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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
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