Q:

What does $i^i $ equal and why? 

I've been reading up on why the value of 0^0 is controversial (see Zero to the zero power - is $0^0=1$?)  and I wondered: is it possible for $i^i$ to have a value?  I plugged it into a TI-83 calculator and it returned 0.2078795764 (!)  How is this possible and why is the result a real number not a complex number? 
Update I realize how to use Euler's Identity of $e^{i\theta} = \cos(\theta) + i\sin(\theta)$ and I understand the oldest answer's value of $\theta = \pi/2$ to solve for $i^i$, but doesn't this mean I can pick any odd multiple of $\pi/2$ (such as $3\pi/2, 5\pi/2$, etc.)  as a value of $\theta$?  Does that mean that $i^i$ is somehow "periodic" like the cosine curve is? 

A:

$i=e^{i(\pi/2+2k\pi)}$ therefore $i^i=(e^{i(\pi/2+2k\pi)})^i=e^{-\pi/2-2k\pi}$ 
for $k=0$ you get $0.207879576350761908...$
for $k=1$ you get $0.000388203203926766...$
....
Or 
$i=1\times i$, then $i^i=1^i\times i^i$: as tomasz commented, this post may be helpful: What is the value of $1^i$?