Using the principal branch $\arg(\xi(z))\in\left(-\dfrac{7\pi}{4}, \dfrac{\pi}{4}\right]$ for $$f(z)=\log(z^2-1)=\log(\xi(z)),$$ what's its branch cut's equation/how does it look like?
I was doing the previous exercise for the principal branch $\arg(\xi(z))\in\left(-\pi, \pi\right]$ (which is simpler), and I was wondering how'd the branch cut look for a principal branch different from the usual $(-\pi,\pi]$ one and the $(0,2\pi]$ one.
I guess we'd need to impose $\arg(\xi(z))=\dfrac{\pi}{4}+2\pi n$, meaning that $\mathfrak{Re}(\xi(z))\geq 0$ and $\mathfrak{Im}(\xi(z))\geq 0$: $$ \begin{aligned} &\mathfrak{Re}(\xi(z)) =x^2-y^2-1\geq 0\\ &\mathfrak{Im}(\xi(z))=2xy\geq 0. \end{aligned} $$ In brief, we'd get that either $x,y\geq 0$ or either that $x,y\leq 0$, and that $y^2\leq x^2-1$. How do we interpret this? Did I do this correctly?