We have, z=1−3icosθ3+2icosθ,θ∈(0,2π)
⇒z=1−3icosθ3+2icosθ×1+3icosθ1+3icosθ
⇒z=1+9cos2θ(3+2icosθ)(1+3icosθ)
⇒z=1+9cos2θ(3−6cos2θ+11icosθ)
Now, Re(z)=1+9cos2θ3−6cos2θ=0
⇒3−6cos2θ=0
⇒cos2θ=63
⇒cos2θ=21
⇒cosθ=±21
⇒θ=4π,∵θ∈(0,2π)
Hence, sin23θ+cos2θ
=sin2(43π)+cos2(4π)
=(sin(π−4π))2+(cos(4π))2
=(sin(4π))2+(cos(4π))2=1
Since, sin2θ+cos2θ=1