I want to find the set of $z$'s such that $\{z: e^z=-1\}$. Then this just mean that I have to solve $\cos(-iz)+i\sin(-iz)=-1$ which is equivalent to having $\cos(-iz)=-1$ and $\sin(-iz)=0$ Then I find that the set of solutions is such that $-iz=\pi + 2k\pi$ or in other words, $z=(1+2k)\pi i$
Should I also consider the possibility that $\sin(-iz)=i$ and $\cos(-iz)=0$ or is it irrelevant to take this possibility into account? I am not sure. Thx.