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Another fallacy

  1. Jun 14, 2005 #1
    HI, since e^[i*pi]=-1,then e^[2i*pi]=1.
    taking lin of both sides,2*i*pi=0,that makes i=0
    squaring both sides,you get -1 to equal 0.
    Is there a problem with what i have done,and what is it?
     
  2. jcsd
  3. Jun 14, 2005 #2

    Curious3141

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    When complex outputs are allowed, logarithms become multivalued. In this case, the correct formulae are :

    [tex]e^{2k\pi i} = 1, \ln(1) = 2k\pi i[/tex] where [itex]k[/itex] ranges over the integers.

    In your case, you're confusing the result for k = 0 with k = 1, hence the apparent paradox.
     
    Last edited: Jun 14, 2005
  4. Jun 14, 2005 #3

    Zurtex

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    Yes and:

    [tex](-1)^2 = 1 \quad 1^2 = 1 \quad 1 = - 1 \quad 2 = 0[/tex]

    It's the same things really :rolleyes:
     
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