Let S = \ z = x + iy: (2z - 3i)/(4z + 2i)) is a real number . Then...

We have,

$Z = \dfrac{2z - 3i}{4z + 2i}$

$⇒ Z = \dfrac{2x + i\left(2y - 3\right)}{4x + i\left(4y + 2\right)}$

$⇒ Z = \dfrac{2x + i\left(2y - 3\right)}{4x + i\left(4y + 2\right)} \times \dfrac{4x - i\left(4y + 2\right)}{4x - i\left(4y + 2\right)}$

$⇒ Z = \dfrac{8{x}^{2} + \left(4y + 2\right)\left(2y - 3\right) + i\left[4x\left(2y - 3\right) - \left(8xy + 4x\right)\right]}{16{x}^{2} + {\left(4y + 2\right)}^{2}}$

Since, $Z$ is purely real, so $\dfrac{4x\left(2y - 3\right) - \left(8xy + 4x\right)}{16{x}^{2} + {\left(4y + 2\right)}^{2}} = 0$

$⇒ - 12x + 8xy - \left(8xy + 4x\right) = 0$

$⇒ x = 0$

And,

$4x + i\left(4y + 2\right) \ne 0$

$⇒ i\left(4y + 2\right) \ne 0$

$⇒ y \ne - \dfrac{1}{2}$

Hence this is the required option.