Say I have to solve the following equation $$ \operatorname{Log}(z) = -2+3i $$ where this is the principal value of the logarithm. I can think of two ways to solve it; first is just raising both sides from $e$ to get $$ z = e^{-2+3i} \stackrel{?}{=} e^{-2+3i+2i\pi k} $$ Where I'm not sure if the second equality is necessary. Second is noting $$ \operatorname{Log}(z) = \operatorname{Log}|z|+i\operatorname{Arg}(z) = -2+3i $$ where $\operatorname{Arg}(z)$ is the principl; branch. This means for polar coordinates $r = e^{-2}$ and $\theta = 3$ and the solution $e^{-2+3i}$ is unique. (mini-question: would this mean if my imaginary part was 5 there would be no solution because this is outside $(-\pi,\pi]$?)
Of course, if this was $\log(z)$ then I would have $\arg(z) = 3+2\pi k$ and there would be multiple solutions. I think the unique solution is correct, but at the same time is $e^{-2+3i+2i\pi}$ not also in the domain of $\operatorname{Log}(z)$? I thought $\operatorname{Log}(z)$ had a domain of everything except $(-\infty,0]$.
There is no difference between your approaches. Remember that the exponential function is periodic with period $2\pi i$. Therefore, $e^{-2+3i+2i\pi k}$ is always the same number, no matter the choice of $k\in\mathbb Z$.