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E to the pi i

  1. Feb 4, 2005 #1
    ok, here it goes, why is e^(pi.i)=-1 ?
     
    Last edited: Feb 4, 2005
  2. jcsd
  3. Feb 4, 2005 #2

    Integral

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    It comes form Eulers relationship

    [tex] e^{ix} = \cos(x) + i \sin(x) [/tex]

    Edit: LOL, I fixed it already!
    Now I am moving this to Math.
     
    Last edited: Feb 4, 2005
  4. Feb 4, 2005 #3
    Any complex number [tex] a + bi [/tex] can be written in the form [tex] r(\cos{\theta} + i\sin{\theta}) [/tex]. De Moivre's theorem states that this in turn can be written in the form [tex] re^{i\theta} [/tex]. That is, [tex] re^{i\theta} = r(\cos{\theta} + i\sin{\theta}) [/tex]. Plug in [tex] \theta = \pi [/tex] and behold the magic!
     
  5. Feb 4, 2005 #4
    Was it Euler? well, in that case u would have to use Taylor's expansion series and yes, it would work. Why is this important at all then?
     
  6. Feb 4, 2005 #5
    It is commonly regarded to be one of the most (ahem) beautiful and elegant mathematical relationships in our universe (yes, our :biggrin: ). C'mon, wouldn't you agree that it is beautiful, succintly relating the 5 most important numbers in mathematics?!?! ([tex] e^{\pi i} + 1 = 0 [/tex])
     
  7. Feb 5, 2005 #6

    Integral

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    In addition it uses each of the fundamental mathematical operations, addition, multiplication, exponentiation, and equality. It is consider mathematical poetry.
     
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