જોz = frac{{1 - isqrt 3 }}{{1 + isqrt 3 }},તો | vedclass

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Question

(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

Vedclass · Accepted Answer

$240$

Vedclass · Answer

(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 $arg(z) = {	an ^{ - 1}}\frac{y}{x} = {	an ^{ - 1}}\sqrt 3 = \frac{\pi }{3} = {60^{o.}}$
Since the complex number lies in $III$ quadrant, therefore
$arg\,(z)$ is ${180^o}$ + ${60^o} = {240^o}$
Aliter : $arg\left( {\frac{{1 - i\sqrt 3 }}{{1 + i\sqrt 3 }}} ight) = arg(1 - i\sqrt 3 ) - arg(1 + i\sqrt 3 )$
$ = - {60^o} - {60^o} = - {120^o}$or ${240^o}$.

જો$z = \frac{{1 - i\sqrt 3 }}{{1 + i\sqrt 3 }},$તો $arg(z) = $ ............. $^\circ$

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