# Homework Help: Branch points and cuts of multivalued function

1. Nov 14, 2012

### elgen

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 $f(z)=Log(1+z^\alpha)$ where $\alpha$ is a rational number and $z\in \mathcal{C}$.

2. Relevant equations

Obviously, for the function $z^\alpha$, it has two branch points at $z=0$ and $z=\infty$. Also, for the function $Log(z)$, it has two branch points at $z=0$ and $z=\infty$ 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$f(z)=Log(1+z^\alpha)$. Let $z=r e^{j\theta}$. we have
$Log[1+r^\alpha e^{i\alpha(2\pi k+\theta)})=Log(1+r^\alpha\cos(\alpha(2\pi k+\theta))+i r^\alpha\sin(\alpha(2\pi k+\theta))]$.

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

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

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

2. Nov 14, 2012

### hedipaldi

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

3. Nov 14, 2012

### elgen

Thank you for the pointer. For the branch point at $0$, I got $r=1$ and $\theta=\frac{\pi}{a}+2\pi k$. For the branch point at $\infty$, I got $z=\infty$.

I tried $a=2/3$ 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 $Log(1+(r e^{i\theta})^a)$ and it has a branch cut starting at 0 and along the negative real axis.

The bottom one uses $Log(1+r^a e^{i\theta a})$. It gives one branch cut predicted by the calculation and another branch cut along the positive real axis.

Last edited: Nov 14, 2012
4. Nov 14, 2012

### jackmell

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

$$1+z^{m/n}=0$$

to obtain:

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

Won't that do it?

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

$$f(z)=\log(1+\sqrt{z})$$

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
5. Nov 14, 2012

### hedipaldi

the function z^α also contributes branch points

6. Nov 14, 2012

### elgen

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?

7. Nov 14, 2012

### jackmell

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:

$$f(z)=1+z^{1/2}=1+r^{1/2} e^{i\arg(z)}$$
where now the quantity $arg(z)$ 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];
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.

#### Attached Files:

• ###### log sheet.jpg
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Last edited: Nov 15, 2012
8. Nov 15, 2012

### 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.