What is i^i

Math Online Tom Circle

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

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

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Author: tomcircle

Math amateur

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