If $\frac{3}{2+\cos \theta +i\sin \theta }=x+iy$ then $(x-1)(x-3)=$

$\frac{3}{2+\cos \theta +i\sin \theta }=x+iy$ 

$\Rightarrow x=\frac{3\left(2+\cos \theta \right)}{{\left(2+\cos \theta \right)}^2+\left({\sin }^2\theta \right)}=\frac{6+3\cos \theta }{5+4\cos \theta }$

$\Rightarrow y=-\frac{\left(3\sin \theta \right)}{{\left(2+\cos \theta \right)}^2+{\sin }^2\theta }=\frac{-3\sin \theta }{5+4\cos \theta }$

$(x-1)(x-3)=\frac{(1-\cos \theta )(-9-9\cos \theta )}{{(5+4\cos \theta )}^2}$

$=-\frac{9\left({\sin }^2\theta \right)}{{\left(5+4\cos \theta \right)}^2}=-\left(y^2\right)$

$\therefore \left(x-1\right)\left(x-3\right)=-y^2$