1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Branch points and cuts of multivalued function

  1. Nov 14, 2012 #1
    1. The problem statement, all variables and given/known data

    This problem is on how to identify the branch points and the branch cuts of a multivalued function. Consider the following function [itex]f(z)=Log(1+z^\alpha)[/itex] where [itex]\alpha[/itex] is a rational number and [itex]z\in \mathcal{C}[/itex].

    2. Relevant equations

    Obviously, for the function [itex]z^\alpha[/itex], it has two branch points at [itex]z=0[/itex] and [itex]z=\infty[/itex]. Also, for the function [itex]Log(z)[/itex], it has two branch points at [itex]z=0[/itex] and [itex]z=\infty[/itex] and the negative real axis is often selected to be the branch cut.

    3. The attempt at a solution

    My difficulty lies in identifying the branch points of the cascaded function[itex]f(z)=Log(1+z^\alpha)[/itex]. Let [itex]z=r e^{j\theta}[/itex]. we have
    [itex]Log[1+r^\alpha e^{i\alpha(2\pi k+\theta)})=Log(1+r^\alpha\cos(\alpha(2\pi k+\theta))+i r^\alpha\sin(\alpha(2\pi k+\theta))][/itex].

    If I let
    [itex] 1+r^\alpha\cos(\alpha(2\pi k+\theta)) < 0[/itex] and
    [itex]r^\alpha\sin(\alpha(2\pi k+\theta)) =0 [/itex]. I have [itex]\theta=\frac{m\pi}{\alpha}-2\pi k[/itex] and [itex]r^\alpha\cos(m\pi)<-1[/itex].

    I get [itex]m[/itex] must odd, and [itex] r>1 [/itex]. This seems to suggest that a branch cut starting at [itex]r>1[/itex] and an angle [itex]\theta=\frac{m\pi}{\alpha}-2\pi k[/itex].

    When I plot the function in Mathematica, the branch cut starts at z=0 and extends along the negative real axis. This is in contradiction with my prediction.

    How could I identify the branch cut and branch points of this function?

    Thank you.

  2. jcsd
  3. Nov 14, 2012 #2
    The branch points of Log(z) are 0 and ∞ so find the points where 1+z^α=0 or ∞.
  4. Nov 14, 2012 #3
    Thank you for the pointer. For the branch point at [itex]0[/itex], I got [itex]r=1[/itex] and [itex]\theta=\frac{\pi}{a}+2\pi k[/itex]. For the branch point at [itex]\infty[/itex], I got [itex]z=\infty[/itex].

    I tried [itex]a=2/3[/itex] and used Mathematica to plot the riemann sheet. Interestingly, the following two commands give different branch cuts.


    The top one uses [itex]Log(1+(r e^{i\theta})^a)[/itex] and it has a branch cut starting at 0 and along the negative real axis.

    The bottom one uses [itex]Log(1+r^a e^{i\theta a})[/itex]. It gives one branch cut predicted by the calculation and another branch cut along the positive real axis.
    Last edited: Nov 14, 2012
  5. Nov 14, 2012 #4
    For the non-zero finite branch points, why not just solve:


    to obtain:

    [tex]z=(-1)^{n/m}=\large{e^{\frac{ni}{m}(\pi+2k\pi)},\quad k=0,1,2,\cdots,m-1}[/tex]

    Won't that do it?

    Also, Mathematica by default is only plotting the principal branch. Take for example, the function:


    which has a finite branch-point at 0 and 1. However, the branch point at z=1 is not part of the principal branch and so Mathematica only shows the branch-cut eminating from z=0 along the negative real axis (the default branch cut for the principal-valued log function) when you just do a standard plot of the function.
    Last edited: Nov 14, 2012
  6. Nov 14, 2012 #5
    the function z^α also contributes branch points
  7. Nov 14, 2012 #6
    Thank you for the assistance.

    That would equally do.

    I think my confusion is more at the conceptual level. when we have a function y=f(z), we have "z" associated with the domain, and y associated with a co-domain (range).

    a) is the branch point a point in the domain or the co-domain?

    b) Why do you say that there are two finite branch point at 0 and 1? How do you know that the branch point at z=1 is not part of the principle branch?
  8. Nov 14, 2012 #7
    Branch-points are part of the domain. I meant there is a branch-point at zero and a branch-point at 1. Just wanted to distinguish those from the branch-point at infinity. Take the quantity:

    [tex]f(z)=1+z^{1/2}=1+r^{1/2} e^{i\arg(z)}[/tex]
    where now the quantity [itex]arg(z)[/itex] is the analytically-continuous version of the argument function and let the radius be greater than one and start at z=2 and go around in the positive sense. Plot the arg(f(z)) as z goes around a 2pi rotation using Manipulate in Mathematica:

    Code (Text):

    f[z_] := 1 + 2 Exp[I t/2];
     ParametricPlot[{Re[f[z]], Im[f[z]]} /. z -> 2 Exp[I t], {t, 0, tval},
       PlotRange -> {{-3, 3}, {-3, 3}}], {tval, 0.01, 2 \[Pi]}]
    The arg(f(z)) only reaches the principal branch cut of pi when z has made a complete revolution and so does not encounter the branch-cut eminating from z=1 on this first revolution but would encounter it on the second revolution.

    That's probably confussing. Sorry. Here's a plot of Im(1+z^{1/2}). The red surface is the principal branch that you get with the standard Mathematica plot function.

    Attached Files:

    Last edited: Nov 15, 2012
  9. Nov 15, 2012 #8
    I think I am convinced that although each branch has a domain defined over the complex plane, the domain can have a different branch cut for each branch.

    Thank you for the helpful discussion.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook