Let 0 = rcosθ and 2√3 = rsinθ
By squaring and adding, we get
(0)2 + (2√3)2 = (rcosθ)2 + (rsinθ)2
⇒ 0+(2√3)2 = r2 (cos2θ + sin2θ)
⇒(2√3)2 = r2
⇒ r = 2√3
∴ cosθ = 0 and sinθ = 1
Since, θ lies in first quadrant, we have
θ = π/2
Since, θ ∈ (-π ,π] it is principal argument.