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Math Help - branch cuts / complex logs

  1. #1
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    Question branch cuts / complex logs

    Hi. So I'm new to branch cuts and I think I get the general idea, but I'm stuck with a couple homework questions along the same lines as:

    I'm given the branch cut, eg (x=0, y>0)
    and told for example log(-1) = -i(pi) on this branch
    Then asked to find the value of some other simple complex logs on the same branch

    I guess I'm just confused as to how the given value of log(-1) (or log(1) in another question) is relevent... do I need to use it to find a value of k in
    log(z)=Log(r) + i(Argz + 2kpi) ?
    but then I can't see how I would use the value of k to answer the problem... does the 2kpi even come into it once you have a branch cut? Am I looking at this totally wrong?? v. confused...

    sorry if this isn't too coherent, i'm tired.. but any help appreciated!
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  2. #2
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    Remember that the imaginary part of the logarithm is i times the argument. If for example you are told that log(-1) = -iπ, it means that you are supposed to take -π (as opposed to +π or any other odd multiple of π) as the argument for -1. To stay on the same branch, you must ensure that the argument varies continuously as you move from -1 to any other point.

    To take up your example: given the branch cut x=0, y≥0, and told that log(-1) = -iπ on this branch, suppose that you want to find log(1) on this branch. To get from -1 to 1 without crossing the cut, you have to go round below the origin. As you do so, the argument will increase from -π to 0, which means that you must take 0 as the logarithm of 1.

    On the other hand, if (with that same branch cut) you were told to take log(-1) = iπ, that means that you must take π as the argument of -1. As you go continuously from -1 to 1, passing below the origin, the argument again increases, this time from π to 2π. So you would have to take log(1) = 2πi.
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  3. #3
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    First, you way better at this than me ok Opalg. However, I'd like to present a different approach to this in hopes it may be useful to others.

    The key to understanding much about branch-cuts is the logarithmic multi-sheet. Get that and many of your problems with branch cuts are over. We know the log function is defined as:

    \log(z)=ln|r|+i \arg(z)

    What's the imaginary part of that function in terms of a real function of two real variables? It's v(r,\theta)=\arg(z);\;z=re^{i\theta} with  \arg(z)=\text{Arg}(z)+2k\pi. What's a plot of that function look like? How about mine below? It's a helical sheet extending both up and down and radially without bounds although I've just clipped a small (r=3) portion of three turns of it. The overlapping of the sheets cause of course the multi-values since for any (x,y) pair, each sheet has a value. I've shown three values of \text{Im}(\log(2-2i)) as the blue, green, and red dot. However if we wish to define a single-valued analytic logarithm, we . . . cut . . . the multi-sheet so that we don't have any overlapping. I've color-coded three such non-overlapping ``branches'' of the log function in the figure. Each of these branches or sheets can be identified by:

    \begin{aligned}\log_0(z)&=ln|r|+i\Theta;\quad -\pi\leq\Theta<\pi;&\quad\text{(red)}\\<br />
\log_{-1}(z)&=ln|r|+i\Theta;\quad -3\pi\leq\Theta<-\pi;&\quad\text{(blue)}\\<br />
\log_{-2}(z)&=ln|r|+i\Theta;\quad \pi\leq\Theta<3\pi;&\quad\text{(green)}<br />
\end{aligned}

    Can you see why once I pick a branch, I can then identify a single value of \log(z) for any z?
    Attached Thumbnails Attached Thumbnails branch cuts / complex logs-logsheet.jpg  
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  4. #4
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    Quote Originally Posted by Squiggles View Post
    Hi. So I'm new to branch cuts and I think I get the general idea, but I'm stuck with a couple homework questions along the same lines as:

    I'm given the branch cut, eg (x=0, y>0)
    and told for example log(-1) = -i(pi) on this branch
    Then asked to find the value of some other simple complex logs on the same branch
    Remember that to define a complex logarithm we need to pick a particular way for defining \arg z since they are many ways.

    The 'principal' logarithm is defined by \log z = \ln |z| + i \arg z where \arg z is the angle of the complex number given by -\pi < \theta \leq \pi . For example, if we want to find \log (-1) then we need to find \theta which in this case is \pi. Therefore, \log (-1) = \ln |-1| + i \pi = \pi i. It turns out that \log z is analytic everywhere except on (-\infty,0] - this is called its branch. The reason is that the \arg z function jumps values on this line, note it jumps from -\pi to \pi and vice-versa, thus it is not even continous there.

    The logarithm along the positive axis is defined by \arg z having values 0\leq \theta < 2\pi. Therefore, \log (i) = \ln |i| + i \arg (i) = \tfrac{\pi}{2}i. It is exactly the same as above except \theta is different. Compare what would have happened if we did the same computation using the principal logarithm, we would get a different result. Note that on [0,\infty) this logarithm jumps values from 0 to 2\pi. It turns out that this logarithm is analytic everywhere except on its branch.

    The other thing we need to result is that if instead of defining 0\leq \theta < 2\pi we defined 0 < \theta \leq 2\pi it would not matter since e^{i\phi} = e^{i(\phi + 2\pi)}. The point we need to realize here is that when defined a value for \arg z it does not matter if we choose (a,a+2\pi] or [a,a+2\pi) since they gives the same function for \arg z.

    When the problem says to define logarithm along the branch [0,i\infty) it is asking to define a logarithm so that \log is analytic everywhere except on the line [0,i\infty). Looking at the above two examples it should be clear. We ought to define \arg in such a way so that it jumps values from a to a+2\pi and vice-versa. If we pick a=-\tfrac{3\pi}{2} it works. Therefore we define \arg z by \tfrac{-\pi}{2}\leq \theta < \tfrac{\pi}{2}. Therefore \log (-1) = \ln |-1| + i \arg (-1) = -\pi i.
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  5. #5
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    Quote Originally Posted by shawsend View Post
    Nice picture of a Riemann surface! It might be worth pointing out explicitly that what a branch cut does is to define the line that forms the boundary between one sheet of the helix and the next (differently-coloured) one. If you want to stay on the same branch then you have to keep to one of the coloured sheets.
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  6. #6
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    Here's something else I think can be helpful to readers struggling to understand branch cuts: Review the Mathematica code below and understand why it's generating the picture I made. Now, I created the cuts along the negative real axis. Change the code so that the branch cut is now along the positive imaginary axis (re-define the three branche functions \log_0,\; \log_1,\;\log_{-1} corresponding to this change accordingly). Change the points so now the set is somewhere else. Generate a plot of the real component of \log(z). What's different about that plot? Plot both the real and imaginary components of \log(a+bi) side by side (using GraphicsGrid[{{realplot,imagplot}}]). Just some ideas if any reader is interested. It helps me understand it and perhaps will you.

    Code:
    f[z_, k_] := Log[Abs[z]] + 
           I*(Arg[z] + 2*k*Pi); 
    z1 = 2 - 2*I; 
    point1 = Graphics3D[{{PointSize[0.04], 
               Green, Point[{{Re[z1], Im[z1], 
                     Im[f[z1, 0]]}}]}, {PointSize[0.04], 
               Blue, Point[{{Re[z1], Im[z1], 
                     Im[f[z1, 1]]}}]}, {PointSize[0.04], 
               Red, Point[{{Re[z1], Im[z1], 
                     Im[f[z1, -1]]}}]}}]; 
    
    p1 = ParametricPlot3D[
         {Re[z], Im[z], Im[f[z, 0]]} /. 
           z -> r*Exp[I*t], {r, 0, 3}, 
         {t, -Pi, Pi}, PlotStyle -> Red]
    p2 = ParametricPlot3D[
         {Re[z], Im[z], Im[f[z, -1]]} /. 
           z -> r*Exp[I*t], {r, 0, 3}, 
         {t, -Pi, Pi}, PlotStyle -> Blue]
    p3 = ParametricPlot3D[
         {Re[z], Im[z], Im[f[z, 1]]} /. 
           z -> r*Exp[I*t], {r, 0, 3}, 
         {t, -Pi, Pi}, PlotStyle -> Green]
    Show[{p1, p2, p3, point1}, 
       PlotRange -> {{-3, 3}, {-3, 3}, 
           {-3*Pi, 3*Pi}}, BoxRatios -> {1, 1, 1}]
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  7. #7
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    thanks, that makes a lot more sense to me now! (well, enough to get by with my homework anyway )
    cheers
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