# Choice of branch-cuts

1. Oct 28, 2013

### dipole

I understand that at a branch-point a function is multi-valued in a neighborhood of that point and hence will be discontinuous across the branch-cut. However, the choice of branch cut is arbitrary, and I believe should have an affect on the value of the function at various points, not just at the branch cut. I'm not quite seeing it though... Can someone give me (or link) a simple example where the choice of one specific branch cut or another affects the value of the function either on the branch cut or at some point off of it?

Or, perhaps better, if you can give me an example of a function, with two different choices of a branch-cut, and a point which is off of both those branches, and whose value is affected at that point because of the difference in the choice of branches?

Also, I don't quite understand when you should include the branch point in your cut or not, and I've seen lots of examples where the branch points are or are not included... clearly I'm not quite understanding something, so any comments on that would be appreciated. :)

2. Oct 28, 2013

### jackmell

Always start with the lowly $\sqrt{z}$ to study branching geometry. But first, you need to learn how to plot a multifunction. Here's the code for plotting the full imaginary sheet and two specific single-valued determinations in Mathematica:

Code (Text):

p1 = ParametricPlot3D[{Re[z], Im[z], Im[Sqrt[z]]} /.
z -> r Exp[I t], {r, 0, 1}, {t, -\[Pi], \[Pi]}, PlotStyle -> Red];
p2 = ParametricPlot3D[{Re[z], Im[z], Im[-Sqrt[z]]} /.
z -> r Exp[I t], {r, 0, 1}, {t, -\[Pi], \[Pi]}, PlotStyle -> Blue];
point1 = Graphics3D[{Green, PointSize[0.05],
Point[{-1/2, -1/2, Im[Sqrt[-1/2 - I/2]] + 0.1}]}];
point2 = Graphics3D[{Black, PointSize[0.05],
Point[{-1/2, -1/2, Im[-Sqrt[-1/2 - I/2]] + 0.1}]}];
p3 = Show[{p1, p2, point1, point2}, PlotRange -> All]
GraphicsGrid[{{p3, p1, p2}}]
http://img6.imageshack.us/img6/3046/r6zm.jpg [Broken]

$$|r|^{1/2} e^{i/2 t},\quad -\pi<t\leq \pi,\quad \text{Red}$$
$$|r|^{1/2} e^{i/2 t},\quad \pi<t\leq 3\pi\quad \text{Blue}$$

But I could have segregated the multisheet other ways by placing the branch-cut elsewhere. I could even chop it up into a hundred pieces with 100 branch cuts but if we want to obtain the largest single-valued piece, I'd use a single branch-cut.

Now notice the two points, one on the blue sheet, the other on the red sheet. These points correspond to the two values of the imaginary part of $\sqrt{-1/2-1/2 i}$. And depending on which single-valued sheet I choose, I'll get one or the other (imaginary) value of $\sqrt{-1/2-1/2i}$ corresponding to the green or orange point. And even where the imaginary sheet cross, the value of the function is different because the real sheet would not cross there which you could verify by modifying my code to plot the real sheet of this function.

Now, using this model, see if you can answer the rest of your questions.

Last edited by a moderator: May 6, 2017
3. Oct 29, 2013

### dipole

Thanks for the post, I had a midterm so I couldn't respond sooner. I looked at the code, but since I'm not able to change the branch-cut Mathematica chooses I don't really quite understand. In your example you use $-\sqrt{z}$ and $+ \sqrt{z}$ but that's not really a branch-cut is it, since even if you just restrict yourself to $+ \sqrt{z}$ the function is still multi-valued...

If I were calculating the function $\sqrt{z}$ by hand, how would my choice of branch-cut come into play?

For example, using your point, say I have the cut $(-\infty, 0 )$ or the cut $(0, i\infty)$ both of which are fine and make the function single-valued. For $p = -1/2 - 1/2i$ you get,

$r = 1/\sqrt{2}$, $\theta = 5\pi /4$

and so,

$w = \sqrt{-1/2 - 1/2i } = 2^{-1/4}e^{(5\pi /8)i}$

How does a different choice of branch cut affect this?

Last edited: Oct 29, 2013
4. Oct 29, 2013

### Office_Shredder

Staff Emeritus
dipole, for the square root function picking a branch cut is equivalent to just picking a way of saying what the argument of your complex number is. I know that
$$\sqrt{ r e^{i\theta}} = \sqrt{r}e^{i\theta/2}$$
so all I need is a well defined way of telling you what theta is... for example, is the argument if i pi/2, 5pi/2, 9pi/2, etc.? I could make a choice of branch cut where all my arguments are between 0 and 2pi, in which case
$$\sqrt{i} = e^{i\pi/4}$$
If I choose all my arguments to be between pi and 3pi, then
$$\sqrt{i} = e^{i 5\pi/4}$$

5. Oct 29, 2013

### jackmell

http://https://www.physicsforums.com/attachment.php?attachmentid=63433&d=1383086600 [Broken]Let's just take this part first:

In brief: Mathematica will use the principal branch cut for multivalued functions but you can easily hard-code any branch cut you wish. As a matter of coding convenience, I just drew the principal branch:

$$\text{Principal branch}=|r|^{1/2}e^{i/2\theta},\quad -\pi<\theta\leq \pi$$

and the $n=1$ branch:

$$|r|^{1/2}e^{I/2(\theta+2\pi)}=-|r|^{1/2}e^{i/2\theta},\quad -\pi<\theta\leq \pi$$

and since Mathematica interprets $\sqrt{z}$ as the principal branch, then it will interpret $-\sqrt{z}$ as the $n=1$ branch.

I then just plotted them together to obtain the imaginary part of the full multivalued function.

And this is the important point:

Easy of those branches is single-valued. Obviously just by looking at the plots and noticing the blue and red sheets, taken separately do not overlap.

Now, it's easy to create any other branch. For example, I could excise a piece of the multi-function with a branch cut along the positive imaginary axis:

$$\text{mybranch}=|r|^{1/2}e^{i/2 t},\quad \pi/2<t \leq 5\pi/2$$

with the code:
Code (Text):

p1 = ParametricPlot3D[
Evaluate[{Re[z], Im[z], Im[Sqrt[z]]} /. z -> r Exp[I t]], {r, 0,
5}, {t, \[Pi]/2, \[Pi]}]
p2 = ParametricPlot3D[
Evaluate[{Re[z], Im[z], Im[-Sqrt[z]]} /. z -> r Exp[I t]], {r, 0,
5}, {t, -\[Pi], \[Pi]/2}]
Show[{p1, p2}, PlotRange -> All, BoxRatios -> {1, 1, 1}]

Now try and edit my code to place the branch cut at the negative imaginary axis and plot the results. In this case, you would want theta to go from $-\pi/2$ all the way round up to $\pi$ on the principal branch, and then go from $\pi$ to $3\pi/2$ on the n=1 branch, but since Mathematica will only plot the principal branch, we plot instead $-\sqrt{z}$ from $-\pi$ to $-\pi/2$. Ok, that's probably really confusing. Sorry.

Here's the imaginary sheet of a perfectly-fine, single-valued branch of the square root function analytic everywhere except at the 5 branch cuts. When you can explain how, then I think you'll have a good understanding of the geometry of multivalued functions and how to chop them up into various single-valued "branches". :)

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• ###### sqrt root with 5 branch cuts.jpg
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Last edited by a moderator: May 6, 2017
6. Oct 31, 2013

### dipole

Alright thanks to the both of you, that helps a lot. So basically a branch cut is really a restriction on the values of the argument of the function.

Let's say I have the function $w(z) = \sqrt{z^2 - 1}$ instead, which has branch points at $\pm 1$.

One valid branch cut is to go from $-1$ along the real axis through infinity, and then back around to $+1$ without passing through the origin. In this case, at the origin there's no issues.

So, would the restriction on $\theta$ then look something like this?

$$0 < \theta < \pi \qquad \textrm{for} \qquad r \geq 1$$

and no restriction on $\theta$ for $r < 1$?

7. Nov 1, 2013

### WWGD

Just a comment: another way of looking at the branch cut, say B is that you want $$\mathbb C -B$$ (the complement of B in $$\mathbb C$$ to not have curves that

wind around the branch points, since not winding around branch points avoids multi-valuedness.

If you work with a branch $log$, you can use that $\sqrt {z^2-1}=e^{(1/2)log(z^2-1)}$

and then you want to avoid having the values of $z^2-1$ fall in the branch cuts; e.g., if you use

the standard branch $$-\pi< \theta \leq \pi$$ , then you want to avoid landing on the negative Real axis.

8. Nov 2, 2013

### jackmell

I do not feel that is adequate enough to completely describe the branch you're defining, that is, one with branch cuts $(-\infty,-1]$ and $[1,\infty)$.

Rather, the function has two such branches, one of which could be defined as:

$$\text{mybranch(z)}=\begin{cases} r^{1/2}e^{i/2(\text{Arg}(f))}, && 0<\theta\leq \pi/2 \\ r^{1/2}e^{i/2(\text{Arg}(f)+2\pi)},&& \pi/2<\theta\leq \pi\\ r^{1/2}e^{i/2(\text{Arg}(f))}, && \pi<\theta\leq 3\pi/2\\ r^{1/2}e^{i/2(\text{Arg}(f)+2\pi)}, && 3\pi/2<\theta\leq 2\pi \end{cases}$$

with $f=z^2-1$ and $-\pi<\text{Arg}(u)\leq \pi$.

The real part of this branch is shown below along with a black contour reflecting $\text{mybranch}(z)$ with $z=2e^{it}$ as t goes from 0 to $2\pi$.

#### Attached Files:

• ###### mybranch.jpg
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Views:
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Last edited: Nov 2, 2013