Argument of frac{1-i sqrt{3}}{1+i sqrt{3}} is | vedclass

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

Vedclass · Accepted Answer

$240$

Vedclass · Answer

(C) Let $z = \frac{1-i \sqrt{3}}{1+i \sqrt{3}}$.Multiply the numerator and denominator by the conjugate of the denominator $(1-i \sqrt{3})$:$z = \frac{(1-i \sqrt{3})(1-i \sqrt{3})}{(1+i \sqrt{3})(1-i \sqrt{3})} = \frac{1 - 2i \sqrt{3} + i^2(3)}{1^2 + (\sqrt{3})^2} = \frac{1 - 2i \sqrt{3} - 3}{1 + 3} = \frac{-2 - 2i \sqrt{3}}{4} = -\frac{1}{2} - i \frac{\sqrt{3}}{2}$.Since the complex number $z = -\frac{1}{2} - i \frac{\sqrt{3}}{2}$ lies in the third quadrant,the argument is given by $	ext{Arg}(z) = -\pi + 	an^{-1}\left|\frac{	ext{Im}(z)}{	ext{Re}(z)}ight|$ or $\pi + 	an^{-1}\left(\frac{\sqrt{3}/2}{1/2}ight)$.$	ext{Arg}(z) = 180^{\circ} + 	an^{-1}(\sqrt{3}) = 180^{\circ} + 60^{\circ} = 240^{\circ}$.

Argument of $\frac{1-i \sqrt{3}}{1+i \sqrt{3}}$ is (in $^{\circ}$)

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