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##\int \frac{dz}{z} ## along a line on the complex plane

  1. Sep 11, 2015 #1
    Some time ago I stumbled upon the integration ##\int \frac{dz}{z} ## along a line on the complex plane. I was confused because ##Ln(z)## is a multivalued function but apparently the way you do it is by only considering the principal branch from ##[-\pi,\pi]##.

    But I don't understand this at all. I can't find the theory behind this methodology anywhere and was hoping someone could clarify it for me or at least point me towards a source.

    Thanks.
     
  2. jcsd
  3. Sep 11, 2015 #2

    andrewkirk

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    I think it means that you always restrict the imaginary part of the log to the range ##[-\pi,+\pi]##.

    There's a formula for the principal value in the wikipedia article on Complex Logarithm.
     
  4. Sep 11, 2015 #3
    Yeah it does mean that. But why choose that branch?
     
  5. Sep 11, 2015 #4

    fzero

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    Suppose that the path we're supposed to take the integral over starts at the point ##w_0## and ends at ##w##. Obviously the path should not pass through the origin. We can take ##w_0 = r_0 e^{\theta_0}##, with ##\theta_0 \in (-\pi,\pi)##. Similarly, we will have ##w= r e^{\theta}##, with ##\theta \in (-\pi,\pi)##.

    We are going to deform the original path in the following way. First we take a circular arc ##L_1## from ##w_0## to the positive real axis at ##r_0##. Then we take a real interval ##L_2## from ##r_0## to ##r##. Finally, we have another circular arc ##L_3## from the positive real axis at ##r## to the point ##w##. The arcs ##L_{1,3}## have to chosen in such a way that the deformation of the original path doesn't cross the origin (a little sketching will make this clear).

    The contribution to the integral from ##L_2## is ##\pm \ln r/r_0##. The contribution from ##L_1## and ##L_3## include a term ##i\arg w_0## and ##i\arg w## respectively, but may also include some contributions ##\pm i \pi## representing the branch of the log that we had to choose depending on the relation between the original path, the origin and the part of the real axis we used for ##L_2##.

    I can't upload a sketch right now, but hopefully you can follow along and make your own to see how this works.
     
  6. Sep 12, 2015 #5
    I've sketched what you said but I don't see where the "contributions ##\pm i\pi ## " come from.

    Doing the integral ##\int_{L_1} \frac{dz}{z} ## yields ##\ln(r_o)-\ln(r_o)-i\arg w_o ##.

    What am I seeing wrong? Also I don't understand how this resolves the preference towards the branch ##(-\pi,\pi)##.
     
  7. Sep 12, 2015 #6

    fzero

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    Let me choose a specific example to illustrate how we can have different contributions to the integral depending on the original path and the location of the origin.

    Let ##w_0 = e^{-i3\pi/4}## and ##w = 2 e^{i\pi/4}##. So we integrate from the 3rd quadrant to the 1st quadrant and let's say we use a path on the right side of the origin. Then ##L_1## involves the counterclockwise arc ## z = e^{it}## where ## -3\pi/4 \leq t \leq 0##, so we get
    $$\int_{L_1} \frac{dz}{z} = i \int_{-3\pi/4}^0 dt = 3i\pi/4. $$
    ##L_2## gives us ##\ln 2##, while ##L_3## gives ##i\pi/4##. So
    $$I = \ln 2 + i \pi.$$

    However, if the path was on the left side of the origin, we need to take ##L'_1## to be the clockwise arc from ## -3\pi/4## to ##0##. So now we get
    $$\int_{L'_1} \frac{dz}{z} = i \int_{-3\pi/4}^{-\pi} dt + i \int_\pi^0 dt = -5i\pi/4. $$
    The other integrals are the same, so we get
    $$I = \ln 2 - i \pi.$$
    So this is a different branch of the log. It's apparently not ##\pm i \pi##, but ##\pm 2i\pi##.

    Note that the reason for the different answers is that the 2nd path crosses the negative real axis, so we have to choose a different branch of the log. If we put the branch cut on the negative real axis, then we have the ##(-\pi,\pi)## branch and we don't allow any integration paths to cross the negative real axis. As long as we keep to this rule, then we will only get a single value for the integration. I think this is what you wanted to see.
     
  8. Sep 12, 2015 #7
    But I thought that the branch ##(-\pi,\pi)## was the "true" one. Are all branches valid then and its just a matter of convention to choose ##(-\pi,\pi)##?
     
  9. Sep 12, 2015 #8
    I mean the integral as a sort of Riemann sum should strictly be single valued shouldn't it?
     
  10. Sep 12, 2015 #9

    fzero

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    Given a specified path, the integral is single-valued for any path that can be smoothly deformed to the original path. In the example, the second path cannot be smoothly deformed to the first path because we're not allowed to move the path through the origin. The first path doesn't cross the negative real axis, so the ##(-\pi,\pi)## branch is a valid choice. The second path does cross the negative real axis, so the integral that I did selects a different branch.

    Whether or not you can use the principal branch will depend on the nature of the problem. If you can specify the problem so as to avoid the branch cut on the negative real axis, then you can probably use the principal branch. However, it may be that you can't avoid the negative real axis, like when the second path is specified in the example.
     
  11. Sep 13, 2015 #10
    Okay okay I think I get it now. I was told the branch ##(-\pi,\pi)## avoided all these problems but I don't see why, since not all paths can be smoothly deformed such that ##(-\pi,\pi)## is a valid branch. I.e. Maybe you're forced to cross the negative real axis... Right?
     
  12. Sep 13, 2015 #11

    fzero

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    Yes, that is what I meant. Here's an example of a problem where it is easiest to put the branch cut of the logarithm on the positive real axis: https://en.wikipedia.org/wiki/Methods_of_contour_integration#Example_.28IV.29_.E2.80.93_branch_cuts
     
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