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

Complex Numbers and Quadratic Equations

Solution:

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