Understanding branch cuts for functions with multiple branch points

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.