Please, i need an immediate answer with proof

If an imaginary number i is raised to i, i.e,i^i.., its value is 0.207689... ¡i = 0.207689...

But if a real number or a variable like x is raised to i, how many real numbers, variables exists..?

Or, is it possible, like ¡i , 2^i or x^i does it exists...??

Please i need this immediately, I'm eager to give 10 points

2^i = e^(ln(2^i))
= e^(i ln(2))
= cos(ln(2)) + i sin(ln(2))

I'm pretty sure that's the way it works.

I just checked a few values on my TI-83+, so here is what I think:

Numbers in this interval when raised to the power of i will result in numbers with complex parts:

(-∞,∞) \ { -1,0,1}

Whereas -1 and 1 raised to the power of i will result in real numbers, and 0 raised to the power of i is undefined.

So I would say if real number is raised to i, only 2 real numbers would exist. I don't know how to prove it though, maybe proof by contradiction, hmm...

EDIT- Here is a website that has info for your question:
http://www.math.toronto.edu/mathnet/ques…

A=angle in radians
R=distance from Origin
any number but zero can be express as such:

e^(iA+R)=R(cos(A)+isin(A))

with this general equation you can find any complex number to any complex power or figure out how to get a real from a complex via powers

by euler: i=ei(pi)/2
therifore ii = (ei(pi)/2)
i=ei^2(pi)/2=e-(pi)/2 ~0.208....

i'm not sure but you can use log if its a equation.....