i^i solution | i^i is real number


Using the representation that i=eiπ/2i=eiπ/2, we have ii=(eiπ/2)i=ei2π/2=eπ/2ii=(eiπ/2)i=ei2π/2=e−π/2.

i=eiπ/2i=eiπ/2 comes from the representation that eiθ=cos(θ)+isin(θ)eiθ=cos⁡(θ)+isin⁡(θ), which for θ=π/2θ=π/2 gives us eiπ/2=cosπ/2+isinπ/2=0+i1=ieiπ/2=cos⁡π/2+isin⁡π/2=0+i⋅1=i.