Question

For any integer $n,$ the argument of $z=\frac{(\sqrt{3}+i{)}^{4n+1}}{(1-i\sqrt{3}{)}^{4n}}$ is

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

Answer 1

The correct option is A $\frac{\pi }{6}$

$z=\frac{(\sqrt{3}+i{)}^{4n+1}}{(1-i\sqrt{3}{)}^{4n}}=\frac{i^{4n}(\sqrt{3}+i{)}^{4n+1}}{(i+\sqrt{3}{)}^{4n}}=\sqrt{3}+i$
$\tan \theta =\frac{1}{\sqrt{3}}$
As the complex number lies in the first quadrant, so
$\arg \left(z\right)=\theta =\frac{\pi }{6}$