Branch cut of the principle value log

[bmxicle]

Homework Statement

Find the branch points of g(z) = log(z(z+1)/(z-1)) and defining a branch of g as the principle branch of the logarithm find the location of the branch cuts.

Homework Equations

The Attempt at a Solution

Since $$g(z) = log(z) + log(z+1) - log(z-1)$$ the branch points are 0, 1, -1 and infinity.

To find the branch cuts we need to find where the argument is $$arg(z) \in (-\pi,\pi]$$

putting each branch point into polar coordinates gives:
$$z1 = r_1e^{i\theta_1}$$
$$z2 = r_2e^{i\theta_2} -1$$
$$z3 = r_3e^{i\theta_3} +1$$
$$z = ln(\left|{\dfrac{r_1 r_2}{r_3}}\right|) +i(\theta_1 + \theta_2- \theta_3)$$

let $$c = ln(\left|{\dfrac{r_1 r_2}{r_3}}\right|)$$ and then:

$$(-\infty, -1) \Rightarrow g(z) = c + \pi i$$
$$(-1, 0) \Rightarrow g(z) = c + 2\pi i$$
$$(0, 1) \Rightarrow g(z) = c + \pi i$$
$$(1, \infty) \Rightarrow g(z) = c$$

So this gives $$(0, 1)$$ as the point where the branch cut isn't defined, but the solution I was given says that $$(1, \infty)$$ is also not defined, so I'm not sure where I'm going wrong.

 

[mighty2000]

One possible explanation could be that the branch cut for the branch point at infinity is actually defined as the entire complex plane, since the argument of infinity is not well-defined. Therefore, the branch cut would be the entire complex plane minus the point (0,1). This would make sense since the function g(z) would have a singularity at infinity, so the branch cut would have to include that point.