Branch cut of $\sqrt{z^2-1}$. I was reading something that defined the function $f(z)=\sqrt{z^2-1}$ on $\mathbb{C}\setminus [-1,1]$ where the branch cut is such that the argument of $z$ and $\sqrt{z^2-1}$ are in the same quadrant. I think I understand what this means and I think it corresponds to the usual branch of the square root.
Later, they say that $\sqrt{z^2-1}\leq 0$ for $z< -1$. I don't understand why this should be true. I have tried taking limits from above and below the imaginary axis but confused myself.
This is the way I'm understanding it: taking a point in the second quadrant slightly above the real axis, we can write 
$z=re^{i(\pi-\epsilon)}$ for $r>1$. Then $z^2=r^2e^{i2\pi-2\epsilon}$, i.e a complex number with argument almost $2\pi$. When you subtract one, you decrease the argument but for $\epsilon$ small it should still be nearly $2\pi$. There are two complex numbers which square to this one, one is just above the negative real axis, the other is just below the positive real axis.
To have a continuous function, we must choose the first. 

Let $f(z)=\sqrt{z^2-1}$ for $z\in \mathbb{C}\setminus[-1,1]$, with the branch cut on  $[-1,1]$ such that $\arg(z)$ and $\arg(\sqrt{z^2-1})$ are in the same quadrant.
Branch points of $f(z)$ are at $z=-1$ and $z=1$.  Corresponding branch cuts are contours that begin at $z=-1$ and $z=1$ and end at the point at infinity.


Example branch cuts include rays on the real axis from $(i)$ $z=-1$ to $z=-\infty$ and $z=1$ to $-\infty$, $(ii)$ $z=-1$ to $z=-\infty$ and $z=1$ to $\infty$, and (iii) $(i)$ $z=-1$ to $z=\infty$ and $z=1$ to $\infty$.
But the branch cuts need not be straight line paths.  For example, we could choose the branch cut from $z=1$ to be hyperbolic path $\text{Im}(z)=\frac1{\text{Re}(z)}-1$ from $z=1$ to $z=i\infty$ in the first quadrant.


In terms of set equivalence (See this answer and this one for references), we can write for any value of $f(z)$ as
$$\sqrt{z^2-1}=\sqrt{z-1}\sqrt{z+1}$$
for some value of $\sqrt{z-1}$ and some value of $\sqrt{z+1}$.  We choose, therefore, to cut the plane from $-1$ to $\infty$ and from $1$ to $\infty$, both along the real axis, so that
$$\begin{align}
\sqrt{z^2-1}&=\sqrt{|z+1|}e^{i\arg(z+1)/2}\sqrt{|z-1|}e^{i\arg(z-1)/2}\\\\
&=\sqrt{|z^2-1|}e^{i(\arg(z+1)+\arg(z-1))/2}
\end{align}$$
where $0<\arg(z+1)\le 2\pi$ and $0<\arg(z-1)\le 2\pi$.  Then, $0<\arg(\sqrt{z^2-1})\le 2\pi$,
Note with these choices of branches for $\sqrt{z+1}$ and $\sqrt{z-1}$, we satisfy the requirement that $\arg(z)$ and $f(z)=\arg(\sqrt{z^2-1})$ are in the same quadrant.
Moreover, along the real axis for which $\text{Re}(z)>1$, $f(z)$ is continuous.  Hence, we have now defined a function $f(z)$ that is single-valued on $\mathbb{C}\setminus[-1,1]$ and $\arg(z)$ and $\arg(f(z))$ are in the same quadrant.

Finally, note that the $\text{Re}(z)<-1$, we have $\arg(z+1)=\arg(z-1)=\pi$, $\arg(f(z))=\pi$, and $\sqrt{z^2-1}=-\sqrt{|z^2-1|}$.
