Originally Posted by **nairbdm** okay, now I got:
Cn= [(-2in)(e^(-inpi)-e^(inpi))] / ((-2in)^2 + 1)
can that simplifty any more? |

The exponentials will simplify. Note that $\displaystyle e^{i n \pi} = \left( e^{i \pi} \right)^n$ and $\displaystyle e^{-i n \pi} = \left( e^{- i \pi} \right)^n$ ......