If x+iy = frac{3}{2+cos theta + i sin theta}, | vedclass

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(A) Given that $x+iy = \frac{3}{2+\cos 	heta + i \sin 	heta}$.Taking the modulus on both sides,we have $|x+iy| = \left| \frac{3}{2+\cos 	heta + i \

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$4x-3$

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(A) Given that $x+iy = \frac{3}{2+\cos 	heta + i \sin 	heta}$.Taking the modulus on both sides,we have $|x+iy| = \left| \frac{3}{2+\cos 	heta + i \sin 	heta} ight|$.Since $|x+iy| = \sqrt{x^2+y^2}$,we get $\sqrt{x^2+y^2} = \frac{3}{|2+\cos 	heta + i \sin 	heta|}$.Calculating the modulus of the denominator: $|2+\cos 	heta + i \sin 	heta| = \sqrt{(2+\cos 	heta)^2 + \sin^2 	heta} = \sqrt{4 + 4\cos 	heta + \cos^2 	heta + \sin^2 	heta} = \sqrt{5+4\cos 	heta}$.Thus,$\sqrt{x^2+y^2} = \frac{3}{\sqrt{5+4\cos 	heta}}$,which implies $x^2+y^2 = \frac{9}{5+4\cos 	heta}$.Now,consider $4x-3$. From $x+iy = \frac{3(2+\cos 	heta - i \sin 	heta)}{(2+\cos 	heta)^2 + \sin^2 	heta} = \frac{3(2+\cos 	heta) - 3i \sin 	heta}{5+4\cos 	heta}$,we have $x = \frac{3(2+\cos 	heta)}{5+4\cos 	heta}$.Then $4x-3 = \frac{12(2+\cos 	heta)}{5+4\cos 	heta} - 3 = \frac{24+12\cos 	heta - 15 - 12\cos 	heta}{5+4\cos 	heta} = \frac{9}{5+4\cos 	heta}$.Therefore,$x^2+y^2 = 4x-3$.

If $x+iy = \frac{3}{2+\cos \theta + i \sin \theta}$,then $x^2+y^2 =$

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