I was entering random things in my calculator and came across this. It makes no sense to me conceptually. How can the square root of negative one raised to the power of the square root of negative one be a positive decimal?
4 Answers

a^x = e ^(x log a)
so i^i = e^(i log i)
log(z) is by definition log (z) + i argument(z)
where z is a complex number
so log(i) = log(i) + i argument (i) = 0 + ipi/2
then ilog(i) = i i * pi/2 = pi/2
and thus
i^i = e^(i ln i) = e^pi/2 =

you have to be introduced to the subject of complex variables and the complex logarithm to understand it.
let z = i^i
now take the complex log
log z = i*log(i)
(same rule as real logs applies here log a^b = b*log a)
now the complex log is defined as:
Log[w] = ln[w]+i(Arg[w]+2Pi*k), for any integer k
the capital L version means the general case for all k,
with lower case l the log uses the principle value
let's look at this piece by piece
ln[w] means the natural log of the magnitude of the complex number w. when w=i, w = 1, and ln(1)=0
Arg(w) means the angle with respect to the positive real axis the complex number w makes. i is 90 degrees from it, but it must be in radians: Pi/2
thus we have
log i = i*Pi/2
and ilog i = ii*Pi/2 = Pi/2
but we want to know what i^i is and we so far calculated the log of i^i, so all we have to do is take the antilog of this (i.e. raise e to the power of log(i^i) since e^(log w) = w),
so i^i = e^(log(i)) = e^(Pi/2)
and you can evaluate this on your calculator and get:
approximately .2079

Complex numbers do strange things. The formula e^ix = cos x + i sin x (if I have remembered it right) obviously leads to a real result for appropriate values of x. Also, we have e^i pi = 1. and e^pi is a real number.

We have the identity:
e^(ix) = cosx + isinx
e^(iÏ/2) = cos(Ï/2) + isin(Ï/2) = 0 + i1 = i
Take the natural log of both sides.
i*Ï/2 = ln i
Multiply both sides by i.
iÂ²(Ï/2) = i ln i = ln(i^i)
1(Ï/2) = ln(i^i)
Ï/2 = ln(i^i)
Exponentiate both sides.
e^(Ï/2) = i^i
i^i = e^(Ï/2) â 0.2078795