Originally Posted by

**princeps** According to Euler's formula : $\displaystyle e^{ix}=\cos x + i\cdot \sin x$ we may write :

$\displaystyle e^{i\cdot \frac{\pi}{2}}=i \Rightarrow \left(e^{i\cdot \frac{\pi}{2}}\right)^{i}=i^{i}\Rightarrow e^{\frac{-\pi}{2}}=i^{i}$

So,

$\displaystyle e^{\pi}=\left(e^{\frac{\pi}{2}}\right)^2=\left( \frac{1}{e^{\frac{-\pi}{2}}}\right)^2=\left(\frac{1}{i^{i}}\right)^2= \left(i^{-i}\right)^2=(-1)^{-i}$

Since$\displaystyle e^{\pi}$ is proven by Gelfond–Schneider theorem to be transcendental number it follows that $\displaystyle (-1)^{-i}$ is a transcendental number. So,my question is :

Are numbers : $\displaystyle (-1)^{i} , 1^{-i} , 1^{i}$ transcendental numbers ?