Is $-e^{i\pi} = 1$?

Question

Since $e^{i\pi} = \cos \pi + i\sin \pi = -1,$ a suspicious argument is to proceed to conclude that $$-e^{i\pi} = 1.$$ However, this leads to $$-e^{i\pi} = e^{0}.$$ Is the above reasoning wrong?

Answer 1

That's correct. There's nothing wrong with the above reasoning. Is it equally wrong that $e^{2\pi n i}=1$ for all $n\in\mathbb{Z}$? The Euler formula you quoted shows that the exponential function, as a complex function, is periodic. Namely, it is non-injective, or $e^z=e^w$ does not imply $z=w$.

Answer 2

There's nothing suspicious if you remember that $$e^{i\pi} = -1 \iff -(e^{i\pi}) = 1$$ and being clear that this is not to say $$(-e)^{i\pi} = 1$$

So there is nothing wrong with $$-(e^{i\pi}) = e^{0}.$$ But that is not to say that $i\pi = 0$.