1. The problem statement, all variables and given/known data Given that, x+yi=3/[2+cos θ+i(sin θ)] , prove that x^2+y^2=4x-3 2. Relevant equations r^2=x^2+y^2 3. The attempt at a solution since we know that x=r sinθ , y=r cosθ, i multiply r on the right hand side to equal 3r/[2r+x+yi], then multiply its comjugate (2r+x)-yi to equal (3rx+6r^2 -3ryi) / [(2r+x)^2 + y^2] . then i equal the imaginary part on left handside, yi= -3ryi/[(2r+x)^2 + y^2], (2r+x)^2 + y^2 = -3r, 4r^2+x^2+ 4rx +y^2 = -3r, from the Relevant equations, 5r^2+4rx= -3r ,divide both sides by r 5r=-4x-3, r=-(4x+3)/5, x^2+y^2= [-(4x+3)/5]^2, until here, i dont know how to proceed the question... anyone can help me?