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

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Aicharya · Accepted Answer

$\begin{aligned} z & =\frac{-2}{1+\sqrt{3} i} \ \Rightarrow & =\frac{-2(1-\sqrt{3} i)}{(1+\sqrt{3} i)(1-\sqrt{3} i)} \ & =\frac{-2(1-\sqrt{3} i)}{(1)^2-(\sqrt{3} i)^2} \ & =\frac{-2(1-\sqrt{3} i)}{1-3 i^2} \ & =\frac{-2(1-\sqrt{3} i)}{1+3} \quad:\left[i^2=-1ight]\end{aligned}$ $\begin{aligned} & =\frac{-2}{4}(1-\sqrt{3} i) \ & =\frac{-1}{2}(1-\sqrt{3} \mathrm{i}) \ & 	herefore \quad z=\frac{-1}{2}+\frac{\sqrt{3} i}{2} \end{aligned}$ $\begin{aligned} 	herefore \quad \arg (z) & =	an ^{-1}\left(\frac{b}{a}ight) \ & =	an ^{-1}\left(\frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}}ight) \ & =	an ^{-1}(-\sqrt{3}) \ & =\pi-	an ^{-1}(\sqrt{3}) \ & =\pi-\frac{\pi}{3} \ & =\frac{2 \pi}{3}\end{aligned}$

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

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