# Gelfond's constant and complexity

• June 1st 2011, 01:40 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, 02:29 AM
FernandoRevilla
$(-1)^i=e^{i\log (-1)}=e^{i(0+\pi i)}=e^{-\pi}\neq e^{\pi}$
• June 1st 2011, 02:39 AM
Stro
Thank you!!