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

Given: $$z = x + iy$$, $$\frac{2z - 3i}{2z + i}$$ is purely imaginary.

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

Rationalize by multiplying by conjugate of denominator:

$$= \frac{[2x + i(2y-3)][2x - i(2y+1)]}{|2x + i(2y+1)|^2}$$

Numerator real part: $$4x^2 + (2y-3)(2y+1) = 4x^2 + 4y^2 - 4y - 3$$

For the expression to be purely imaginary, the real part = 0:

$$4x^2 + 4y^2 - 4y - 3 = 0$$

Given $$x + y^2 = 0$$, so $$x = -y^2$$:

$$4y^4 + 4y^2 - 4y - 3 = 0$$

We need $$y^4 + y^2 - y$$. From the equation:

$$4(y^4 + y^2 - y) = 3$$

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

The correct answer is Option 3: $$\frac{3}{4}$$.