Q. If $P$ is the point in the Agrand diagram corresponding to the complex number $\sqrt{3} + i$ And if $OPQ$ is an isosceles right angled triangle, right angled at $'O'$ then $Q$ represents the complex number
A) $- 1i\sqrt{3}$ or $1 - i\sqrt{3}$
B) $1 \pm i\sqrt{3}$
C) $\sqrt{3} - i$ or $1 - i\sqrt{3}$
D) $- 1 \pm i\sqrt{3}$
Explanation
(a)
Let $z = \sqrt{3} + i$
$\therefore\ \ \ \ \ \ \ \arg{(z) = \tan^{- 1}{\left( \frac{1}{\sqrt{3}} \right) = 30{^\circ}}}$
For making a right angled $\mathrm{\Delta}OPQ$, point $Q$ either in IInd quadrant or IVth quadrant
If the point $Q$ is in IInd quadrant, then we take
$\theta = 120{^\circ}$
$\therefore\ \ \tan{120{^\circ} = - \cot{30{^\circ} = \frac{\sqrt{3}}{- 1}}}$
$\therefore$ Point $Q$ is $( - 1,\ \sqrt{3})$ and if the point $Q$ is in IVth quadrant then we take
$\theta = - 60{^\circ}$
$\therefore\ \ \ \ \ \ \tan{( - 60{^\circ}) = - \tan{60{^\circ} = - \frac{1}{\sqrt{3}}}}$
$\therefore$ Point $Q$ is $(1,\ \sqrt{3})$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.