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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

$i \log \left( \frac{x - i}{x + i} \right)$ is equal to $(x \in R)$

What is the square root of $\sqrt{50} + \sqrt{48}$?

$\sinh^{-1}(-2) + \operatorname{cosech}^{-1}(-2) + \coth^{-1}(-2) = $

The value of $\sqrt{12\sqrt{5} + 2\sqrt{55}}$ is:

$\sec h^{-1}\left(\frac{1}{2}\right)-\operatorname{cosec} h^{-1}\left(\frac{3}{4}\right)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module