Is it true that $\log(z^2-1)=\log(z-1)+\log(z+1)$ for all complex numbers?

Question

I know that Log$(z^2-1)\neq $Log$(z+1)+$Log$(z-1)$ for all complex numbers, but is it true that:

$\log(z^2-1)=\log(z-1)+\log(z+1)$, $\forall z\in\mathbb{C}-${$-1,1$}?

My ultimate goal is to separate the original function into these parts and take their different branches as to make the function analytic on my desired domain.

Answer 1

Yes, as long as you take the multivalued log, then it is true. That is, $\log(ab) = \log(a) + \log(b)$ always for the multivalued log. This is why it's often better to do computations with the multivalued log, as long as you realize that you are not dealing with one number but with an infinite set of numbers.