Is i^i real or complex? plz explain

i think it is not possible for this to exist. AS,

let y=i^i taking log both sides

log y= i* log i

log i= log (root(-1))= 1/2* log (-1) which is not defined as log (-1) is not defined.

so it doesnot exist

it is real.

i^i=e^(i*pie/2)^i                      {as i = cos 0 + i*sin 0 = e^(i*pie/2) }

    =e^(i*i*pie/2)

    =e^(-pie/2) which is real