If $Amp (\frac{z + 2}{z - 4 i}) = \frac{π}{2}$ 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 D.

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Detailed Solution

$Amp (\frac{z + 2}{z - 4 i}) = \frac{π}{2}, z = x + iy Amp (\frac{(x + 2) + iy}{x + i (y - 4)}) = \frac{π}{2} {Tan}^{- 1} \frac{y}{x + 2} - {Tan}^{- 1} \frac{y - 4}{x} = \frac{π}{2} \frac{\frac{y}{x + 2} - \frac{y - 4}{x}}{1 + \frac{y}{x + 2} . \frac{y - 4}{x}} = \infty \Rightarrow \frac{xy - xy - xy + 4 x + 8}{x (x + 2) + y (y - 4)} = \infty \Rightarrow x (x + 2) + y (y - 4) = 0 and - 2 y + 4 x + 8 > 0 \Rightarrow x^{2} + y^{2} + 2 x - 4 y = 0 and 2 x - y + 4 > 0$