((e^(i*pi))^x)-((e^(-i*pi))^x)=0? how?

powerplayer

Can someone help explain this? Wolfram says it is zero but I don't know why?

Whovian

Perhaps this will help: http://en.wikipedia.org/wiki/Euler's_formula

powerplayer

Ok I know eulers but how does 1^x - (-1)^x = 0?

Whovian

That's ##\left(-1\right)^x-\left(-1\right)^x##, after some simplification.

Mentallic

[tex]e^{ix}=\cos(x)+i \sin(x)[/tex] hence after plugging [itex]x=-\pi[/itex] we get [tex]e^{-i\pi}=\cos(-\pi) +i \sin(-\pi)[/tex] and recall that [itex]\cos(-x)=\cos(x)[/itex] and [itex]\sin(-x)=-\sin(x)[/itex] thus we have [tex]e^{-i\pi}=\cos(\pi)-i\sin(\pi)=-1-0i=-1[/tex] while similarly, [tex]e^{i\pi}=\cos(\pi)+i\sin(\pi)=-1+0i=-1[/tex]

powerplayer

Ok I see now thx

haruspex

((e^(i*pi))^x)-((e^(-i*pi))^x) does not reduce to 1^x - (-1)^x.
It reduces to (-1)^x - (-1)^x.