Q. if $\sqrt{x + iy} = \pm (a + ib)$, then $\sqrt{- x - iy}$ is equal to
A) $\pm (b + ia)$
B) $\pm (a - ib)$
C) $(ai + b)$
D) $\pm (b - ia)$
Explanation
(d)
Given, $\sqrt{x + iy} = \pm (a + ib)$
$\Rightarrow \ \ \ x + iy = a^{2} - b^{2} + 2iab$
$\Rightarrow \ \ \ \ x = a^{2} - b^{2},\ \ \ y = 2ab$
$\therefore\ \ \ \ \sqrt{- x - iy} = \sqrt{- \left( a^{2} - b^{2} \right) - 2iab}$
$= \sqrt{b^{2} - a^{2} - 2iab} = \pm (b - ia)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.