Question

If $z=\frac{-2}{1+i\sqrt{3}}$, then the value of arg (z) is
(a) π
(b) $\frac{\mathrm{\pi }}{3}$
(c) $\frac{2\mathrm{\pi }}{3}$
(d) $\frac{\mathrm{\pi }}{4}$

Answer 1

(c) $\frac{2\pi }{3}$
z = $\frac{-2}{1+i\sqrt{3}}$

Rationalising z, we get,

$$\begin{gathered} z=\frac{-2}{1+i\sqrt{3}}\times \frac{1-i\sqrt{3}}{1-i\sqrt{3}} \\ \Rightarrow z=\frac{-2+i2\sqrt{3}}{1+3} \\ \Rightarrow z=\frac{-1+i\sqrt{3}}{2} \\ \Rightarrow z=\frac{-1}{2}+\frac{i\sqrt{3}}{2} \end{gathered}$$
$$\begin{gathered} \tan \alpha =\left|\frac{\mathrm{Im}\left(z\right)}{\mathrm{Re}\left(z\right)}\right| \\ =\sqrt{3} \\ \Rightarrow \alpha =\frac{\mathrm{\pi }}{3} \\ \mathrm{Since},z\mathrm{lies}\mathrm{in}\mathrm{the}\mathrm{second}\mathrm{quadrant}. \\ \mathrm{Therefore},\arg \left(z\right)=\mathrm{\pi }-\frac{\mathrm{\pi }}{3} \\ =\frac{2\mathrm{\pi }}{3} \end{gathered}$$