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

  • #1
Can someone help explain this? Wolfram says it is zero but I don't know why?
 

Answers and Replies

  • #3
Ok I know eulers but how does 1^x - (-1)^x = 0?
 
  • #4
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That's ##\left(-1\right)^x-\left(-1\right)^x##, after some simplification.
 
  • #5
Mentallic
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Ok I know eulers but how does 1^x - (-1)^x = 0?

[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]
 
  • #6
[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]
Ok I see now thx
 
  • #7
haruspex
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Ok I know eulers but how does 1^x - (-1)^x = 0?

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

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