We have , z = - 1 - i $\sqrt{3}$ Let - 1 = r cos θ ... (i) and $-\sqrt{3}$ = r sin θ ... (ii) Squaring and adding (i) and (ii), we get $r^2(\cos^2\theta + \sin^2\theta)$ = 1 + 3 ⇒ $r^2$ = 4 ⇒ r = $\sqrt{4}$ = 2 ∴ 2 cos θ = - 1 , 2 sin θ = - $\sqrt{3}$ [By (i) and (ii)] cos θ = $-\frac{1}{2}$ , sin θ = $-\frac{\sqrt{3}}{2}$ ⇒ cos θ = - cos $\left(\frac{\pi}{3}\right)$ , sin θ = - sin $\left(\frac{\pi}{3}\right)$ Both cosθ and sinθ are negative. ∴ θ lies in the third quadrant. ∴ θ = - $\left(\pi - \frac{\pi}{3}\right)$ = - $\left(\frac{3\pi - \pi}{3}\right)$ = $\frac{-2\pi}{3}$ ∴ Modulus is 2 and argument is $\frac{-2\pi}{3}$