What is i^i

$Latex i^{i } = 0.207879576...$
$latex i = \sqrt{-1}$

If a is algebraic and b is algebraic but irrational then $latex a^b $ is transcendental. (Gelfond-Schneider Theorem)

Since i is algebraic but irrational, the theorem applies.

1. We know
$latex e^{ix}= \cos x + i \sin x$

Let $latex x = \pi/2 $

2. $latex e^{i \pi/2} = \cos \pi/2 + i \sin \pi/2 $

$latex \cos \pi/2 = \cos 90^\circ = 0 $

$latex \sin 90^\circ = 1 $
$latex i \sin 90^\circ = (i)*(1) = i $

3. Therefore
$latex e^{i\pi/2} = i$
4. Take the ith power of both sides, the right side being $latex i^i $ and the left side =
$latex (e^{i\pi/2})^{i}= e^{-\pi/2} $
5. Therefore
$latex i^{i} = e^{-\pi/2} = .207879576...$