Let, $$z = x + iy$$

$$\therefore$$ $$\overline z = x - iy$$

Given, $$\overline z - {z^2} = i(\overline z + {z^2})$$

$$ \Rightarrow (x - iy) - {(x + iy)^2} = i\left[ {(x - iy) + {{(x + iy)}^2}} \right]$$

$$ \Rightarrow (x - iy) - ({x^2} - {y^2} + 2ixy) = i[x - iy + {x^2} - {y^2} + 2ixy]$$

$$ \Rightarrow (x - {x^2} + {y^2}) - iy(1 + 2x) = xi - {i^2}y + {x^2}i - i{y^2} + 2{i^2}xy$$

$$ \Rightarrow (x - {x^2} + {y^2}) - iy(1 + 2x) = xi + y + i{x^2} - i{y^2} - 2xy$$

$$ \Rightarrow (x - {x^2} + {y^2}) - iy(1 + 2x) = y(1 - 2x) + i(x + {x^2} - {y^2})$$

Comparing both sides real part we get,

$$x - {x^2} + {y^2} = y - 2xy$$

$$ \Rightarrow x - {x^2} + {y^2} - y + 2xy = 0$$ ..... (1)

And comparing both sides imaginary part we get,

$$ - y(1 + 2x) = x + {x^2} - {y^2}$$

$$ \Rightarrow - y - 2xy = x + {x^2} - {y^2}$$

$$ \Rightarrow x + {x^2} - {y^2} + y + 2xy = 0$$ ...... (2)

Adding equation (1) and (2) we get,

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

$$ \Rightarrow 2x + 4xy = 0$$

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

$$\therefore$$ $$x = 0$$ or $$1 + 2y = 0 \Rightarrow y = - {1 \over 2}$$

Case 1 : When $$x = 0$$ :

Put $$x = 0$$ at equation (1), we get

$${y^2} - y = 0$$

$$ \Rightarrow y(y - 1) = 0$$

$$ \Rightarrow y = 0,1$$

$$\therefore$$ $$z = 0 + 0i$$ or $$0 + i$$

Case 2 : When $$y = - {1 \over 2}$$ :

Put $$y = - {1 \over 2}$$ in equation (1), we get

$$x - {x^2} + {1 \over 4} + {1 \over 2} - x = 0$$

$$ \Rightarrow {x^2} = {1 \over 4} + {1 \over 2}$$

$$ \Rightarrow {x^2} = {3 \over 4}$$

$$ \Rightarrow x = \, \pm \,{{\sqrt 3 } \over 2}$$

$$\therefore$$ $$z = {{\sqrt 3 } \over 2} - {i \over 2}$$ or $$z = - {{\sqrt 3 } \over 2} - {i \over 2}$$

$$\therefore$$ Number of distinct $$z = 4$$