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

Transcendental Number

Transcendental numbers: e, Π, L...

What about $Latex e^{e}, \pi^{\pi} ,\pi^{e} $ ?

Aleksander (Alexis) Osipovich Gelfond (1906-68):

Gelfond-Schneider Theorem

$Latex a^ b $ transcendental if
a is algebraic, not 0 or 1
b irrational algebraic number

Examples:
$latex \sqrt{6}^{\sqrt{5}}, 3^{\sqrt{7}}$
Hilbert Number: $latex 2^{\sqrt {2}}$ (Hilbert Problem proven by Gelfond}

Is log 2 transcendental ?
[log = logarithm Base 10]

Proof:
$latex 10^{log 2} = 2$

1) Sufficient to prove log 2 irrational
Assume log 2 rational
log 2= p/q, p and q integers
$latex 10^ {log 2} = 2 = 10^ {p/q}$
raise power q
$latex 2^{q} = 10^{p} = (2.5)^{p}$
$latex 2^{q} = 2^{p}.5^{p}$

Case 1: p>q
$latex 1 = 2^{p-q}.5^{p}$
=> False

Case 2: q>p
$latex 2^{q-p}= 5^{p}$
Left is even : $latex 2^{m} \text { = even} $
Right is odd: $latex 5^{n} \text {= ....5} $
=> False

Therefore p,q do not exist,
=> log 2 irrational

Reference: Top 15 Transcendental Numbers: http://sprott.physics.wisc.edu/pickover/trans.html