Let $\overline{z}$ 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 $\overline{z}-z^2=i(\overline{z}+z^2)$ is _____________.

The correct answer is 4.

We are given the complex equation:

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

Let us rearrange terms to group $\overline{z}$ and $z^2$:

$\overline{z}-z^2=i\overline{z}+i{z}^2$

$\overline{z}(1-i)=z^2(1+i)$

Dividing both sides by $1+i$, we get:

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

To simplify $\frac{1-i}{1+i}$, multiply the numerator and denominator by the complex conjugate of the denominator, which is $1-i$:

$\frac{1-i}{1+i}=\frac{{(1-i)}^2}{1-i^2}=\frac{1-2i+i^2}{1-(-1)}=\frac{1-2i-1}{2}=\frac{-2i}{2}=-i$

Thus, the equation simplifies to:

$z^2=-i\overline{z}$

Now, let us take the modulus on both sides of the equation:

$\left|z^2\right|=|-i\overline{z}|$

Using the properties $\left|z^2\right|={\left|z\right|}^2$, $|-i|=1$, and $\left|\overline{z}\right|=\left|z\right|$:

${\left|z\right|}^2=\left|z\right|$

${\left|z\right|}^2-\left|z\right|=0$

$\left|z\right|(\left|z\right|-1)=0$

This yields two cases:

Case 1: $\left|z\right|=0$

This implies $z=0$. Substituting $z=0$ into the original equation, we get $0=0$, which is true. Therefore, $z=0$ is 1 distinct root.

Case 2: $\left|z\right|=1$

Since $\left|z\right|=1$, we know that $z\overline{z}={\left|z\right|}^2=1$, which gives $\overline{z}=\frac{1}{z}$.

Substitute $\overline{z}=\frac{1}{z}$ into $z^2=-i\overline{z}$:

$z^2=-i\left(\frac{1}{z}\right)$

$z^3=-i$

The equation $z^3=-i$ has 3 distinct roots in the complex plane (specifically, $z=i$, $z=e^{i7\pi /6}=-\frac{\sqrt{3}}{2}-\frac{1}{2}i$, and $z=e^{\mathrm{iInternal}11\pi /6}=\frac{\sqrt{3}}{2}-\frac{1}{2}i$).

Combining all cases, the total number of distinct roots is $1+3=4$.