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

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

$x = 0 such that y > 2 / 3$

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

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

answer is A.

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Detailed Solution

$z (z + 3) = 2 \Rightarrow z z + 3 z - 2 = 0$

$| z |^{2} + 3 z = 2 \Rightarrow x^{2} + y^{2} + 3 x + i 3 y = 2 comparing real, imaginary parts both sides x^{2} + y^{2} + 3 x = 2 and y = 0$

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