
Taylor Series
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.

Re: Taylor Series
The sum for $\displaystyle \cos$ runs over even terms, and the corresponding one for $\displaystyle \sin$ of odd terms. So you just have to check that the coefficients of $\displaystyle e^{iz}$ for odd/even indexes match.

Re: Taylor Series
Start with the $\displaystyle e^{iz}$ series.
Break it down by what happens to that $\displaystyle (i)^n$ factor.
That will break the $\displaystyle e^{iz}$ series into the sum of four series via the values of n mod 4.
Collect real and imaginary terms.
Simpify for the real and imaginary parts (this takes some care  gotta manipulate the series and their indices and such carefully).
That'll do it.