If z = frac{3}{2 + cos theta + i sin theta},t | vedclass

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(B) Given $z = \frac{3}{2 + \cos 	heta + i \sin 	heta}$.Rearranging the terms,we get $2 + \cos 	heta + i \sin 	heta = \frac{3}{z}$.$\cos 	heta +

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a circle having centre on $x$-axis

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(B) Given $z = \frac{3}{2 + \cos 	heta + i \sin 	heta}$.Rearranging the terms,we get $2 + \cos 	heta + i \sin 	heta = \frac{3}{z}$.$\cos 	heta + i \sin 	heta = \frac{3}{z} - 2 = \frac{3 - 2z}{z}$.Taking the modulus on both sides,$|\cos 	heta + i \sin 	heta| = \left| \frac{3 - 2z}{z} ight|$.Since $|\cos 	heta + i \sin 	heta| = 1$,we have $1 = \frac{|3 - 2z|}{|z|}$,which implies $|z| = |3 - 2z|$.Let $z = x + iy$. Then $|x + iy| = |3 - 2(x + iy)| = |(3 - 2x) - i(2y)|$.Squaring both sides: $x^2 + y^2 = (3 - 2x)^2 + (-2y)^2$.$x^2 + y^2 = 9 - 12x + 4x^2 + 4y^2$.$3x^2 + 3y^2 - 12x + 9 = 0$.Dividing by $3$: $x^2 + y^2 - 4x + 3 = 0$.This is the equation of a circle with centre at $(2, 0)$,which lies on the $x$-axis.

If $z = \frac{3}{2 + \cos \theta + i \sin \theta}$,then the locus of $z$ is :-

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