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(D) Let $z = x + iy$. The expression $\frac{2z - 3i}{2z + i}$ is purely imaginary,so its real part is $0$.Let $w = \frac{2(x + iy) - 3i}{2(x + iy) + i

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$\frac{3}{4}$

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(D) Let $z = x + iy$. The expression $\frac{2z - 3i}{2z + i}$ is purely imaginary,so its real part is $0$.Let $w = \frac{2(x + iy) - 3i}{2(x + iy) + i} = \frac{2x + i(2y - 3)}{2x + i(2y + 1)}$.Multiply numerator and denominator by the conjugate of the denominator: $2x - i(2y + 1)$.The real part is $\frac{4x^2 + (2y - 3)(2y + 1)}{4x^2 + (2y + 1)^2} = 0$.Thus,$4x^2 + 4y^2 - 4y - 3 = 0$.Given $x + y^2 = 0$,we have $x = -y^2$.Substituting $x^2 = y^4$ into the equation: $4y^4 + 4y^2 - 4y - 3 = 0$.This implies $4(y^4 + y^2 - y) = 3$.Therefore,$y^4 + y^2 - y = \frac{3}{4}$.

Let the complex number $z = x + iy$ be such that $\frac{2z - 3i}{2z + i}$ is purely imaginary. If $x + y^2 = 0$,then $y^4 + y^2 - y$ is equal to:

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