Question

Question #615a2

Answer 1

`i^i = e^(-pi/2)`

Explanation:

Using Euler's formula , `e^(itheta) = cos(theta)+isin(theta)` we have

`i = 0 + i*1 = cos(pi/2)+isin(pi/2) = e^(ipi/2)`

Substituting that in for the base in in our problem, and applying the property `(x^a)^b = x^(ab)`, we get

`i^i = (e^(ipi/2))^i = e^(ipi/2*i) = e^(i^2pi/2) = e^(-pi/2)`

Thus, we find the interesting result that `i^i` is a real number.