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

1. May 24, 2012

powerplayer

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

2. May 24, 2012

Whovian

3. May 24, 2012

powerplayer

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

4. May 24, 2012

Whovian

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

5. May 24, 2012

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, $$e^{i\pi}=\cos(\pi)+i\sin(\pi)=-1+0i=-1$$

6. May 24, 2012

powerplayer

Ok I see now thx

7. May 26, 2012

haruspex

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