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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)/21/2 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
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