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(B) Let $\sqrt{-8 - 6i} = x + iy$.Squaring both sides,we get $-8 - 6i = (x + iy)^2 = x^2 - y^2 + 2xyi$.Equating real and imaginary parts,we have $x^2

Vedclass · Accepted Answer

$\pm(1 - 3i)$

Vedclass · Answer

(B) Let $\sqrt{-8 - 6i} = x + iy$.Squaring both sides,we get $-8 - 6i = (x + iy)^2 = x^2 - y^2 + 2xyi$.Equating real and imaginary parts,we have $x^2 - y^2 = -8$ $(i)$ and $2xy = -6$ $(ii)$.We know that $(x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2$.$(x^2 + y^2)^2 = (-8)^2 + (-6)^2 = 64 + 36 = 100$.Since $x^2 + y^2$ must be positive,$x^2 + y^2 = 10$ $(iii)$.Adding $(i)$ and $(iii)$,$2x^2 = 2 \implies x^2 = 1 \implies x = \pm 1$.Subtracting $(i)$ from $(iii)$,$2y^2 = 18 \implies y^2 = 9 \implies y = \pm 3$.From $(ii)$,$2xy = -6$,so $x$ and $y$ must have opposite signs.Thus,the square roots are $\pm(1 - 3i)$.

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

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