Let the complex number z = x + iy be such that (2z - 3i)/(2z + i)...

Given,

The complex number $z = x + iy$ be such that $\dfrac{2z - 3i}{2z + i}$ is purely imaginary,

Now putting the value of $z = x + iy$ in $\dfrac{2z - 3i}{2z + i}$ we get,

$\dfrac{2\left(x + iy\right) - 3i}{2\left(x + iy\right) + i} = \dfrac{\left(2x + i\left(2y - 3\right)\right)}{\left(2x + i\left(2y + 1\right)\right)}\dfrac{\left(2x - i\left(2y + 1\right)\right)}{\left(2x - i\left(2y + 1\right)\right)}$

Now taking real part as,

$\dfrac{4{x}^{2} + \left(2y - 3\right)\left(2y + 1\right)}{4{x}^{2} + {\left(2y + 1\right)}^{2}} = 0$

$⇒ 4{x}^{2} + 4{y}^{2} - 4y - 3 = 0$

$⇒ {x}^{2} + {y}^{2} - y - \dfrac{3}{4} = 0$

Now using,

$x + {y}^{2} = 0$$⇒ x = - {y}^{2}$ we get,

${x}^{2} + {y}^{2} - y - \dfrac{3}{4} = 0$

$⇒ {y}^{4} + {y}^{2} - y - \dfrac{3}{4} = 0$

$⇒ {y}^{4} + {y}^{2} - y = \dfrac{3}{4}$