Given that, x+yi=3/[2+cos θ+i(sin θ)] , prove that x^2+y^2=4x-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
x^2+y^2= [-(4x+3)/5]^2, until here, i dont know how to proceed the question... anyone can help me?