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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.

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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,

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

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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…

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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