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If $z (z + 3) = 2$ then the locus of $z = x + iy$ is

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$x = 0$ and $y > 2 / 3$

$x^{2} + y^{2} + 3 x - 2 = 0, y = 0$

$x^{2} + y^{2} + 2 x - 4 y = 0$ such that $2 x - y + 4 > 0$

$x^{2} + y^{2} + 2 x - 4 y = 0$ such that $y < 0, x^{2} + y^{2} > 1$

answer is A.

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Detailed Solution

$z (z + 3) = 2, z = x + iy \Rightarrow x + iy (x - iy + 3) = 2 \Rightarrow (x + iy) (x - iy) + 3 (x + iy) = 2 \Rightarrow (x^{2} + y^{2} + 3 x - 2) + i (3 y) = 0 \Rightarrow x^{2} + y^{2} + 3 x - 2 = 0 a n d y = 0$

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