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If $\frac{3}{2 + \cos θ + i \sin θ} = x + i y$ then $(x - 1) (x - 3)$ =

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$y^{2}$

$- y^{2}$

answer is B.

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Detailed Solution

$\frac{3}{x + i y} - 2 = \cos θ + i \sin θ \frac{(3 - 2 x) - 2 i y}{x + i y} = \cos θ + i \sin θ$

taking modulus both sides

${(3 - 2 x)}^{2} + 4 y^{2} = x^{2} + y^{2} x^{2} - 4 x + 3 = - y^{2}$

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