As we know that,

arg (z₁z₂) = arg(z₁)+arg(z₂) so,

arg (z₁z₂z₃) = arg(z₁)+arg(z₂)+arg(z₃)

arg((1+√3 i)(1+i)(cos θ+ i sin θ)

= arg(1+√3 i)+arg(1+i)+arg(cos θ + i sin θ).......... (1)

∴arg(z₁ ) = θ

z₂ = arg(1+i)

z₃ = arg(cos θ + i sin θ)

⇒1(cos θ + i sin θ)

arg(z₃ ) = θ

∵ In r(cos θ + i sin θ) , arg(z) = θ

By putting the value of all arg in 1, we get