If we want to find, say, the inverse $\tan$ function, $\tan^{-1}$, in terms of (complex) logarithm function we start with the equation $z=\tan w =\frac{\sin w}{\cos w}=\frac{1}{i}\frac{e^{iw}-e^{-iw}}{e^{iw}+e^{-iw}}$, and solve it for $w$, which gives $e^{2iw}=\frac{i-z}{i+z}$.
I have a doubt exactly at this stage: In the last equation if we take the logarithm of both sides, then we get $\log (e^{2iw})=\log\left(\frac{i-z}{i+z}\right)$. From this point on I saw in many books saying that
$$2iw=\log\left(\frac{i-z}{i+z}\right)\tag1$$
(and so that $w=\frac {1}{2i}\log\left(\frac{i-z}{i+z}\right)$). However in general we have that $\log (e^{2iw})\neq 2iw$. My question is that how do we get the equality $(1)$ from $\log (e^{2iw})=\log\left(\frac{i-z}{i+z}\right)$.
You must keep in mind that $\log$ is not a function. In fact, every $z\in\mathbb{C}\setminus\{0\}$ has infinitely many logarithms. To be more precise, if $w$ is a logarithm of $z$, then$$\{\text{logarithms of }z\}=\{w+2k\pi i\,|\,k\in\mathbb Z\}.$$
In particular $\log e^z=\log w$ simply means that $z$ itself is a logarithm of $w$.