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Where's the fallacy

  1. Jul 9, 2007 #1
    This one has me stumped.

    [tex]e^{\pi i} = -1[/tex]
    [tex]e^{2 \pi i} = (-1) ^ 2 = 1 [/tex]
    [tex]ln(e^{2 \pi i}) = ln(1) = 0 [/tex]
    [tex]2 \pi i = 0 [/tex]

    Or is 2 pi i actually 0, and this does not actually imply that either pi = 0 or i = 0?
     
    Last edited: Jul 9, 2007
  2. jcsd
  3. Jul 9, 2007 #2

    Integral

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    Since [itex] 2 \pi i [/itex] is an imaginary number you must use the definition of LN in the complex plane. You are using the definition for the real number line.

    in the complex plane we have:

    [tex] logz = log(r) + i \theta [/tex]

    edit (changed my definition of z)
    [tex] r = |z| [/tex]
    and
    [tex] \theta = arg(z) [/tex]
     
    Last edited: Jul 9, 2007
  4. Jul 9, 2007 #3

    matt grime

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    More than that - logs of complex numbers are only defined up to multiples of 2pi. One can ought to take the principal branch - this is just a slightly more complicated variation on the square root 'fallacies'.
     
  5. Jul 9, 2007 #4
    The mistake is here:
    [tex] \ln e^z = z[/tex]
    This is not true for complex numbers.

    Note: the other way around:
    [tex] \exp (\ln z) = z, \ z\not =0[/tex]
    Is true.
     
  6. Jul 10, 2007 #5
    Ok thanks. As you can see, I haven't taken my complex variables class yet.
     
  7. Jul 10, 2007 #6

    mathwonk

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    its sort of like saying, (2^2 =4 and (-2)^2 = 4 so 2 = -2.)
     
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