10.35 For what real values of $x$ and $y$ are the complex numbers $z_{1}=4 i-2 x y-x y i$ and $z_{2}=y^{2} i-x^{2}+3$ equal?
10.35 For the complex numbers $z_{1}=-2 x y+(4-x y) i$ and $z_{2}=\left(3-x^{2}\right)+y^{2} i$ to be equal, the following system of equations must hold:

$\left\{\begin{array}{l}3-x^{2}=-2 x y, \\ y^{2}=4-x y\end{array} \Rightarrow\left\{\begin{array}{l}x^{2}-2 x y=3, \\ y^{2}+x y=4\end{array} \Rightarrow\right.\right.$

$\Rightarrow\left\{\begin{array}{l}4 x^{2}-8 x y=12, \\ 3 y^{2}+3 x y=12\end{array} \Rightarrow\left\{\begin{array}{l}4 x^{2}-8 x y=3 y^{2}+3 x y, \\ y^{2}+x y=4\end{array} \Rightarrow\right.\right.$

$\Rightarrow\left\{\begin{array}{l}4 x^{2}-11 x y-3 y^{2} \\ y^{2}+x y=4\end{array}=0,\left\{\begin{array}{l}4\left(\frac{x}{y}\right)^{2}-11 \frac{x}{y}-3=0, \\ y^{2}+x y=4 .\end{array}\right.\right.$

From the first equation, we find $\frac{x}{y}=3$ or $\frac{x}{y}=-\frac{1}{4}$.

Thus, we need to solve the two systems

$\left\{\begin{array}{l}\frac{x}{y}=3, \\ y^{2}+x y=4\end{array}\right.$ and $\left\{\begin{array}{l}\frac{x}{y}=-\frac{1}{4}, \\ y^{2}+x y=4 .\end{array}\right.$

The solution to the first system gives the pairs of numbers $x_{1}=3, y_{1}=1$ and $x_{2}=-3, y_{2}=-1$, and the solution to the second system gives the pairs

$$
x_{3}=\frac{\sqrt{3}}{3}, y_{3}=-\frac{4 \sqrt{3}}{3} \text { and } x_{4}=-\frac{\sqrt{3}}{3}, y_{4}=\frac{4 \sqrt{3}}{3}
$$

Answer: $\quad x_{1}=3, y_{1}=1 ; \quad x_{2}=-3, y_{2}=-1 ; \quad x_{3}=\frac{\sqrt{3}}{3}, \quad y_{3}=-\frac{4 \sqrt{3}}{3}$;

$$
x_{4}=-\frac{\sqrt{3}}{3}, y_{4}=\frac{4 \sqrt{3}}{3}
$$