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 $\frac{\pi }{6}$

We have, $z=\frac{{(\sqrt{3}+i)}^{4n+1}}{{(1-i\sqrt{3})}^{4n}}=\frac{{\left(2e^{i\frac{\pi }{6}}\right)}^{4n+1}}{{\left(2e^{-i\frac{\pi }{3}}\right)}^{4n}}$

$=\frac{2^{4n+1}.e^{i(4n+1)\frac{\pi }{6}}}{2^{4n}.e^{-i4n\frac{\pi }{3}}}$$=2.e^{i(2n+1)\frac{\pi }{6}}$

$=2.e^{2n\pi i}.e^{\frac{\pi i}{6}}=2.e^{\frac{\pi i}{6}}(\because e^{2n\pi i}=1)$

$\therefore arg\left(z\right)=\frac{\pi }{6}$