Q:

ln(x) in real and ln(x) in complex.

Given 
$$e^{e^x}=1$$
Wolfram gave me the answer
$$x=\ln{(2i\pi n_1)}+2i\pi n_2\quad n_1,\,n_2 \in \Bbb N$$
doesn't that mean if i have $e^x=1$, The answer is $x=\ln(1)+2i\pi n, n\in \Bbb N$. 
But honestly, it bothers me. Why we add an extra $2i\pi n$ after we take the logarithm? Is it complex logarithm or what? Cz, yes, i know $e^x=0$ has no solutions in real plane. But i have a feeling it's a complex logarithm.
Please explain to me with easy language. Thanks in advance.

A:

The complex logarithm is a multivalued function. This is the reason of your inifnitely many solutions. More precisely, given $z=\rho e^ {i \theta} \in \mathbb{C} $, we have:
$$ \ln z:= \ln \rho + i (\theta + 2 n \pi) $$
with $n \in \mathbb{N}$. If, for instance, $z=1$, the complex logarithms are:
$$ \ln 1 = 0 +2n \pi i = 2 n \pi i $$ 

