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question mark

what is the

square root of -i

argument of -i

Samarth Khandelwal , 13 Years ago
Grade 11
anser 4 Answers
Chetan Mandayam Nayakar

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

where N is ANY integer

Last Activity: 13 Years ago
Akash Kumar Dutta

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

Last Activity: 13 Years ago
sandy sandy

arg is 270 ad 1is the root

Last Activity: 13 Years ago
Harsh Gupta

Argument if -i is -π/2.

Last Activity: 12 Years ago
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