Understanding branch cuts for functions with multiple branch points 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 $(-\infty,-1)$?
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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, 
$$f(x) = \sqrt{(-1-x)e^{i (-\pi)} (1-x)e^{i (0)}} = e^{-i \pi /2} \sqrt{x^{2}-1} = -i \sqrt{x^{2}-1} .$$

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. 
