-8-6 i-A- 1 - 3 iB- -1-3 iC- -1-3 i D- -3- i

The correct option is B ±(1 3i)Let √( 8 6i)=x+iy=z ⇒ 8 6i=(x+iy)^2 ∴ x^2 y^2 = 8 ...(i) 2xy= 6 ...(ii) Now x^2+y^2 = √(64+36)=± 10 ...(iii) From (i) and ...

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Byju's Answer

Standard XI

Mathematics

Square Root of a Complex Number

√-8-6 i=A. 1 ...

Question

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

\pm

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\pm

(1-3i)

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\pm

(1+3i)

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\pm

(3-i)

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Solution

The correct option is B $\pm$ (1-3i)
Let $\sqrt{- 8 - 6 i} = x + i y = z$

$\Rightarrow - 8 - 6 i = (x + i y)^{2}$

$∴ x^{2} - y^{2} = - 8 . . . (i) 2 x y = - 6 . . . (i i)$

Now $x^{2} + y^{2} = \sqrt{64 + 36} = \pm 10 . . . (i i i)$

From (i) and (ii),we have $x = \pm 1 a n d y = \pm 3$

Hence $z = \pm (1 - 3 i)$

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√−8−6i=

The correct option is B ±(1-3i)Let √−8−6i=x+iy=z ⇒−8−6i=(x+iy)2 ∴x2−y2=−8 ...(i)2xy=−6 ...(ii) Now x2+y2=√64+36=±10 ...(iii) From (i) and (ii),we have x=±1 and y=±3 Hence z=±(1−3i)

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