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Some contradiction I can't explain about e^(ix)

  1. Feb 22, 2006 #1
    Let k be a real number (not necessarily an integer).
    [tex]e^{i\cdot2\pi\cdot k}=\cos(2\pi k) + i\sin(2\pi k)=[/tex] some complex number on the unit circe.
    [tex]e^{i\cdot2\pi\cdot k}=(e^{i2\pi})^k=1^k=1[/tex]

    so if take [itex]\tilde{k}=2\pi k[/itex] then [itex]1=e^{i\tilde{k}}=[/itex]every other number on the unit circle.

    How can this be explained??
  2. jcsd
  3. Feb 22, 2006 #2
    solved... I'm an idiot.
    [tex]1^k[/tex] with complex numbers are all the unit's [tex]k^{-1}[/tex] roots...
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