$Latex i^{i } = 0.207879576…$

$latex i = sqrt{-1}$

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

$latex e^{ipi/2} = i$

4. Take the **i***th* power of both sides, the right side being $latex i^i $ and the left side =

$latex (e^{ipi/2})^{i}= e^{-pi/2} $

5. Therefore

$latex i^{i} = e^{-pi/2} = .207879576…$