Square roots of -i- where i-1- are

Let x=√( 1)=( 1)^1/2 x= [ cos ( π2 )+i sin ( π2 ) ]^1/2 x= [ cos ( 2nπ π2 )+i sin ( 2nπ π2 ) ]^1/2,n=0,1 x= [ cos(4n 1)π2+i sin(4n 1)π2 ]^1/2,n=0,1 x=cos(4 ...

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Byju's Answer

Standard XII

Mathematics

Properties of Cube Root of a Complex Number

Square roots ...

Question

Square roots of $- i$ , where $i = \sqrt{- 1}$ , are

i, - i

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c o s (- \frac{π}{4}) + i s i n (- \frac{π}{4}), c o s (\frac{3 π}{4}) + i s i n (\frac{3 π}{4})

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c o s (\frac{π}{4}) + i s i n (\frac{3 π}{4}), c o s (\frac{3 π}{4}) + i s i n (\frac{π}{4})

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c o s (\frac{3 π}{4}) + i s i n (- \frac{3 π}{4}), c o s (- \frac{3 π}{4}) + i s i n (\frac{3 π}{4})

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Solution

The correct option is B $c o s (- \frac{π}{4}) + i s i n (- \frac{π}{4}), c o s (\frac{3 π}{4}) + i s i n (\frac{3 π}{4})$
Let $x = \sqrt{- 1} = (- 1)^{1 / 2}$
$x = {[c o s (\frac{- π}{2}) + i s i n (- \frac{π}{2})]}^{1 / 2}$
$x = {[c o s (2 n π - \frac{- π}{2}) + i s i n (2 n π - \frac{π}{2})]}^{1 / 2}, n = 0, 1$
$x = {[c o s (4 n - 1) \frac{π}{2} + i s i n (4 n - 1) \frac{π}{2}]}^{1 / 2}, n = 0, 1$
$x = c o s (4 n - 1) \frac{π}{4} + i s i n (4 n - 1) \frac{π}{4}, n = 0, 1$
$x = c o s (- \frac{π}{4}) + i s i n (- \frac{π}{4}), c o s (\frac{3 π}{4}) + i s i n (\frac{3 π}{4})$

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Square roots of −i, where i=√−1, are

Square roots of $- i$ , where $i = \sqrt{- 1}$ , are