Use taylor series to verify $\displaystyle e^{-iz}= cos z - isinz$

i know cos z =$\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^nz^{2n}}{2n!}$

and sin z = $\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^nz^{2n+1}}{2n+1!}$

and $\displaystyle e^{-iz} =\sum_{n=0}^{\infty}\frac{(-iz)^n}{n!} $

but i cant get the steps to get there.