To compute this, we’ll make use of Euler’s formula cis(x) = eix = cos(x) + i·sin(x): ei(π/2) = cos(π/2) + i·sin(π/2) = i, and exponentiating by i we get ii = (ei(π/2))i = ei·i(π/2) = e-π/2 ∈ ℝ. But we also have cis(2πk + π/2) = i, k ∈ ℤ. Thus by the same logic, we get ii = e-(2πk + π/2) for k ∈ ℤ. Infinitely many evaluations, so ii doesn't have a distinct/unique value? This seems like a discrepancy. I don't have any strong arguments, but I have this W|A computation telling me that ii is indeed distinct, since none of e-5π/2, e-9π/2, e-13π/2, etc are equal. Also, since the exponential function from ℝ onto ℝ+ is certainly injective, each of these numbers e-(2πk + π/2) must be distinct. If ii is distinct, why is its value e-π/2 = 0.2078795... ? Can anyone explain this?