How many options are there for branch cuts for $f(z) = \log(z-1) - \log(z+1)$? In this function: 
$$f(z) = \log(z-1) - \log(z+1)$$
it seems to me like there is only one option for the branch cut for this problem, between $z=1$ and $z=-1$ but I’m seeing other resources that indicate that there are several options.
On page 13 of this resource it indicates that there is more than one option for the branch cut.  
Here’s my logic:
-   since $z=1$ and $z=-1$ are the only two branch points 
-   and since we must draw branch cuts such that it is impossible to draw a circle around any one branch point without crossing the branch cut
-   and since $z=\infty$ is not a branch point
-   then it seems to me like the line connecting $z=1$ and $z=-1$ is the only option for the branch cut.
But this resource indicates that there are these two options for branch cuts:
$f(z)=\log(z+1)-\log(z-1)$">
But I don’t understand why the rightmost option is even an option.  There’s no branch point at $z=\infty$
 A: In the extended complex numbers, $\mathbb{C}\,\cup\,\{\infty\}$, the point at infinity is the stereographic projection from the point $(0,0,1)$ on the Riemann Sphere onto the complex plane $\mathbb{C}$.  
Hence, any curve $C_1$ that begins at $z_1\in\mathbb{C}$ and ends at the point at $\infty$ (i.e., at $(0,0,1)$ on the sphere) joins at the point at $\infty$ ($(0,0,1)$ on the sphere) with any curve $C_2$ that begins at $z_2\in \mathbb{C}$ and ends at infinity.
As just one example, for $z_1=-1$ and $z_2=1$, choose the curve $C_1$ to be the set of points $\{z|\text{Im}(z)=0, \text{Re}(z)\le -1\}$ and choose the curve $C_2$ to be the set of points $\{z|\text{Im}(z)=0, \text{Re}(z)\ge 1\}$.  These curves meet at the point at infinity (i.e., $(0,0,1)$ on the sphere).
That said, we could choose to cut the plane with the any curve that begins at $z=-1$ and ends at $z=1$.  This curve need not be the straight line adjoining the branch points.  Or we could have any curves, not only linear paths on the real axis), that begin at $\pm 1$ and terminate at the point at $\infty$. 
