(c)If $z = \frac{{1 – i\sqrt 3 }}{{1 + i\sqrt 3 }} = \frac{{(1 – i\sqrt 3 )(1 – i\sqrt 3 )}}{{(1 + i\sqrt 3 )(1 – i\sqrt 3 )}}$
$ = \frac{{1 – 3 – 2i\sqrt 3 }}{{1 + 3}} = \frac{{ – 2 – 2i\sqrt 3 }}{4} = – \frac{1}{2} – i\frac{{\sqrt 3 }}{2}$
Thus $arg(z) = {\tan ^{ – 1}}\frac{y}{x} = {\tan ^{ – 1}}\sqrt 3 = \frac{\pi }{3} = {60^{o.}}$
Since the complex number lies in $III$ quadrant, therefore
$arg\,(z)$ is ${180^o}$ + ${60^o} = {240^o}$
Aliter : $arg\left( {\frac{{1 – i\sqrt 3 }}{{1 + i\sqrt 3 }}} \right) = arg(1 – i\sqrt 3 ) – arg(1 + i\sqrt 3 )$
$ = – {60^o} – {60^o} = – {120^o}$or ${240^o}$.