If x+iy=3cos⁡θ+isin⁡θ+2 than  4x−x2−y2

1x+iy=cos⁡θ+23+isin⁡θ3⇒x−iyx2+y2=cos⁡θ+23+isin⁡θ3⇒xx2+y2=cos⁡θ+23Also x2+y2=9(2+cos⁡θ)2+sin2⁡θ=95+4cos⁡θ∴4x−x2+y2=4(cos⁡θ+2)3−1x3+y3=4cos⁡θ+53⋅95+4cos⁡θ=3