8 6i . Half the coefficent of i is 3 whose factors #160; are 3 and 1 #160; and since #160; the real part #160; is 8 , we take #160; the factor #16 ...

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Byju's Answer

Standard XII

Mathematics

Iota

√-8-6i =

Question

$\sqrt{- 8 - 6 i}$ =

Open in App

Solution

-8-6i . Half the coefficent of i is 3 whose factors are 3 and 1 and since the real part is -8 , we take the factor as 1 and 3 such that $1^{2} - 3^{2} = - 8$
$∴ \sqrt{8 - 6 i} = \pm (1 - 3 i)$
$∴ \sqrt{- 8 + 6 i} = \pm (1 + 3 i)$

Suggest Corrections

√−8−6i =

-8-6i . Half the coefficent of i is 3 whose factors are 3 and 1 and since the real part is -8 , we take the factor as 1 and 3 such that 12−32=−8∴√8−6i=±(1−3i)∴√−8+6i=±(1+3i)

$\sqrt{- 8 - 6 i}$ =

-8-6i . Half the coefficent of i is 3 whose factors are 3 and 1 and since the real part is -8 , we take the factor as 1 and 3 such that $1^{2} - 3^{2} = - 8$
$∴ \sqrt{8 - 6 i} = \pm (1 - 3 i)$
$∴ \sqrt{- 8 + 6 i} = \pm (1 + 3 i)$