Given that, x+iy=2/(3+costheta+isintheta).
:. (x+iy)(3+costheta+isintheta)=2, i.e.,
(3+costheta)x+ixsintheta+(3+costheta)iy+i^2ysintheta=2, or,,
(3+costheta)x-ysintheta+i{xsintheta+(3+costheta)y}=2+0i.
Comparing the Real and Imaginary Parts, we have,
(3+costheta)x-ysintheta=2, and, xsintheta+(3+costheta)y=0.
Solving these eqns. for (3+costheta) and sintheta, we get,
3+costheta=(2x)/(x^2+y^2) and sintheta=-(2y)/(x^2+y^2), or,
costheta=(2x)/(x^2+y^2)-3 and sintheta=-(2y)/(x^2+y^2).
But, cos^2theta+sin^2theta=1,
rArr {(2x)/(x^2+y^2)-3}^2+{-(2y)/(x^2+y^2)^2}=1.
rArr (4x^2)/(x^2+y^2)^2-(12x)/(x^2+y^2)+9+(4y^2)/(x^2+y^2)^2=1.
rArr (4x^2+4y^2)/(x^2+y^2)^2=(12x)/(x^2+y^2)-8, i.e.,
(4(x^2+y^2))/(x^2+y^2)^2=4{(3x)/(x^2+y^2)-2}, or,
1/(x^2+y^2)=(3x)/(x^2+y^2)-2.
rArr 2=(3x-1)/(x^2+y^2)," giving,"
2(x^2+y^2)=3x-1, as desired!
Q.E.D.
Enjoy Maths.!