Question

square root of 1-iota

Answer 1

square root of 1-iota=√iota
Let us assume √ i =x+ iy, as we are expecting square root of iota to be complex no.

Then, (√ i)^2 =(x+iy)^2 =x^2 -y^2 +2xyi

Comparing both side : x^2 - y^2 =0 and, 2xy=1

Similarly, we can find x^2 +y^2 =+1 or -1

then, x = y = -(1/√ 2) or +(1/√ 2)

So, √ i = +/- (1/√2 ) ( 1 + i ), which is x complex no. so our assumption was correct.

And √ i lies on the complex plane. Cheers ;
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Answer 2

Answer:

Here is the answer..

Hope it helps..