Sum of squares of modulus of all the complex numbers $$z$$ satisfying $$\bar{z} = iz^2 + z^2 - z$$ is equal to

Given,

$$\bar z=iz^2+z^2-z$$

Let

$$z=x+iy$$

Then,

$$\bar z=x-iy$$

Substituting,

$$x-iy=(1+i)(x+iy)^2-(x+iy)$$

Comparing real and imaginary parts,

$$2x=x^2-y^2-2xy$$

and

$$x^2-y^2+2xy=0$$

Using the second equation in the first,

$$2x=-4xy$$

$$x(1+2y)=0$$

Hence,

$$x=0\quad \text{or}\quad y=-\frac12$$

Case 1:

$$x=0$$

Then from

$$x^2-y^2+2xy=0$$

we get

$$y=0$$

Hence,

$$z=0$$

Case 2:

$$y=-\frac12$$

Substituting in

$$x^2-y^2+2xy=0$$

$$x^2-\frac14-x=0$$

$$4x^2-4x-1=0$$

$$2x-1=\pm\sqrt2$$

$$x=\frac{1\pm\sqrt2}{2}$$

Thus,

$$z=\frac{1+\sqrt2}{2}-\frac{i}{2}$$

or

$$z=\frac{1-\sqrt2}{2}-\frac{i}{2}$$

Now,

$$|0|^2=0$$

$$\left|\frac{1+\sqrt2}{2}-\frac{i}{2}\right|^2 = \frac{(1+\sqrt2)^2+1}{4}$$

$$\left|\frac{1-\sqrt2}{2}-\frac{i}{2}\right|^2 = \frac{(1-\sqrt2)^2+1}{4}$$

Therefore,

$$0+\frac{(1+\sqrt2)^2+1}{4}+\frac{(1-\sqrt2)^2+1}{4}=2$$

Hence,

$$\boxed{2}$$