Question

How does π^(ie)+1=0 work? Thnx

Answer 1

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

Answer 2

yes, sry, I mixed it up :S

Answer 3

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.

Answer 4

thanks, that makes sense :)