Q. If $z = \frac{1 - i 3}{1 + i 3}$ , then $ar g (z) =$

2089 179 Complex Numbers and Quadratic Equations Report Error

A

$6 0^{\circ}$

33%

B

$12 0^{\circ}$

33%

C

$24 0^{\circ}$

33%

D

$30 0^{\circ}$

0%

Solution:

$z = \frac{1 - i 3}{1 + i 3} = \frac{2 e ^{- i \frac{π}{3}}}{2 e ^{i \frac{π}{3}}} = e^{- i \frac{2 π}{3}}$
$- \frac{2 π}{3} = - 12 0^{\circ} \approx 24 0^{\circ}$