The square roots of the complex number (-5-12 | vedclass

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(A) Let $\sqrt{-5-12i} = a+ib$,where $a, b \in \mathbb{R}$.Squaring both sides,we get $(a+ib)^2 = -5-12i$.$(a^2-b^2) + i(2ab) = -5-12i$.Equating real

Vedclass · Accepted Answer

$\pm(2-3i)$

Vedclass · Answer

(A) Let $\sqrt{-5-12i} = a+ib$,where $a, b \in \mathbb{R}$.Squaring both sides,we get $(a+ib)^2 = -5-12i$.$(a^2-b^2) + i(2ab) = -5-12i$.Equating real and imaginary parts: $a^2-b^2 = -5$ and $2ab = -12$,which implies $ab = -6$.Since $b = -6/a$,substituting into the first equation gives $a^2 - (-6/a)^2 = -5$.$a^2 - 36/a^2 = -5 \implies a^4 + 5a^2 - 36 = 0$.Factoring the quadratic in $a^2$: $(a^2+9)(a^2-4) = 0$.Since $a \in \mathbb{R}$,$a^2 = 4$,so $a = \pm 2$.If $a = 2$,$b = -6/2 = -3$. If $a = -2$,$b = -6/(-2) = 3$.Thus,the square roots are $\pm(2-3i)$.

The square roots of the complex number $(-5-12i)$ are

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