We have , z = - √3 + i
Let - √3 = r cos θ ... (i) and 1 = r sin θ ... (ii)
Squaring and adding (i) and (ii), we get
r2(cos2θ+sin2θ) = 3 + 1 ⇒ r2 = 4 ⇒ r = 2
Substituting the value of r in (i) and (ii), we get 2 cos θ = - √3 , 2 sin θ = 1
⇒ cos θ =

−√3

2

, sin θ =

1

2

⇒ cos θ = - cos (

π

6

) , sin θ = sin (

π

6

)
Here, cosθ is negative where sinθ is positive.
∴ θ lies in the second quadrant.
∴ θ = (π−

π

6

) =

5π

6

∴ Modulus is 2 and argument is

5π

6