If Z = frac{-2}{1 + sqrt{3}i},where i = sqrt{ | vedclass

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(A) Given $Z = \frac{-2}{1 + \sqrt{3}i}$.Rationalizing the denominator:$Z = \frac{-2(1 - \sqrt{3}i)}{(1 + \sqrt{3}i)(1 - \sqrt{3}i)}$$Z = \frac{-2(1 -

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$\frac{2\pi}{3}$

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(A) Given $Z = \frac{-2}{1 + \sqrt{3}i}$.Rationalizing the denominator:$Z = \frac{-2(1 - \sqrt{3}i)}{(1 + \sqrt{3}i)(1 - \sqrt{3}i)}$$Z = \frac{-2(1 - \sqrt{3}i)}{1^2 - (\sqrt{3}i)^2}$$Z = \frac{-2(1 - \sqrt{3}i)}{1 + 3} = \frac{-2(1 - \sqrt{3}i)}{4}$$Z = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$.Here,the real part $a = -\frac{1}{2}$ and the imaginary part $b = \frac{\sqrt{3}}{2}$.Since $a  0$,the complex number lies in the second quadrant.$\arg(Z) = \pi - 	an^{-1}\left|\frac{b}{a}ight|$$\arg(Z) = \pi - 	an^{-1}\left|\frac{\sqrt{3}/2}{-1/2}ight|$$\arg(Z) = \pi - 	an^{-1}(\sqrt{3})$$\arg(Z) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.

If $Z = \frac{-2}{1 + \sqrt{3}i}$,where $i = \sqrt{-1}$,then the value of $\arg(Z)$ is

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