Gelfond's constant and complexity

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June 1st 2011, 12:40 AM
Stro

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!
June 1st 2011, 01:29 AM
FernandoRevilla

$(-1)^i=e^{i\log (-1)}=e^{i(0+\pi i)}=e^{-\pi}\neq e^{\pi}$
June 1st 2011, 01:39 AM
Stro

Thank you!!

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