I previously solved the equation $e^{e^z}= 1$. The result I get is
$$ z = \ln( 2 \pi ) + \ln(i) + \ln(k) = \ln(2 \pi) + {\pi \over 2}i + \ln(k)$$
where $k \neq 0$.
I was told that I needed to make a case distinction for $k < 0$ and $k > 0$ but it is not clear to me why.
If $k$ is negative then $\ln(k) = \ln(-1) + \ln(|k|) = i\pi + \ln(|k|)$. But the $i \pi$ is already contained in $\ln(k)$ if $k$ is negative.
Why does one need to make a case distinction in the solution of this equation?