Are both square roots of -1 valid in Euler's Identity?

Question

In every case I've ever seen, Euler's Identity is written as

$e^{i\pi} + 1 = 0$

with the "positive" $\sqrt{-1}$. However, my understanding is that both $i$ and $-i$ are valid for $\sqrt{-1}$.

Does this mean that

$e^{-i\pi} + 1 = 0$

is also a valid identity?

Answer 1

Yes.

$$e^{-i\pi} =\cos (-\pi)+i\sin(-\pi)=-1$$

Answer 2

Since $\ker(\exp)=2\pi i \Bbb Z$, we also have $e^{i\pi-2\pi i}=e^{i\pi}$.

Answer 3

Note that if $\theta \in \mathbb R$ then $$ e^{-i\theta} =\cos(-\theta) + i\sin(-\theta) = \cos(\theta) - i\sin(\theta). $$ Plugging in $\theta = \pi$ yields $$ e^{-i\pi} = -1. $$