Step 1: Aim of the problem. We are given a complex number equal to $x+iy$ and we want the product $(x-1)(x-3)$. First we must find $x$ and $y$ separately. Step 2: Rationalise the denominator. Multiply top and bottom by the conjugate $2+\cos\theta-i\sin\theta$ to make the bottom real: \[ x+iy=\frac{3(2+\cos\theta-i\sin\theta)}{(2+\cos\theta)^2+\sin^2\theta}. \] Step 3: Simplify the bottom. Expand: $(2+\cos\theta)^2+\sin^2\theta=4+4\cos\theta+\cos^2\theta+\sin^2\theta=5+4\cos\theta.$ So \[ x=\frac{3(2+\cos\theta)}{5+4\cos\theta},\qquad y=\frac{-3\sin\theta}{5+4\cos\theta}. \] Step 4: Compute $x-1$ and $x-3$. \[ x-1=\frac{3(2+\cos\theta)-(5+4\cos\theta)}{5+4\cos\theta}=\frac{1-\cos\theta}{5+4\cos\theta}, \] \[ x-3=\frac{3(2+\cos\theta)-3(5+4\cos\theta)}{5+4\cos\theta}=\frac{-9-9\cos\theta}{5+4\cos\theta}=\frac{-9(1+\cos\theta)}{5+4\cos\theta}. \] Step 5: Multiply them. \[ (x-1)(x-3)=\frac{-9(1-\cos\theta)(1+\cos\theta)}{(5+4\cos\theta)^2}=\frac{-9(1-\cos^2\theta)}{(5+4\cos\theta)^2}=\frac{-9\sin^2\theta}{(5+4\cos\theta)^2}. \] Step 6: Compare with $y^2$. Since $y^2=\dfrac{9\sin^2\theta}{(5+4\cos\theta)^2}$, we get $(x-1)(x-3)=-y^2$. \[ \boxed{-y^2} \]