# functions with multiple branch points

I'm having some difficulties with branch points.

For example, $\displaystyle \sqrt{1-z^{2}}= \sqrt{1+z} \sqrt{1-z}$.

Say you chose two different branches of the logarithm so that $- \pi \le \arg(1+z) < \pi$ and $0 \le \arg(1-z) < 2 \pi$. I understand why there would be a cut long $(-\infty,-1]$ but why would there be a cut along $(-\infty,1]$ and not $[1,\infty)$? It seems backwards.

And I'm having a bit of trouble understanding why these choices result in $\sqrt{1-z^{2}}$ being continuous along $(-\infty,1)$ and the cut only being on $[-1,1]$.

EDIT:

Starting on page 142 in the solutions manual to Gamelin's textbook, there are multiple diagrams of the cut for $\log_{0} (1-z)$ being on $(-\infty,1]$

http://www.sb.hik.se/personal/sanno/dokument/Chapter_7.pdf

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This thread may help you as well as the link there. –  Raymond Manzoni Mar 11 '13 at 11:05
You're (almost) right. I can't compute. Your formula isn't correct either though since $\arg_0 \in (0,2\pi)$. Your rhs is in $(\pi,3\pi)$. I'll draw you a new picture and fix it, but I don't have time to do it right now. If nobody else answers, I'll return later. –  mrf Mar 11 '13 at 11:27
How do we adjust $\arg_{0}(1-z) = \pi + \arg_{0}(z-1)$ so that it's technically correct? But I think I understand now why I keep seeing the cut going in the seemingly wrong direction. –  Random Variable Mar 11 '13 at 15:42
So if we approach the cut to the right of $-1$ from above it appears that the the argument of $\sqrt{1-z^{2}}$ is $\frac{3 i \pi}{2} = -i$ And if approach from the below the argument is $-\frac{\pi}{2} = -i$. So we actually do seem to have continuity along $(\infty,-1)$. Am I making making sense, or talking nonsense? –  Random Variable Mar 11 '13 at 15:58
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## 3 Answers

There are no "natural" or "canonical" "cuts" associated to a "multivalued function", despite much tradition that may accidentally give this impression. Rather, what we are doing in telling a "cut" is restricting the function(s) to a slightly smaller domain on which there are well-defined holomorphic (and continuous) "single-valued" versions of the function.

For explicit functions such as $\log(z)$ and $\sqrt{1-z^2}$, we understand the "bad points" (in these cases $0,\infty$ and $\pm 1$, respectively). In the case of $\log$ we have to prevent ourselves making a loop around $0$ that returns to the same point. In the case of $\sqrt{1-z^2}$, there are perhaps-surprising options. Namely, following a loop around either $\pm 1$ flips the sign of the square root. This is "bad". How to prevent this? Well, we can prevent all loops around both $\pm 1$ by cuts $(-\infty,-1]$ and $[1,\infty)$. Or cuts vertically to $+i\infty$ from both $\pm 1$, for that matter.

Or we can simultaneously prevent such sign-flips by slitting along some (non-self-intersecting) path from $-1$ to $+1$, for example, $[-1,+1]$. Or an arc of the unit circle going from $-1$ to $+1$. Edit: to be clearer! ... the point is that if we constrain ourselves to go around either none, or both the bad points, the net sign flip is ... null.

(Edited...) A sillier example, to amplify the point, is about $\sqrt{(1-z^2)(4-z^2)}$. Here there are 4 bad points, $\pm 1,\pm 2$. Any combination of slits connecting one of these to another... such as $1$-to-$2$ and $-1$-to-$-2$, or $1$-to-$-2$ and $-1$-to-$2$, causes the sign-flips created by travelling loops to be "doubled", that is, no net sign flip, thus, a well-defined function.

The issue is not about supposed relations between args, really, but, rather, to kill off enough of the homology of the domain so that under the Monodromy Map all that's left maps to $\{1\}$ in the permutation of function elements under analytic continuation...

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Maybe I'm taking nonsense. But if you want a single cut on $[-1,1]$ you can't just will it there. You need to choose branches for $\ln(1+z)$ and $\ln(1-z)$ so that together you actually have continuity along $(-\infty,-1)$ and $(1,\infty)$. And that is brought about by the above branch choices as I attempted to point out. –  Random Variable Mar 14 '13 at 0:41
Ah, perhaps this is the key trick: one doesn't need continuity of both pieces individually, but only of the aggregate. That is, a product of two functions neither individually well-defined, may (fairly amazingly!) be well-defined, etc. –  paul garrett Mar 14 '13 at 0:53
But if asked to choose the branch of $\ln(1-z)$ so that the cut is on $(-\infty,1]$ which branch would you choose? The fact that it's $\ln(1-z)$ and not $\ln(z-1)$ makes it a bit tricky. That's why I was trying to come up with a relationship between arguments. –  Random Variable Mar 14 '13 at 1:24
I understand that it could be perceived as such, but, in fact, there is no "heavy lifting" required: for either $\log(1-z)$ or $\log(z-1)$, pick a non-self-intersecting path connecting 1 to $\infty$ (the two bad points), and choose a value at any other point, and, by analytic continuation, you have a well-defined "branch" of log-of-whatever on what's left after removing the branch. (School-math scenarios may explicitly demand far more, but all that is illusory, no matter what books or instructors say.) –  paul garrett Mar 14 '13 at 1:52
It is exactly, visibly true, just by rotating and translating. Perhaps the way the question was posed made it seem to be asking a larger question. Sorry for misunderstanding. –  paul garrett Mar 14 '13 at 15:59
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I would recommend chapter 2.3 in Ablowitz, but I can try to explain in short.

Let

$$w : = (z^2-1)^{1/2} = [(z+1)(z-1)]^{1/2}.$$

Now, we can write

$$z-1 = r_1\,\exp(i\theta_1)$$ and similarly for $$z+1 = r_2\,\exp(i\theta_2)$$ so that

$$w = \sqrt{r_1\,r_2}\,\exp(i(\theta_1+\theta_2)/2).$$ Notice that since $r_1$ and $r_2$ are $>0$ the square root sign is the old familiar one from real analysis, so just forget about it for now.

Now let us define $$\Theta:=\frac{\theta_1+\theta_2}{2}$$ so that $w$ can be written as

$$w = \sqrt{r_1r_2} \exp(\mathrm{i}\Theta).$$

Now depending on how we choose the $\theta$'s we get different branch cuts for $w$, for instance, suppose we choose both $$\theta_i \in [0,2\pi),$$ then if you draw a phase diagram of $w$ i.e. check the values of $\Theta$ in different regions of the plane you will see that there is a branch cut between $[-1,1].$ This is because just larger than $1$ and above the real line both $\theta$s are $0$ hence $\Theta = 0$, while just below both are $2\pi$ hence $\Theta = 4\pi/2=2\pi$ which implies that $w$ is continuous across this line (since $e^{i2\pi} = e^{i\cdot 0}$). Similarly below $-1$ same analysis shows that $w$ is continuous across $x<-1$.

Now for the part $[-1,1]$, you will notice that just above this line $\theta_1 = \pi$ while $\theta_2 = 0$ so that $\Theta = \pi/2$ hence $$w = i\,\sqrt{r_1r_2}.$$

Just below we still have $\theta_1 = \pi$ but $\theta_2 = 2\pi$ so that $\Theta = 3\pi/2 (= -\pi/2)$ hence $w = -i\,\sqrt{r_1r_2}$ is discontinuous across this line. Hope that helped some.

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I think what you actually looked at was $(z^{2}-1)^{\frac{1}{2}}$ and not $(1-z^{2})^{\frac{1}{2}}$. But correct me if I'm wrong. –  Random Variable Mar 11 '13 at 15:17
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Well, here's my potentially flawed understanding.

$\arg_{0}(1-z) = \arg_{0} \Big((-1)(z-1)\Big) = \arg_{0}(-1) + \arg_{0}(z-1) = \pi + \arg_{0}(z-1)$

But as mrf pointed out this is not quite right.

So I choose a different branch on the right side of the equation to make it "right."

$\arg_{0}(1-z) = \pi + \arg_{\pi}(z-1)$

So the cut is where the cut for $\arg_{\pi}(z-1)$ is located (i.e., on ($-\infty,1])$. Approaching the cut from above, $\arg_{0}(1-z)$ approaches $2 \pi$. While approaching the cut from below, $\arg_{0}(1-z)$ approaches $0$.

Combining both branch choices we have continuity along $(-\infty,1)$ since when approaching the negative real axis to the left of $-1$ from above, the argument of $\sqrt{1-z^{2}}$ approaches $e^{\frac{i \pi}{2}} e^{\frac{2 i \pi}{2}} = e^{\frac{3 i \pi}{2}} = i$, while when approaching from below, the argument approaches $e^{-\frac{i \pi}{2}} e^{\frac{0 i \pi}{2}} = e^{-\frac{i \pi}{2}} = i.$

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