Solving $e^z=\pm1$ in $\bf C$ given $e^z = -1$
we have: $e^z = 1*e^{i\pi}*e^{2ki\pi}$
and taking $ln$ both sides yields: $z = ln(1)+i\pi+2ki\pi$
and if given $e^z = 1$
we have : $e^z = 1*e^{2ki\pi}$ and taking $ln$ both sides yields: $z = ln(1)+2ki\pi$
is this analogy correct?, thanks for any help that you can offer
i referenced the following post but just need clearity on this (finals have me shook up)
[Can someone please explain the following definition of $\ln(e^z)$
and 
[solving for z in $|e^z| = 2$

Avoid logs.
Note that $e^z=\pm1$ iff $e^{2z}=1$ iff $2z = 2k\pi i$ iff $z=k\pi i$.
