Question

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. $e^{\pi /3}$
• E. $\frac{\pi }{3}$

Answer 1

Answer: (E)

Given, $z=\sqrt{3}+i,$ $\arg (z^2{e}^{z-i})=\arg [(3-1+2\sqrt{3}i)e^{\sqrt{3}}]$
$=\arg [(2+2\sqrt{3}i)e^{\sqrt{3}}]$
$={\tan }^{-1}\left[\frac{2\sqrt{3}}{2}\right]$
$=\frac{\pi }{3}$