WHAT IS THE SQUARE ROOT OF -i?

WRITE THE ARGUMENT OF -i.

find sq root by the simple step... root (-i)=a+ib...............
and arg of -i= -pi/2 or 5pi/2

sqr root of -i is +or-(1-i)/2^1/2 and argument is -∏/2

let (a+ib)^2=-i then calculate and b to get the z=√-i

which is +-1 which turn to get arg(z)=pie\4