What is \(i^i\)? (Here, i is the complex number.)
Starting with: ei·x = cos(x) + i·sin(x)
If x = pi/2 ---> ei·(pi/2) = cos( pi/2 ) + i·sin( pi/2 )
---> ei·(pi/2) = 0 + i·1 = i ---> i = ei·(pi/2)
So: ii = ( ei·(pi/2) )i = ei^2 · pi/2 = e-1·pi/2 = e- pi/2