# Question #615a2

Oct 7, 2016

${i}^{i} = {e}^{- \frac{\pi}{2}}$

#### Explanation:

Using Euler's formula , ${e}^{i \theta} = \cos \left(\theta\right) + i \sin \left(\theta\right)$ we have

$i = 0 + i \cdot 1 = \cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right) = {e}^{i \frac{\pi}{2}}$

Substituting that in for the base in in our problem, and applying the property ${\left({x}^{a}\right)}^{b} = {x}^{a b}$, we get

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

Thus, we find the interesting result that ${i}^{i}$ is a real number.