Question

If z = –1 +$\sqrt{–3}$, then arg (z) = ____________.

Answer 1

$$\begin{gathered} z=-1+i\sqrt{–3} \\ x=-1\mathrm{and}y=\sqrt{3} \\ \theta ={\tan }^{-1}\frac{\sqrt{3}}{-1} \\ ={\tan }^{-1}\left|-\sqrt{3}\right| \\ \theta =\frac{\mathrm{\pi }}{3} \end{gathered}$$,
Since z lies in IV quadrant.
⇒ argument of z is π $-\frac{\mathrm{\pi }}{3}=\frac{2\mathrm{\pi }}{3}$.