Wow, how was that visual proof?<p>For those who aren't looking forward to a lame video, the basic idea is this:<p>e^(pi)(i) = -1.<p>Why?<p>Because you can express e^x as a taylor series:<p>e^x = 1 + x + x^2/2! + x^3/3! + ...
thus
e^ix = 1 + ix + ix^2/2! + ix^3/3! + ..
e^ix = (1 - x^2/2! + x^4/4! + ..) + i(x - x^3/3!+x^5/5!+...)
implies
e^ix = cosx + isinx<p>because those parenthetical quantities are the taylor series for cosx and sinx respectively<p>So, e^i(pi) = cos(pi) + isin(pi).
= -1 + i(0)
= -1