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Math Help - Gelfond's constant and complexity

  1. #1
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    Gelfond's constant and complexity

    Gelfond's constant contains really odd complex math to me.

    Consider that e^{-i\pi} = -1 is also proven by Taylor series.

    So we can show that e^{\pi} = (e^{-i\pi})^{i}} = (-1)^{i} and e^{\pi} = (e^{i\pi})^{-i}} = (-1)^{-i} which goes against my normal logic.

    Anyone know why it's shown like it is?

    Thanks!
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    (-1)^i=e^{i\log (-1)}=e^{i(0+\pi i)}=e^{-\pi}\neq e^{\pi}
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  3. #3
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    Thank you!!
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