How to find branch cut of $\ln(\sin(z))$? I know that $\ln(z)$ is undefined when $z=0$ or $z=\inf$ , so I manged to find the critical points to be $z=\\{k*\pi,\inf*i, -\inf*i\\}$ , for every integer k, but I don't know how to proceed from here to find the branch cuts.

First: $$\sin(x+iy)= \sin(x)\cosh(y)+i\cos(x)\sinh(y)$$ Now it depends how you choose the branch cut of the logarithm, I'll pick it along the negative real axis. To find points on the branch cut, we must thus have: $$ \sin(x)\cosh(y)<0\\\ \cos(x)\sinh(y)=0 $$ The first condition gives $\sin(x)<0$. The second one gives either $y=0$ or $\cos(x)=0$. We have two cases:

1. $\sin(x)<0,\cos(x)=0$ implies $x=-\pi/2+2k\pi$. Therefore, on all verical lines $\Re(z)=-\pi/2+2k\pi$, we have a branch cut.
2. $\sin(x)<0,y=0$ defines intervals on the real axis of the form $((2k-1)\pi,2k\pi)$, giving the second branch cut condition: $z\in ((2k-1)\pi,2k\pi)$.



The whole set is then: $$ B=\\{z\in\mathbb{C}:\exists k\in\mathbb{Z}.(\Re(z)=-\pi/2+2k\pi\,\,\lor\,\,z\in ((2k-1)\pi,2k\pi))\\} $$