TITLE: Why is $2\pi i \neq 0?$
QUESTION [14 upvotes]: We know that $e^{\pi i} = -1$ because of de Moivre's formula. ($e^{\pi i} = \cos \pi + i\sin \pi = -1).$
Suppose we square both sides and get $e^{2\pi i} = 1$(which you also get from de Moivre's formula), then shouldn't $2\pi i=0$? What am I missing here?

REPLY [3 votes]: Actually, the following is true:
$ e^{2n\pi i} = 1  \forall  n \in \mathbb{Z} $
As a special case of $n = 0$ this gives what you know: $e^0 = 1$, but as mentioned, this is only a special case of the general formula given above.
You could as well assume that $2\pi = 0$ because of
$\sin 2\pi = \sin 0 \land \cos 2\pi = \cos 0$.
As others already mentioned, this stuff just isn't injective, so conclusions relying on the injectivity can be wrong.