I am revising Complex Analysis and am confused about how to approach this question.
I want to define a holomorphic branch of $f(z)=\log(z^2-1)$ on the cut plane $$\Bbb{C}\backslash\{(-\infty,-1]\cup[1,\infty)\}$$
But I don't know how to do this? Am I needing to doubly restrict the $\operatorname{Log}(z)=\log(|z|)+i\arg(z)$ funtion, so that the argument function takes values between $(-\pi,0)$ and $(0,\pi)$? I find branch cuts very confusing so any help appreciated. I also further want to show that the branch is single valued as we cross the real axis away from the cut, which I guess can only happen between $(-1,1)$.
Best Answer
$h(z) = z^2-1$ maps $D = \Bbb{C}\backslash\{(-\infty,-1]\cup[1,\infty)\}$ to $S = \Bbb{C}\backslash[0,\infty)$.
On $S$ you can define a holomorphic branch of the logarithm as $$ \operatorname{Log}(w)=\log(|w|)+i\arg(w) \text{ with } 0 < \arg(w) < 2 \pi \, . $$
The composition $f = \operatorname{Log} \circ h$ is the desired function.