Remember that we haven't actually defined exponentiation for any two complex numbers: only integer powers of any a+bi (squared, cubed,...), and e to any complex number. So i^i doesn't really make sense. How should we define it?

Well, we might as well try to reduce to one of the cases where we did define things. When exponentiating real numbers, we have the identity a^b = e^(b ln(a)), so let's try to set i^i = e^(i ln(i)). This isn't really progress yet -- what's ln(i)? It should be a complex number z such that e^z = i. Euler's formula gives e^{i pi/2} = i, so z=i pi/2 seems as good a guess as any. Then i^i = e^(i ln(i)) = e^(i*i pi/2) = e^(-pi/2) = 1/sqrt(e^pi) = 0.20787..., the number the calculator gave.

The catch is that ln(i) isn't really uniquely defined. e^(i pi/2 + 2 pi k i) = i for any integer value of k, and there's nothing particularly special about the i*pi/2 that we chose. This means that i^i doesn't have a single value; any other choice of natural log would also have a reasonable claim to give i^i . This non-uniqueness of a logarithm is why we don't usually define the exponential of any two complex numbers in general, and instead stick with the two cases mentioned above.