How to solve $e^x = 2$ I know that $\ln(x)$ is the inverse of the exponential function $a^x$.
So I thought that
$$
e^x=2 \Leftrightarrow x = \ln(2)
$$
but my calculator says $x = \ln(2) + 2 i \pi n$, where $N \in \mathbb{Z}$. What have $e^x$ and $\ln(x)$ to do with the unit circle?

In Complex Analysis one defines for $z=a+bi\in\mathbb{C}$
$$
e^{z}:=e^{a}(\cos(b)+i\sin(b))
$$
Then you can prove that for any $z_{1},z_{2}\in\mathbb{C}$
$$
e^{z_{1}}\cdot e^{z_{2}}=e^{z_{1}+z_{2}}
$$
Note that by this definition 
$$
e^{2\pi in}=1
$$
for all $n\in\mathbb{N}$.
Hence 
$$
e^{\log(2)+2\pi in}=e^{\log(2)}\cdot e^{2\pi in}=2\cdot1=2
$$
