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Trigonometric form and 1=-1

  1. Dec 2, 2007 #1
    Hi,
    as I was studying complex numbers today I came across this, and I couldn't explain it:
    [tex]1=e^{i0}[/tex]
    [tex]1=e^{i2\Pi}[/tex]
    [tex]1^{1/2}=e^{i2\Pi/2}[/tex]
    [tex]1=e^{i\Pi}[/tex]
    [tex]1=-1[/tex]
    Where is the mistake?
    Thank you very much for your help.
     
  2. jcsd
  3. Dec 2, 2007 #2

    Hurkyl

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    Why would you think that the latter equation follows from the former?
     
  4. Dec 2, 2007 #3
    [tex]1^{1/2}=(e^{i2\pi})^{1/2}[/tex]
    Is that wrong?
     
  5. Dec 2, 2007 #4

    HallsofIvy

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    The question really was, "why do you think that 11/2= 1?". In the real number system we take functions to be "single valued". In the complex number system that is no longer possible.

    In particular, in the real number system, we define x1/2 to be "the positive number, a, such that a2= x". Since the complex numbers are not an ordered field that definition cannot be used.
     
  6. Dec 2, 2007 #5
    OK, so basically I would have to take -1, but why not take 1 too?
    And if I take 1, I still get 1=-1, right?
     
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