i(iota)to the power i(iota).

z = e^i@ (representation of complex number Z in ellulers form)

z₁ = iⁱ (given)

taking log both sides

logz₁ = ilogi  .................1

= log(e^ipi/2)

= ipi/2

puting this value of logi in eq 1

logZ₁ = i²(pi/2) = -pi/2  (i² = -1)

Z₁ = e^(-pi/2) , this is the required result

Approved

Hehe log is only defined for non negative real no. :P :D
And you solved it? 

i= e^i(π÷2)Now,i^i= {e^(i(π÷2))}^i= e^(-(π÷2)) which is positive and real. So the upper provided solution is true