We have , z = - $\sqrt{3}$ + i Let - $\sqrt{3}$ = r cos θ ... (i) and 1 = r sin θ ... (ii) Squaring and adding (i) and (ii), we get $r^2(\cos^2 \theta + \sin^2 \theta)$ = 3 + 1 ⇒ $r^2$ = 4 ⇒ r = 2 Substituting the value of r in (i) and (ii), we get 2 cos θ = - $\sqrt{3}$ , 2 sin θ = 1 ⇒ cos θ = $\frac{-\sqrt{3}}{2}$ , sin θ = $\frac{1}{2}$ ⇒ cos θ = - cos $\left(\frac{\pi}{6}\right)$ , sin θ = sin $\left(\frac{\pi}{6}\right)$ Here, cosθ is negative where sinθ is positive. ∴ θ lies in the second quadrant. ∴ θ = $\left(\pi - \frac{\pi}{6}\right)$ = $\frac{5\pi}{6}$ ∴ Modulus is 2 and argument is $\frac{5\pi}{6}$