# solve this other than punching out actual numbers

by kreil
Tags: actual, numbers, punching, series, solve
 P: 518 $$e^{i\pi}=-1$$ I was wondering how on earth this was possible. I know that: $$e^z = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!}+...+\frac{z^n}{n!}$$ So $$e^{i\pi}=1+i\pi+\frac{-\pi^2}{2!}+\frac{-\pi^3i}{3!}+\frac{\pi^4}{4!}...$$ I was wondering if there is any way to solve this other than punching out actual numbers and seeing about where they converge to?
 P: 20 e^ix = cos x + i sin x
 P: 518 thanks, I didn't know about that equation