10.10 For what real values of $x$ and $y$ are the complex numbers $z_{1}=2 x^{2}-3 i-1+y i$ and $z_{2}=y+x^{2} i-3-2 i$ conjugates of each other?
10.10 Let's write these numbers as $z_{1}=\left(2 x^{2}-1\right)+(y-3) i$ and $z_{2}=(y-3)+\left(x^{2}-2\right) i$. For these numbers to be conjugates, the following system must hold:

$\left\{\begin{array}{l}2 x^{2}-1=y-3 \\ y-3=2-x^{2}\end{array}\right.$

From this, it follows that $2 x^{2}-1=2-x^{2}$, or $x^{2}=1$, i.e., $x_{1}=1, x_{2}=-1$. Then, $y_{1}=y_{2}=4$.

Answer: $x_{1}=1, y_{1}=4 ; x_{2}=-1, y_{2}=4$.