What is the branch point structure of $f(z) = \ln\left(5 + \sqrt{\frac{z + 1}{z - 1}}\right)$?
Here is what I have so far.
Starting with analyzing $\sqrt{\frac{z + 1}{z - 1}}$, I realize that $\pm 1$ are branch points. To see this, let $z + 1 = \rho_1 e^{i \theta_1}$ and $z - 1 = \rho_2 e^{i \theta_2}$. We then get $\sqrt{\frac{z + 1}{z - 1}} = \sqrt{\frac{\rho_1}{\rho_2}} e^{i(\theta_1 - \theta_2)/2}$. Replace $\theta_1$ with $\theta_1 + 2\pi$ and we get $\sqrt{\frac{\rho_1}{\rho_2}} e^{i(\theta_1 + 2\pi - \theta_2)/2} = \sqrt{\frac{\rho_1}{\rho_2}} e^{i(\theta_1 - \theta_2)/2} e^{i \pi} = -\sqrt{\frac{\rho_1}{\rho_2}} e^{i(\theta_1 - \theta_2)/2}$, and you can do a similar trick for $\theta_2$, so $\pm 1$ must be branch points, and we will need branch cuts (maybe of the form $1 + \rho$ and $-1 - \rho$, or perhaps the interval $[-1, 1]$). Depending on how we define the branches, we may have either $\text{Re}\left(\sqrt{\frac{z + 1}{z - 1}}\right) \geq 0$ or $\text{Re}\left(\sqrt{\frac{z + 1}{z - 1}}\right) \leq 0$.
How then do I handle this as an argument of $\ln$, which also produces branch cuts? It seems that -5 is a branch point, but I'm unsure how to make that argument.
EDIT: I'm not sure from the context if I'm supposed to solve this for all choices of branches or for standard branches. In the latter case, if I had to guess, it would be that the standard branch for $\sqrt{z}$ is $(-\infty, 0]$, and for $\ln(z)$, $[0,\infty)$.