Why from $e^{i\mu}-e^{-i\mu}=0$ we can conclude that $e^{2i\mu}=1$?

Question

Could you please explain why from $e^{i\mu}-e^{-i\mu}=0$ we can conclude that $e^{2i\mu}=1$ and $2i\mu=2n\pi i$, when $\mu$ is real or complex?

I tried to use Euler's formula but without any success. I feel it must be very basic and simple, but obviously I missed some basics on complex numbers.

Thanks.

Answer 1

$e^{i\mu}-e^{-i\mu}=0\Rightarrow e^{i\mu}=e^{-i\mu} \Rightarrow e^{2i\mu}=1=e^{2in\pi}$ $\big(\text{ as }e^{2in\pi}=\cos(2n\pi)+i\sin(2n\pi)=1\big)$