3B. Determine all complex numbers $z$ for which

$$
z \bar{z}+1=-i(z-\bar{z})
$$
Solution. Let $z=a+ib$, where $a$ and $b$ are real numbers. Since $z \bar{z}=a^{2}+b^{2}$ and $z-\bar{z}=2bi$, the given equation transforms into $a^{2}+b^{2}+1=2b$, from which we get $a^{2}+(b-1)^{2}=0$. However, the sum of squares of real numbers is 0 if and only if the squared numbers are 0, from which we get $a=0$ and $b-1=0$, i.e., $b=1$.

Thus, there is a unique complex number that satisfies the conditions, and that is $z=i$.