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what is the square root of -i argument of -i

what is the


square root of -i


argument of -i

Grade:11

4 Answers

Chetan Mandayam Nayakar
312 Points
8 years ago

sqrt(-i)=e-i((pi/2)+(2N(pi)))

where N is ANY integer

Akash Kumar Dutta
98 Points
8 years ago

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

sandy sandy
10 Points
8 years ago

arg is 270 ad 1is the root

Harsh Gupta
16 Points
8 years ago

Argument if -i is -π/2.

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