If x + iy = frac{3}{2 + cos theta + isin thet | vedclass

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(B) Given $x + iy = \frac{3}{2 + \cos 	heta + i\sin 	heta}$.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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(B) Given $x + iy = \frac{3}{2 + \cos 	heta + i\sin 	heta}$.Taking the modulus on both sides,we have $|x + iy| = \left| \frac{3}{2 + \cos 	heta + i\sin 	heta} ight|$.$|x + iy| = \frac{|3|}{|(2 + \cos 	heta) + i\sin 	heta|} = \frac{3}{\sqrt{(2 + \cos 	heta)^2 + \sin^2 	heta}}$.$|x + iy| = \frac{3}{\sqrt{4 + \cos^2 	heta + 4\cos 	heta + \sin^2 	heta}} = \frac{3}{\sqrt{5 + 4\cos 	heta}}$.Squaring both sides,${x^2} + {y^2} = \frac{9}{5 + 4\cos 	heta}$.From the real part,$x = \frac{3(2 + \cos 	heta)}{5 + 4\cos 	heta} = \frac{6 + 3\cos 	heta}{5 + 4\cos 	heta}$.Thus,$4x - 3 = 4\left( \frac{6 + 3\cos 	heta}{5 + 4\cos 	heta} ight) - 3 = \frac{24 + 12\cos 	heta - 15 - 12\cos 	heta}{5 + 4\cos 	heta} = \frac{9}{5 + 4\cos 	heta}$.Therefore,${x^2} + {y^2} = 4x - 3$.

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

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