Consider $g(z) = \ln(1-z^2)$ defined on $C$ \ $(-\infty, 1]$. Find $g(-i)$ given that $g(i) = \ln(2)$.
I've begun by stating $\ln(1-z^2) = \ln|1-z^2| + i*arg(1-z^2)$.
The real part at $g(-i) = \ln(2)$
To evaluate the argument at $g(-i)$:
$arg(1-z^2) = arg(1-z^2) + \Delta arg(1-z^2)$
where $\Delta arg$ is the change in argument as we travel from $i$ to $-i$ around the branch cut.
$arg(1-z^2) = 0$ because it's only real at $-i$.
I split $\Delta arg(1-z^2) = \Delta arg(1-z) + \Delta arg(1+z)$
I obtained $-2\pi$ after evaluating the above.
So $g(-i) = \ln(2) - 2\pi i$
I cannot obtain the same solution trying to evaluate just $\Delta arg(1-z^2)$. Is it because I do not have the correct solution above? I find that the change in argument is 0. Am I missing something important?
Thanks for any help.