# Solve this other than punching out actual numbers

1. Nov 21, 2004

### kreil

$$e^{i\pi}=-1$$

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?

2. Nov 21, 2004

### CTS

e^ix = cos x + i sin x

3. Nov 21, 2004

### kreil

thanks, I didn't know about that equation

4. Nov 21, 2004

### mathman

If you look at the power series for cos(x), sin(x) and eix, the relationship will be obvious.