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 $|e^z| = 2$

$e^z$ and $-1$ are two complex numbers, which are equals iff their real and imaginary part are. Now
$$
e^z=e^xe^{iy}=e^x\cos y+ie^x\sin y
$$
This is equal to $-1$ iff $\sin y=0$ (i.e. $y=k\pi$) and $-1=e^x\cos k\pi=e^x(-1)^k$  so we take $x=0$ and odd $k$'s.
We finally conclude that $z=x+iy=(2k+1)\pi i$.
