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what is the
square root of -i
argument of -i
sqrt(-i)=e-i((pi/2)+(2N(pi)))
where N is ANY integer
find sq root by the simple step... root (-i)=a+ib............... and arg of -i= -pi/2 or 5pi/2
arg is 270 ad 1is the root
Argument if -i is -π/2.
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