I'm having some trouble grasping the concept of branch cuts. I'm working on the following homework problem.
Consider $f(z)=\ln(z)$. Take the branch cut along the negative imaginary axis and determine where $f(-1)=3\pi$
I understand there are a few different choices when making branch cuts, but I'll work off the one my professor chose in his solution.
Firstly, we let $z=re^{i\theta}$ for $\frac{-3\pi}{2}<\theta < \frac{\pi}{2}$. Why do we choose this specific domain? I understand that the natural log function is discontinuous along the negative imaginary axis, but I'm having some trouble visualizing how constricting $\theta$ to this domain helps with the problem.
I think my second question will be answered with a proper explanation of my first, but I'll ask it anyway. When applying the prescribed value $f(-1)=3\pi$, we use the identity $-1=1*\exp(-\pi i)$. My question is regarding the negative value of $\pi$, why does the domain restriction in my first question give us that the angle is $-\pi$ instead of $\pi$?
Rolling with $-\pi$, we get that $$ f(-1)=\ln(1)+i(-\pi+2\pi k) \ \ \forall \ k \in \mathbb N \\ \text{Letting} \ k=2, \ f(-1)= i(-\pi+4\pi)=-3\pi i$$ Is there something I missed? Why does my answer give $f(-1)=3\pi i$ rather than $f(-1)=3\pi$? Does it have to do with the choice of branch cut?
Thank you.