1. Let $z=x+iy$ Then $2z=2x+2iy$ So, $2z-3i=2x+i(2y-3), \qquad 2z+i=2x+i(2y+1)$

2. Given that $\frac{2z-3i}{2z+i}$ is purely imaginary.

For a quotient $\frac{a+ib}{c+id}$ to be purely imaginary, its real part must be zero.

Let $\frac{2x+i(2y-3)}{2x+i(2y+1)}$ Multiply numerator and denominator by the conjugate of the denominator: $\frac{(2x+i(2y-3))(2x-i(2y+1))}{(2x)^2+(2y+1)^2}$

We only need the real part of the numerator.

3. Compute the real part of the numerator: $(2x+i(2y-3))(2x-i(2y+1))$ Using $(a+ib)(c-id)=ac+bd+i(bc-ad)$ with $a=2x,\quad b=2y-3,\quad c=2x,\quad d=2y+1$ Real part: $ac+bd=(2x)(2x)+(2y-3)(2y+1)$ $=4x^2+(4y^2-4y-3)$ Since the quotient is purely imaginary, $4x^2+4y^2-4y-3=0$ Divide by $4$: $x^2+y^2-y-\frac34=0$

4. Also given: $x+y^2=0 \implies x=-y^2$ Substitute into the previous equation: $(-y^2)^2+y^2-y-\frac34=0$ $y^4+y^2-y-\frac34=0$ Hence, $y^4+y^2-y=\frac34$

5. Therefore the required value is $\boxed{\frac34}$

6. Option check:

• A: $\frac43$ ❌
• B: $\frac32$ ❌
• C: $\frac34$ ✅
• D: $\frac23$ ❌