Q. If $z = \sqrt{3} + i$, then the argument of $z^{2}e^{z - i}$ is equal to
A) $\frac{\pi}{2}$
B) $\frac{\pi}{6}$
C) $e^{\pi/6}$
D) $\pi/3$
Explanation
(d)
Given, $z = \sqrt{3} + i$,
$\arg{(z^{2}e^{z - i})} = \arg{\lbrack\left( 3 - 1 + 2\sqrt{3}i \right)e^{\sqrt{3}}\rbrack}$
$= \arg{\left\lbrack \left( 2 + 2\sqrt{3}i \right)e^{\sqrt{3}} \right\rbrack = \tan^{- 1}\left\lbrack \frac{2\sqrt{3}}{2} \right\rbrack = \frac{\pi}{3}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.