# ((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?

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

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

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

$$e^{ix}=\cos(x)+i \sin(x)$$ 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

haruspex