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 $e^{iz}$?

You are misunderstanding the teacher.  If the function is $f(z)=e^{iz}$ then saying that the period of $f(z)$ is $2\pi$ means that $f(z+2\pi)=f(z)$, i.e. that
$$e^{i(z+2\pi)}=e^{iz}$$
Notice the parentheses in the exponent on the left-hand side, which are missing from the equation in your question.
