## anonymous 4 years ago How does π^(ie)+1=0 work? Thnx

1. anonymous

Did you mean $e^(i \pi)+1=0$?

2. anonymous

yes, sry, I mixed it up :S

3. anonymous

Euler gives us the identity: $e^(ix) = \cos(x) + \sin(x)$ so when you sub pi in as the radian angle it become an evaluation of sin and cos at 180 degrees $\cos(\pi) + \sin(\pi)\ + 1 = 0$ [-1] + [0] + 1 = 0 which is true.

4. anonymous

thanks, that makes sense :)