Let $\bar{z}$ denote the complex conjugate of a complex number $z$ and let $i = \sqrt{-1}$. In the set of complex numbers, the number of distinct roots of the equation $\bar{z} - z^{2} = i(\bar{z} + z^{2})$ is ______.

$\text{Given: } \bar{z} = \text{complex conjugate of } z$

$i = \sqrt{-1}$

$\text{To find: number of distinct roots of }$

$\bar{z} - z^{2} = i(\bar{z} + z^{2})$

$\bar{z} - z^{2} = i(\bar{z} + z^{2})$ $\Rightarrow \ \bar{z} - z^{2} = i\bar{z} + i z^{2}$ $\bar{z}(1 - i) = z^{2}(1 + i)$ $\bar{z} = z^{2} \cdot \dfrac{1+i}{1-i}$ $\dfrac{1+i}{1-i} = \dfrac{(1+i)^{2}}{1 - i^{2}} = \dfrac{2i}{2} = i$ $\bar{z} = i z^{2}$ $\text{Plan: take } |\cdot| \text{ on both sides}$ $|\bar{z}| = |z|, \qquad |i z^{2}| = |z|^{2}$ $|z| = |z|^{2}$ $|z|\left(|z| - 1\right) = 0$ $|z| = 0 \ \Rightarrow \ z = 0$ $\text{or } \ |z| = 1$ $|z| = 1 \ \Rightarrow \ \bar{z} = \dfrac{1}{z}$ $\dfrac{1}{z} = i z^{2}$ $\Rightarrow \ 1 = i z^{3}$ $z^{3} = \dfrac{1}{i} = -i$ $z^{3} + i = 0 \ \Rightarrow \ 3 \text{ distinct roots on } |z| = 1$ $-i = e^{-i\pi/2}$ $\Rightarrow \ z_k = e^{\,i(-\pi/2 + 2k\pi)/3}, \quad k = 0, 1, 2$ $z_0 = e^{-i\pi/6} = \dfrac{\sqrt{3}}{2} - \dfrac{i}{2}$ $z_1 = e^{i\pi/2} = i$ $z_2 = e^{i7\pi/6} = -\dfrac{\sqrt{3}}{2} - \dfrac{i}{2}$ $\text{Roots: } 0, \ \dfrac{\sqrt{3}}{2} - \dfrac{i}{2}, \ i, \ -\dfrac{\sqrt{3}}{2} - \dfrac{i}{2}$ $\text{Total distinct roots} = 1 + 3 = \boxed{4}$