Let $w$ be from $\mathbb{C}$ with $|w| = 1$. I need to show that there exists exactly one $t$ from $[0,2\pi)$ such that $w = e^{it}$.
My approach would be to rewrite $w$ as $a+bi$ and $e^{it}$ as $\cos(t) + i \sin(t)$, and then set up two equations so that $a = \cos(t)$ and $b = \sin(t)$ and then argue with the inverse functions of cos and sin. But how can I proof with those equations that there is only exactly one $t$ which fulfills the requirements? (When I look at the unit circle it's obvious, but that's not a solid proof.)
Furthermore I need to make the proof more universal by finding all $z$ from $\mathbb{C}$ to fulfill the equation $w = e^z$ for any $w$ from $\mathbb{C}$.
Help would be really appreciated.
First of all you already stated yourself the restriction about $[0;\ 2\pi)$, i.e.all mentioned functions $\exp(it)$, $\sin(t)$, and $\cos(t)$ clearly are $2\pi$-periodic.
Now divide that range into open the quadrants $(0;\ \pi/2)$, $(\pi/2;\ \pi)$, $(\pi;\ 3\pi/2)$, and $(3\pi/2;\ 2\pi)$. On each of those the functions $\sin(x)$ and $\cos(x)$ clearly are monotone (as can be seen from their respective derivatives). Moreover the according sign combinations within each of those sub-ranges clearly is unique: $++$, $-+$, $--$, $+-$. Therefore the parameter $t$ indeed is specified uniquely for every point on the unit circle.
(Without your above restriction you clearly would obtain just $t+2\pi\,n$ instead, for any $n\in\mathbb{Z}$.)
--- rk