The original question was: How do you do xi (x to the i power), and how on Earth was it developed? There isn’t really anything to base xi on from previous rules of exponents as it is a completely new idea.
Physicist: Euler, the dude who originally came up with using the imaginary unit, 
Leonhard "Lenny" Euler. A squinty genius.
Euler’s (surprising) equation provides a way to talk about complex exponents (“complex” = “involves
So, for example (using Euler’s equation and some log properties):
So why is this formalism used, instead of some other set of rules? Like everything else in mathematics, it’s a matter of convenience and self consistency. You’d hope that the usual, old rules would apply in a natural, convenient way. For example, you’d want 
![\begin{array}{ll}x^i x^{-i}\\=\left[e^{\ln{(x^i)}}\right]\left[e^{\ln{(x^{-i})}}\right]\\=\left[e^{i\ln{(x)}}\right]\left[e^{-i\ln{(x)}}\right]\\=\left[\cos{(\ln{(x)})} + i\sin{(\ln{(x)})}\right]\left[\cos{(-\ln{(x)})} + i\sin{(-\ln{(x)})}\right]\\=\left[\cos{(\ln{(x)})} + i\sin{(\ln{(x)})}\right]\left[\cos{(\ln{(x)})} - i\sin{(\ln{(x)})}\right]\\=\left[\cos{(\ln{(x)})}\right]^2 - i\cos{(\ln{(x)})}\sin{(\ln{(x)})} + i\cos{(\ln{(x)})}\sin{(\ln{(x)})}-i^2\left[\sin{(\ln{(x)})}\right]^2\\=\left[\cos{(\ln{(x)})}\right]^2-i^2\left[\sin{(\ln{(x)})}\right]^2\\=\left[\cos{(\ln{(x)})}\right]^2 + \left[\sin{(\ln{(x)})}\right]^2\\=1\end{array}](http://s0.wp.com/latex.php?latex=%5Cbegin%7Barray%7D%7Bll%7Dx%5Ei+x%5E%7B-i%7D%5C%5C%3D%5Cleft%5Be%5E%7B%5Cln%7B%28x%5Ei%29%7D%7D%5Cright%5D%5Cleft%5Be%5E%7B%5Cln%7B%28x%5E%7B-i%7D%29%7D%7D%5Cright%5D%5C%5C%3D%5Cleft%5Be%5E%7Bi%5Cln%7B%28x%29%7D%7D%5Cright%5D%5Cleft%5Be%5E%7B-i%5Cln%7B%28x%29%7D%7D%5Cright%5D%5C%5C%3D%5Cleft%5B%5Ccos%7B%28%5Cln%7B%28x%29%7D%29%7D+%2B+i%5Csin%7B%28%5Cln%7B%28x%29%7D%29%7D%5Cright%5D%5Cleft%5B%5Ccos%7B%28-%5Cln%7B%28x%29%7D%29%7D+%2B+i%5Csin%7B%28-%5Cln%7B%28x%29%7D%29%7D%5Cright%5D%5C%5C%3D%5Cleft%5B%5Ccos%7B%28%5Cln%7B%28x%29%7D%29%7D+%2B+i%5Csin%7B%28%5Cln%7B%28x%29%7D%29%7D%5Cright%5D%5Cleft%5B%5Ccos%7B%28%5Cln%7B%28x%29%7D%29%7D+-+i%5Csin%7B%28%5Cln%7B%28x%29%7D%29%7D%5Cright%5D%5C%5C%3D%5Cleft%5B%5Ccos%7B%28%5Cln%7B%28x%29%7D%29%7D%5Cright%5D%5E2+-+i%5Ccos%7B%28%5Cln%7B%28x%29%7D%29%7D%5Csin%7B%28%5Cln%7B%28x%29%7D%29%7D+%2B+i%5Ccos%7B%28%5Cln%7B%28x%29%7D%29%7D%5Csin%7B%28%5Cln%7B%28x%29%7D%29%7D-i%5E2%5Cleft%5B%5Csin%7B%28%5Cln%7B%28x%29%7D%29%7D%5Cright%5D%5E2%5C%5C%3D%5Cleft%5B%5Ccos%7B%28%5Cln%7B%28x%29%7D%29%7D%5Cright%5D%5E2-i%5E2%5Cleft%5B%5Csin%7B%28%5Cln%7B%28x%29%7D%29%7D%5Cright%5D%5E2%5C%5C%3D%5Cleft%5B%5Ccos%7B%28%5Cln%7B%28x%29%7D%29%7D%5Cright%5D%5E2+%2B+%5Cleft%5B%5Csin%7B%28%5Cln%7B%28x%29%7D%29%7D%5Cright%5D%5E2%5C%5C%3D1%5Cend%7Barray%7D&bg=ffffff&fg=000000&s=0 "\begin{array}{ll}x^i x^{-i}\\=\left[e^{\ln{(x^i)}}\right]\left[e^{\ln{(x^{-i})}}\right]\\=\left[e^{i\ln{(x)}}\right]\left[e^{-i\ln{(x)}}\right]\\=\left[\cos{(\ln{(x)})} + i\sin{(\ln{(x)})}\right]\left[\cos{(-\ln{(x)})} + i\sin{(-\ln{(x)})}\right]\\=\left[\cos{(\ln{(x)})} + i\sin{(\ln{(x)})}\right]\left[\cos{(\ln{(x)})} - i\sin{(\ln{(x)})}\right]\\=\left[\cos{(\ln{(x)})}\right]^2 - i\cos{(\ln{(x)})}\sin{(\ln{(x)})} + i\cos{(\ln{(x)})}\sin{(\ln{(x)})}-i^2\left[\sin{(\ln{(x)})}\right]^2\\=\left[\cos{(\ln{(x)})}\right]^2-i^2\left[\sin{(\ln{(x)})}\right]^2\\=\left[\cos{(\ln{(x)})}\right]^2 + \left[\sin{(\ln{(x)})}\right]^2\\=1\end{array}")
(Here I’ve used: Cos(-x) = Cos(x), Sin(-x) = -Sin(x), i2=-1, and the Pythagorean identity.)
Even more important, this technique is used because it recovers the usual rules for real exponents (exponents that don’t involve 
You’ll find (at least, those people who are so inclined will find) that Euler’s equation, and in particular the method for finding imaginary exponents above, is consistent with all the rules of exponentiation. Specifically, 
