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Math Help - Complex number identity

  1. #1
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    Complex number identity

    Show that (-1)^{\frac{1}{\pi}}=e^{(2n+1)i} (n=0, \pm1, \pm2, ...)
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  2. #2
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    Quote Originally Posted by davesface View Post
    Show that (-1)^{\frac{1}{\pi}}=e^{(2n+1)i} (n=0, \pm1, \pm2, ...)

    (-1)^{\frac{1}{\pi}}=e^{\frac{1}{\pi}Ln(-1)}=e^{\frac{1}{\pi}[\ln 1+i\arg(-1)]}=e^{\frac{1}{\pi}(2n+1)\pi i}= ...

    Tonio
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  3. #3
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    These sign errors get me every time. Thanks for the help.
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