Solving $e^z=\pm1$ in $\bf C$

Question

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 $|e^z| = 2$

Answer 1

$e^z$ 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 $z=x+iy=(2k+1)\pi i$.

Answer 2

Avoid logs.

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