One of the math teachers asked me for a nice way to explain how to find 
I’ll put it below in case any of you want to try to work it out.
Starting with Euler’s identity:
We note that
So raising both sides of the equation to the 1/2 power yields:
Of course this method will then let you find 
Continuous Everywhere but Differentiable Nowhere 
Here’s another cool way using the rotation model (think of multiplying by -1 as a 180 degree rotation on the complex plane; thus a 90 degree rotation is the square root of -1). This lets you build the imaginary number line perpendicular to the real number line. If you extend the metaphor, a 45 degree rotation would be the square root of i. Plot a complex number 1 unit from the origin at a 45 degree angle. Now you have a picture like the unit circle, imposed on the complex plane. Using basic trig, both components of the complex number must be root(2)/2.
OK, but, what is i^i? :-)
@Dan: Totally get it! Thanks.
@Kate: its e^(-pi/2).
Since i=e^(i*pi/2), raise both sides to the ith power. You get i^i=(e^(i*pi/2))^i=e^(-pi/2).
It’s a plain, old, ordinary real number. Sigh.
Both sqrt(2)/2 + i sqrt(2)/2 and -sqrt(2)/2 – i sqrt(2)/2 are square roots of i, aren’t they? Is there a way we decide which one of them is meant by sqrt(i) (just as when we write sqrt(9) we know we mean 3 and not -3), which we could use for the square root of any complex number? Maybe the one whose angle theta is between 0 and 180?
@Matt, I think that it’s a matter of symantics…
and 
. But that’s just technical.
Technically, when someone says “the square root of three” there are two answers —
In everyday speak, when someone says “the square root of three” the mean
and when they mean the negative value, they’ll say “negative the square root of three.”
See http://en.wikipedia.org/wiki/Square_root
So I think generally, we can say that
. And we designate the negative solution with the negative values…
Does this make any sense?
While the square root of i has only two possible answers, if I’m not mistaken, i^i has infinitely many. For instance, while i = e^{i pi/2}, it also equals e^{9i pi/2} and e^{-7i pi/2} and e^{17i pi/2} and…
Therefore, at the very least, the elements of the set {e^{-(2k+1) pi/2} | k is an integer} are all answers to the question: what is i^i?
Adam is absolutely right. 0.2078…. is just the principal value (k=0 in Adam’s expression).