Why is $2\pi$ the period of $e^{iz}$? My teacher said that $2\pi$ the period of $e^{iz}$. I tried to show this but I am not sure how to continue 
$$e^{iz+2\pi}=e^{ix+2\pi-y}=e^{2\pi-y}(\cos x+i\sin x)$$
But I am not sure how this equals
$$e^{-y}(\cos x+i\sin x)=e^{-y+i x}=e^{iz}$$
Is $2\pi$ the period of $e^{iz}$?

Let $f(z) = e^{iz}$
The $f(z + 2\pi) = e^{i(z + 2\pi)} = e^{iz + 2\pi i}= e^{iz}e^{2\pi i}$
$e^{2\pi i} = \cos 2\pi + i \sin 2\pi = 1$
So $f(z + 2\pi) = e^{iz} = f(z)$.
So the period of $e^{iz}$ as a function of $z$ is $2\pi$.
But the period of $e^z$ (notice $z$ is not mulitplied by $i$; $z$ is a variable complex number) as a function of $z$ is $2\pi i$
