We have , z = - 1 - i √3
Let - 1 = r cos θ ... (i) and −√3 = r sin θ ... (ii)
Squaring and adding (i) and (ii), we get
r2(cos2θ+sin2θ) = 1 + 3 ⇒ r2 = 4 ⇒ r = √4 = 2
∴ 2 cos θ = - 1 , 2 sin θ = - √3 [By (i) and (ii)]
cos θ = −

1

2

, sin θ = −

√3

2

⇒ cos θ = - cos (

π

3

) , sin θ = - sin (

π

3

)
Both cosθ and sinθ are negative.
∴ θ lies in the third quadrant.
∴ θ = - (π−

π

3

) = - (

3π−π

3

) =

−2π

3

∴ Modulus is 2 and argument is

−2π

3