1. We need to solve $\bar z-z^2=i(\bar z+z^2).$ Let $z=x+iy, \qquad \bar z=x-iy,$ where $x,y\in\mathbb R$.

2. Rearrange the equation: $\bar z-z^2=i\bar z+iz^2$ $\bar z-i\bar z=z^2+iz^2$ $(1-i)\bar z=(1+i)z^2.$

Now divide by $(1-i)$: $\bar z=\frac{1+i}{1-i}z^2.$ Since $\frac{1+i}{1-i}=\frac{(1+i)^2}{(1-i)(1+i)}=\frac{1+2i+i^2}{2}=\frac{2i}{2}=i,$ we get $\bar z=iz^2.$ So we must solve $x-iy=i(x+iy)^2.$

3. Compute $z^2$: $(x+iy)^2=x^2-y^2+2ixy.$ Hence $iz^2=i(x^2-y^2+2ixy)=i(x^2-y^2)-2xy.$ So $iz^2=-2xy+i(x^2-y^2).$

Equating real and imaginary parts in $x-iy=-2xy+i(x^2-y^2),$ we get $x=-2xy \qquad ...(1)$ $-y=x^2-y^2 \qquad ...(2)$

4. Solve the system.

From (1): $x=-2xy$ $x(1+2y)=0.$ So either

• $x=0$, or
• $1+2y=0 \Rightarrow y=-\frac12$.

Case 1: $x=0$

Then from (2): $-y=0-y^2=-y^2$ $y=y^2$ $y(y-1)=0.$ Thus $y=0 \quad \text{or} \quad y=1.$ So the roots are $z=0, \quad z=i.$

Case 2: $y=-\frac12$

Substitute into (2): $-\left(-\frac12\right)=x^2-\left(\frac14\right)$ $\frac12=x^2-\frac14$ $x^2=\frac34.$ Thus $x=\pm \frac{\sqrt3}{2}.$ So the roots are $z=\frac{\sqrt3}{2}-\frac{i}{2}, \quad z=-\frac{\sqrt3}{2}-\frac{i}{2}.$

5. Therefore the distinct roots are: $0,\ i,\ \frac{\sqrt3}{2}-\frac{i}{2},\ -\frac{\sqrt3}{2}-\frac{i}{2}.$ Hence, the number of distinct roots is $\boxed{4}.$

6. Comparison with stored answer: Stored correct answer = $4$. This matches our result.