The equation rearranges to $\bar{z}(1-i)=z^2(1+i)$. Taking modulus on both sides, $|\bar{z}|\cdot \sqrt{2}=|z{|}^2\cdot \sqrt{2}$, hence $|z|=|z{|}^2$. Therefore $|z|=0$ or $|z|=1$.

If $|z|=0$ then $z=0$, which is one solution. For $|z|=1$, write $z=e^{i\theta }$, so $\bar{z}=e^{-i\theta }$ and $z^2=e^{2i\theta }$. Substituting into the relation and using $1-i=\sqrt{2}{e}^{-i\pi /4}$, $1+i=\sqrt{2}{e}^{i\pi /4}$:

$e^{-i\theta }\cdot e^{-i\pi /4}=e^{2i\theta }\cdot e^{i\pi /4}$, giving $-\theta -\frac{\pi }{4}\equiv 2\theta +\frac{\pi }{4}(\mathrm{mod}2\pi )$.

Hence $3\theta =-\frac{\pi }{2}+2n\pi$, i.e., $\theta =\frac{(4n-1)\pi }{6}$, $n\in \mathbb{Z}$. This yields exactly three distinct values of $z$ on the unit circle.

Combined with $z=0$, the total number of distinct roots is $4$.