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Under root -i

  1. Nov 28, 2006 #1
    wat is under root - i ???

    please anser this
     
  2. jcsd
  3. Nov 28, 2006 #2

    Hurkyl

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    Do you mean [itex]\sqrt{-i}[/itex]? You can figure it out yourself! If [itex]a + bi = \sqrt{-i}[/itex], then what does the definition of square root tell you?
     
  4. Nov 28, 2006 #3

    dextercioby

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    HINT:

    [tex] -i=e^{-\frac{i\pi}{2}+2k\pi} , \ k\in\mathbb{Z} [/tex]

    Daniel.
     
  5. Nov 28, 2006 #4

    Gib Z

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    Hey guys, dont post here often, i will more from now on. Anyway, dexter got that from the identity e^(ix)= cos x + i sin x.
     
  6. Nov 29, 2006 #5
    will it be solved by the de mouvies theorm..........i dont think so...
     
  7. Nov 29, 2006 #6

    dextercioby

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    You mean "de Moivre". Yes, it will, since that theorem is a trivial consequence of the fact that

    [tex] \left(e^{ix}\right)^{n}=e^{inx} [/tex]

    Daniel.
     
  8. Nov 29, 2006 #7

    HallsofIvy

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    I personally like Hurkyl's suggestion best but WHY don't you think deMoivre's formula will work? It can be used to find any root of any complex number.
     
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