Question

If $z=\sqrt{3}+i,$ then the argument of $z^2{e}^{z-1}$ is equal to

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{3}$

Answer 1

Answer: (D)

$arg\left(z^2{e}^{z-1}\right)=2argz+0$, (Since $z-i=\sqrt{3})$
$=2\times \frac{\pi }{6}=\frac{\pi }{3}$
$\left[\because argz=ta{n}^{-1}\frac{1}{\sqrt{3}}=\frac{\pi }{6}\right]$