If z = frac{1 - isqrt{3}}{1 + isqrt{3}},then | vedclass

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(C) Given $z = \frac{1 - i\sqrt{3}}{1 + i\sqrt{3}}$.Multiply the numerator and denominator by the conjugate of the denominator $(1 - i\sqrt{3})$:$z =

Vedclass · Accepted Answer

$240$

Vedclass · Answer

(C) Given $z = \frac{1 - i\sqrt{3}}{1 + i\sqrt{3}}$.Multiply the numerator and denominator by the conjugate of the denominator $(1 - i\sqrt{3})$:$z = \frac{(1 - i\sqrt{3})(1 - i\sqrt{3})}{(1 + i\sqrt{3})(1 - i\sqrt{3})} = \frac{1 - 3 - 2i\sqrt{3}}{1 + 3} = \frac{-2 - 2i\sqrt{3}}{4} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$.Since the complex number $z = x + iy$ lies in the third quadrant $(x $arg(z) = 180^\circ + 	an^{-1}\left(\frac{-\sqrt{3}/2}{-1/2}ight) = 180^\circ + 	an^{-1}(\sqrt{3}) = 180^\circ + 60^\circ = 240^\circ$.Alternatively,using the property $arg(\frac{z_1}{z_2}) = arg(z_1) - arg(z_2)$:$arg(z) = arg(1 - i\sqrt{3}) - arg(1 + i\sqrt{3}) = -60^\circ - 60^\circ = -120^\circ$.Since $-120^\circ$ is equivalent to $360^\circ - 120^\circ = 240^\circ$,the correct answer is $240^\circ$.

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

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