Question

For n an integer, 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

Answer: (A)

$\frac{\pi }{6}$

$Z=\frac{(\frac{-1+\sqrt{3}i}{i}{)}^{4n+1}}{(1-i\sqrt{3}{)}^{4n}}$

$Z=\frac{(\frac{2\omega }{i}{)}^{4n+1}}{(-2\omega {)}^{4n}}$

$=\frac{2\omega }{i^{4n+1}}\times (\frac{2\omega }{-2\omega }{)}^{4n}$

$=\frac{2\omega }{i}\times 1$

$=\frac{-1+\sqrt{3}i}{i}$

$=\sqrt{3}+i$
$\therefore arg|z|=ta{n}^{-1}\left(\frac{1}{\sqrt{3}}\right)$
$=\frac{\pi }{6}$