idk 2 i^1/2 = e^(iπ/4) = 1∠45 i^1/2 or sqrt(i) = sqrt(2) + sqrt(2)i [i^x = e^(iπx/2) = 1∠(πx/2) = cos(πx/2) + isin(πx/2)] if we use this function, i^i = e^iπi/2 since i = sqrt(-1) ii = i^2 = -1 so i^i = e^(-1π/2) = e^(-π/2) x^-y = 1/y so i^i = 1/e^(π/2) ≈ 0.207... => i^i ≈ 0.207
shout out to my calculator
