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Math Help - C(omplex) puzzle

  1. #1
    Member courteous's Avatar
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    C(omplex) puzzle

    What's wrong with this?
    {\displaystyle e^{i2\pi n}=1 \text{ ; for } n=0,\pm 1, \pm 2, \hdots }

    \displaystyle {ee^{i2\pi n}=e=e^{1+i2\pi n} }

    e^{1+i2\pi n}=\left\{ e^{1+i2\pi n} \right\}^{(1+i2\pi n)}

    e=e^{(1+i2\pi n)^2}=e^{1+i4\pi n-4\pi^2n^2}=e^{1+i4\pi n}e^{-4\pi^2n^2}

    e^{1+i4\pi n}=e

    e=ee^{-4\pi^2n^2}

    1=e^{-4\pi^2n^2}, which is true only for n=0.
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  2. #2
    Super Member Aryth's Avatar
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    The third line, where for some reason you put that:

    e^{1 + i2\pi n} = \left\{e^{1 + i2\pi n}\right\}^{(1 + i2\pi n)}

    This, of course, is only true when n = 0.
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  3. #3
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    Quote Originally Posted by Aryth View Post
    The third line, where for some reason you put that:

    e^{1 + i2\pi n} = \left\{e^{1 + i2\pi n}\right\}^{(1 + i2\pi n)}

    This, of course, is only true when n = 0.
    This line is correct ( e=e^{1+i2\pi n} \Rightarrow e^{1+i2\pi n}=\left(e^{1+i2\pi n}\right)^{1+i2\pi n}) . The incorrect line is the one after the line you wrote - \left(e^a\right)^b does not necessarily equal e^{ab} when b is not real (and the above argument just shows this leads to a contradiction).
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