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

Can someone help explain this? Wolfram says it is zero but I don't know why?
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 Perhaps this will help: http://en.wikipedia.org/wiki/Euler's_formula
 Ok I know eulers but how does 1^x - (-1)^x = 0?

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

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

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 Quote by powerplayer Ok I know eulers but how does 1^x - (-1)^x = 0?
$$e^{ix}=\cos(x)+i \sin(x)$$ hence after plugging $x=-\pi$ we get $$e^{-i\pi}=\cos(-\pi) +i \sin(-\pi)$$ and recall that $\cos(-x)=\cos(x)$ and $\sin(-x)=-\sin(x)$ thus we have $$e^{-i\pi}=\cos(\pi)-i\sin(\pi)=-1-0i=-1$$ while similarly, $$e^{i\pi}=\cos(\pi)+i\sin(\pi)=-1+0i=-1$$

 Quote by Mentallic $$e^{ix}=\cos(x)+i \sin(x)$$ hence after plugging $x=-\pi$ we get $$e^{-i\pi}=\cos(-\pi) +i \sin(-\pi)$$ and recall that $\cos(-x)=\cos(x)$ and $\sin(-x)=-\sin(x)$ thus we have $$e^{-i\pi}=\cos(\pi)-i\sin(\pi)=-1-0i=-1$$ while similarly, $$e^{i\pi}=\cos(\pi)+i\sin(\pi)=-1+0i=-1$$
Ok I see now thx

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