i\infty[/itex]. This restricts the values of the arg(z) (that is, the phase angle) about those two points. About +i, arg(z) is restricted to [itex][-\frac{3\pi}{2},\frac{\pi}{2})[/itex]. About -i, arg(z) is restricted to [itex][-\frac{\pi}{2},\frac{3\pi}{2})[/itex].

    The other choice of branch cut is to have the branch cut flow from +i down to -i, in which case the values that arg(z) is restricted to is the opposite of what I gave above.

    Note that these are not the only choices of branch cuts: I could have branch cuts from the two branch points go off to infinity in any direction, and they don't even need to be straight lines. What I gave are just convenient choices. I could have had a cut from +i to [itex]+\infty[/itex] and -i to [itex]+\infty[/itex] (both cuts parallel to one another), in which case for both points arg(z) is restricted to [itex][0,2\pi)[/itex].
     
  4. Jun 7, 2009 #3
    so if I wanted to find the phase of the function on each side, what would they be?
     
  5. Jun 7, 2009 #4

    Mute

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    For this particular example: since normally the phase factor is [itex]\exp[i \arg(z)][/itex], for a square root it is [itex]\exp[i\arg(z)/2][/itex]. So on one side of the branch cut you just plug in the value of arg(z) at that boundary, and then on the other side of the branch cut after rotating around the branch point by 2 pi you put in the value of arg(z) defined at the other end of the range.

    So, for the first case I gave above, for the +i point, starting on the left side of the branch cut, the phase factor is [itex]\exp[i3\pi/4][/itex], then going counterclockwise around the point, I arrive at the right side of the cut, where the phase factor is [itex]\exp[i\pi/4][/itex].
     
  6. Jun 7, 2009 #5
    alright, but why doesnt that example work for my previous post, i.e. the mercedes benz contour. Its near the bottom of the calculus and analysis page. Now, I tried restricting the argument to -pi/3, 5pi/3 but the answer i get is always wrong. In short, How would you evaluate this integral with residue calculus?: (1-x^3)^-1/3 from 0 to 1.
     
  7. Jun 7, 2009 #6
    In these more complicated cases it is far more convenient to cut up the contour( in this case in three pieces) and define the function on the contours to be analytical continuations of each other. Then you have much more freedom to choose the branch cuts, so you don't have to waste you time being extremely careful to line up the different branch cuts so that they cancel each other out.

    Now, it is simplest to take the part of the cpontour integral from zero to 1 to be the real integral of (1-x^3)^(1/3). But that then means that you shoud choose the polar agles such that it matches with evaluating (1-x^3)^(-1/3) for real x yielding a real value.

    Now, if we factor:

    1-x^3 = [z1-x][z2 - x][z3 - x]

    Then the choice of z1, z2, and z3, essentially defines our choice of the polar angles. For real x, we want to the product of the tree cube roots not to produce a net phase factor. So, taking z1 = 1, z2= exp(2 pi i/3) and z3 = exp(4 pi i/3) won't do, because then we see that the inverse cube roots a x = 0 become 1, exp(-2/9 pi i) and exp(-4/9 pi i), and we get a net result of exp(-i pi/3) but at x = 0 on the real axis we want to have 1.

    Of course, for real x the two explicitely complex factors should be comlex conjugates of each other, so we choose:

    z1 = 1, z2 = exp(-2 pi i) and z3 = exp(2 pi i)

    Then the inverse cube root of the individual factors multiplied together will produce a real result on the real axis. Of course, you have to make sure that the branch cut won't intersect the first contour. This requirement then fixes the polar angles everywhere on the first contour. On the second contour everything is defined via analytical continuation from the first contour and on the third via analytical continuation from the second, so everything is fixed from this point on.

    Now, you could have chosen z1, z2, and z3, differently, but then the integral from zero to 1 on the contour would not have been the real integral I, there would have been some phase factor relating the two. So, you can save some effort at this point too.
     
  8. Jun 7, 2009 #7
    could we have done it in this method though, i.e. without the analytical continuations, if we did make the different selection of factors? If so, how?
     
  9. Jun 7, 2009 #8
    Yes, what you do is you take the branch cuts from the points
    exp(-2 pi i/3) and the point exp(2 pi/3) that run to the orgin and make them bend at the origin so that they run along the negative real axis, joining the branch cut from the point z = 1. The three branch cuts then combine and effectively cancel. This then makes sure the integrand doesn't have discontinuities on the big circle.
     
  10. Jun 7, 2009 #9
    and can you show me how to bend the branch points? For example, how would this affect the argument? Do you mean the set S={z: z=xexp(2pi*i/3), 0<=x<=1}U{z:z=x, -infinity<x<=0}? Again, how would I restrict the argument?
     
  11. Jun 7, 2009 #10
    Yes, that's correct. The other branch cut starting at exp(-2 pi i/3) is obtained by reflecting this in the real axis.

    About the arguments, what you do is you make a choice about what the value of the function should be on some point of the contour. The contour starts just below the origin moves just below the real axis to just below 1, then encircles the branch point at z = 1, moves back just above the real axis to z = 0 and then we go to z = exp(2 pi i/3) in a similar way and complete the "Mercedes Benz" contour.

    When we go from exp(2 pi i/3) to exp(-2 pi i/3) we cross the origin just to the left so that we cross the three merged branch cuts which then doesn't give rise to a discontinuity there.


    The simplest thing to do is to choose the polar angles such that it yields the real function (1-x^3)^(-1/3) when integrating from x = 0 to 1 just below the real axis. As explained above this means that you want to factor 1-z^3 as:


    1-z^3 = [1-z] [exp(2 pi i/3) - z][exp(-2 pi i/3) - z]

    Because then (1-x^3)^(-1/3) for real x smaller than 1 will be real. If you replace exp(-2pi i/3) by exp(4 pi i/3) then [1-x^3]^(-1/3) for real x smaller than 1 will be proportional to exp(-2 pi i/3).

    But this means that at x = 0 just below the branch cut from z = 1 and just to the right of the other two branch cuts, the argument of 1-z is zero, the argument of exp(2 pi i/3) - z is 2 pi i/3 and the argument of exp(-2 pi i/3) - z is - 2 pi i/3.

    Then having defined the arguments of all three factors at some point and having defined the three branch cuts, everything is fixed.

    In the other thrreat, I think I wrongly took the third branch point to be exp(4 pi i/3), instead of exp(-2 pi i/3). I didn't see this error because I assumed that I had chosen the arguments such that the integral just below the real axis would be real and my computation simply depned in the assumption that this choice had been made.
     
    Last edited: Jun 7, 2009
  12. Jun 7, 2009 #11
    So, the polar angles range from:

    branch point at z = 1: 0 to 2 pi

    branch point at z = exp(2 pi i/3): 2 pi/3 to 8 pi/3

    branch point at z = exp(-2 pi i/3): -8 pi/3 to - 2pi/3
     
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