Branch points and cuts of multivalued function

[elgen]

Homework Statement

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].

Homework 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.

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.Elgen

 

[hedipaldi]

The branch points of Log(z) are 0 and ∞ so find the points where 1+z^α=0 or ∞.

 

[elgen]

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.

http://www.pixhost.org/show/2790/14843468_screenshot.png

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.

 

[jackmell]

For the non-zero finite branch points, why not just solve:

[tex]1+z^{m/n}=0[/tex]

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:

[tex]f(z)=\log(1+\sqrt{z})[/tex]

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.

 

[hedipaldi]

the function z^α also contributes branch points

 

[elgen]

Thank you for the assistance.

For the non-zero finite branch points, why not just solve:

[tex]1+z^{m/n}=0[/tex]

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?

That would equally do.

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

[tex]f(z)=\log(1+\sqrt{z})[/tex]

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.

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?

 

[jackmell]

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?

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:

f[z_] := 1 + 2 Exp[I t/2];
Manipulate[
 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.

 

[elgen]

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.