"Given that $g(z) = \ln(1-z^2)$, defined on $\mathbb{C}\backslash \left(-\infty, 1\right]$, i.e. the branch cut is from $-\infty$ to $1$ along the real axis. Find $g(-i)$ given $g(i) = \ln(2)$"
I tried drawing it out but I'm having trouble making sense of this. I understand how $(-\infty, 0]$ works as a branch cut for $\ln(z)$. If $\ln(z) = a + bi$ we restrict $-\pi<b < \pi $ so there is a single value for $\ln(z)$. This means every time $z$ passes through $(-\infty, 0]$ there is a discontinuous jump in the value of $\ln(z)$.
However I've not been successful for explaining how $-\infty$ to $1$ works for $g(z) = \ln(1-z^2)$.
It seems like if $\left| z\right|^2 > 1$, then making the argument of z go around a circle causes the argument of $1 - z^2$ to go around a circle twice.