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Homework Help: Analytic continuation

  1. Oct 6, 2013 #1
    1. The problem statement, all variables and given/known data
    Choose a branch that is analytic in the circle |z-2|<1. Then analytically continue this branch along the curve indicated in Fig 5.18. Do the new functional values agree with the old?

    [tex]a, 3z^{\frac{2}{3}}[/tex]
    [tex]b, (e^z)^\frac{1}{3}[/tex]

    Fig 5.18 is basically an ellipse like loop starting at z=2, going around the origin and returning to z=2.

    3. The attempt at a solution
    Started with a, well the principle branch satisfies the requirements as the negative real axis isn't included in the circle |z-2| < 1.

    And that's it. Not been able to follow this chapter very well and it feels like I'm not understanding more by reading it more. Not very example heavy as well. How do you go about solving this? They did something similar in an example applying the Monodromy theorem but they used a punctured plane as the domain and I cannot see how a loop can be continuously deformed if there is a hole in the middle. Assuming there's something I'm 100% misunderstanding, thus I'm very much at a loss on how to proceed as that's the only example available.
  2. jcsd
  3. Oct 6, 2013 #2
    You guys draw stuff? Lot easier if you do. Also, try a simpler one first: [itex]w=z^{1/3}[/itex]. Same dif. Expand it as:
    [tex]z^{1/3}=e^{1/3(\ln|z|+i(\text{Arg}(z)+2n\pi))},\quad -\pi< \text{Arg}(z)\leq\pi[/tex]

    Ok, just say the contour is a circle of radius 2 and we start going around the contour over the principal branch, that is, [itex]n=0[/itex]. Then at the point z=2, [itex]\text{Arg}(2)=0[/itex] so that [itex]2^{1/3}[/itex] on this branch is just the real value of the cube root of 2. When you get to -2, we have:

    [tex]e^{1/3(\ln|z|+i(\pi))}=r^{1/3}e^{\pi i/3}[/tex]

    now, as soon as you drop below the negative real axis, the principal argument, Arg(z), drops to [itex]-\pi[/itex] right? Ok, then in order for [itex]e^{1/3(\ln|z|+i(\pi))}[/itex] to remain analytically continuous along this contour across the negative real axis, we need to have the argument vary continuously from what it is at -2, or [itex]\pi/3[/itex], as we drop below the negative real axis. We can do that by "annealing" the branch with [itex]n=1[/itex] to the principal branch at z=-2. Then we have at this point:

    [tex]e^{1/3(\ln|z|+i(-\pi+2\pi))}=r^{1/3}e^{\pi i/3}[/tex]

    which is exactly what it was just above the axis (analytically continuous) and we continue along this branch until we return to the point at 2 which then gives us:

    [tex]e^{1/3(\ln|z|+i(0+2\pi))}=r^{1/3}e^{2\pi i/3}[/tex]

    But recall we started with


    and so does an analytically-continuous path over [itex]|z|=2[/itex] and starting on the principal-branch at z=2, return to the starting point after a [itex]2\pi[/itex] circuit?

    Now do the same analysis with [itex]z^{2/3}[/itex].
  4. Oct 7, 2013 #3
    Thank you, that was incredibly informative. Managed to do both exercises now (the latter I presume doesn't change at all). Think I'll have an easier time trying to properly understand the theory now.
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