# Thread:

1. ## Taylor Series

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

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

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

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

but i cant get the steps to get there.

2. ## Re: Taylor Series

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

3. ## Re: Taylor Series

Start with the $e^{-iz}$ series.
Break it down by what happens to that $(-i)^n$ factor.
That will break the $e^{-iz}$ 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.

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