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My question was poorly worded and thus confusing. I'm going to edit it to make it clearer, and then I'm going to give a brief answer.

Take, for example, the function $$f(z) = \sqrt{1-z^{2}}= \sqrt{(1+z)(1-z)} = \sqrt{|1+z|e^{i \arg(1+z)} |1-z|e^{i \arg(1-z)}}.$$

If we restrict $\arg(1+z)$ to $-\pi < \arg(1+z) \le \pi$, then the half-line $[-\infty,-1]$ needs to be omitted.

But if we restrict $\arg(1-z)$ to $0 < \arg(1-z) \le 2 \pi$, why does the half-line $(-\infty, 1]$ need to be omitted and not the half-line $[1, \infty)$?

And if we define $f(z)$ in such a way, how do we show that $f(z)$ is continuous across $(-\infty,-1)$?

$ $

The answer to the first question is $(1-z)$ is real and positive for $z \in (-\infty,1)$.

And with regard to the second question, to the left of $z=x=-1$ and just above the real axis,

$$f(x) = \sqrt{(-1-x)e^{i (\pi)} (1-x)e^{i (2 \pi)}} = e^{3 \pi i /2} \sqrt{x^{2}-1} = -i \sqrt{x^{2}-1} .$$

While to the left of $z=x=-1$ and just below the real axis,

$$f(x) = \sqrt{(-1-x)e^{i (-\pi)} (1-x)e^{i (0)}} = e^{-i \pi /2} \sqrt{x^{2}-1} = -i \sqrt{x^{2}-1} .$$

Random Variable
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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
  • Be careful. Take the example $f(z)=(z+1)^{a}(z-1)^{b}$, where $a$, $b$ and $a+b$ are not integers, then it is inadequate to cut the plane by the segment $[-1,1]$. On any counterclockwise path enclosing both branch points, we have $(z+1)^{a}(z-1)^{b}\to (z+1)^{a}(z-1)^{b}e^{i2\pi(a+b)}$. As $a + b$ is not an integer, then $e^{i2\pi(a+b)}\ne 1$ and the function is multivalued. We must extend branch cuts from both $0$ and $1$ to infinity, but may not simply connect $-1$ and $1$. For $a=b=1/2$, $a+b=1$ is an integer so we may use $[-1,1]$. – Mark Viola Aug 10 '15 at 04:28
  • @Dr.MV I was unaware of that fact when I posted this question. But I did become aware of it much later. – Random Variable Aug 10 '15 at 15:35
  • @randomvariable Sure. It just seemed that at least one of the answers seemed to indicate that provided both (or neither) branch points are enclosed, everything is cool. But as you know, it might not be. – Mark Viola Aug 10 '15 at 16:30

2 Answers2

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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...

paul garrett
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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 00:41
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    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 00:53
  • 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 01:24
  • 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 01:52
  • There is more than one choice for the branch so that the cut ends up being on $(-\infty,1]$? Just to be clear, by $(-\infty,1]$ I'm talking about along the real axis. – Random Variable Mar 14 '13 at 02:02
  • If you're talking exactly about $\log(z-1)$ or $\log(1-z)$, yes, one could "cut" the plane along the real axis from $1$ to $-\infty$. Or any other non-self-intersecting curve connecting those two points. – paul garrett Mar 14 '13 at 02:38
  • But if I specifically want the cut for $\ln(1−z)$ to be on $(-\infty,1]$, I can't choose the branch $-\pi \le \arg(1−z) < \pi.$ That choice will result in the cut being on $[1,\infty)$. And that was my question. Why? – Random Variable Mar 14 '13 at 03:17
  • $1-z$ avoids $(-\infty,1]$ if and only if $z$ avoids $[0,+\infty)$... ? – paul garrett Mar 14 '13 at 12:29
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    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
  • @paulgarrett I might have misunderstood your post, but if we take the example $f(z)=z^{a}(z-1)^{b}$, where $a$, $b$ and $a+b$ are not integers, then it is inadequate to cut the plane by the segment $[0,1]$. On any counterclockwise path enclosing both branch points, we have $z^{a}(z-1)^{b}\to z^{a}(z-1)^{a}e^{i2\pi(a+b)}$. Inasmuch as $a + b$ is not an integer, then $e^{i2\pi(a+b)}\ne 1$ and the function is multivalued. We must extend the branch cuts from both $0$ and $1$ to infinity, but we may not simply connect $0$ and $1$. – Mark Viola Aug 10 '15 at 04:31
  • @Dr.MV, indeed, I did not mean to suggest that in the more general case you mention that there were (in general) more options. But/and a special case of what you mention does illustrate the point: when neither $a$ nor $b$ are integers, but $a+b$ is, then $z^a(1-z)^b$ is well-defined off $[-1,1]$. Really, the issue is the Monodromy homomorphism. – paul garrett Aug 10 '15 at 12:22
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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.

The Noob
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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