which is greater
i^i or i^-i
i^i^i or i^(pi/2)
Enter these expressions as Google searches if you want them evaluated. For example, a search for i^i will get the response i^i = 0.207879576. In fact, i^i and i^-i are real numbers, so it makes sense to ask which is greater. But i^(i^i) and i^(pi/2) are complex. As pickslides points out, there is no natural sense in which one is greater than the other.
$\displaystyle i^i = \left(e^{\frac{\pi}{2}i}\right)^i$
$\displaystyle = e^{\frac{\pi}{2}i^2}$
$\displaystyle = e^{-\frac{\pi}{2}}$.
Meanwhile $\displaystyle i^{-i} = (i^i)^{-1}$
$\displaystyle = \left(e^{-\frac{\pi}{2}}\right)^{-1}$
$\displaystyle = e^{\frac{\pi}{2}}$.
Clearly $\displaystyle e^{\frac{\pi}{2}} > e^{-\frac{\pi}{2}}$, so $\displaystyle i^{-i} > i^i$.