## Task B-2.6.

Solve the equation $i z^{2}+2 \bar{z}=0$ in the set of complex numbers. Calculate the quotient of the sum of the cubes and the product of all solutions that are different from zero.
## Solution.

Let $z=x+i y, x, y \in \mathbb{R}$. Then we have

$$
\begin{array}{r}
i \cdot(x+i y)^{2}+2(x-i y)=0 \\
i\left(x^{2}+2 x y i-y^{2}\right)+2 x-2 y i=0 \\
i x^{2}-2 x y-i y^{2}+2 x-2 y i=0
\end{array}
$$

By equating the real and imaginary parts, we get the system of equations

$$
\begin{array}{r}
-2 x y+2 x=0 \\
x^{2}-y^{2}-2 y=0
\end{array}
$$

The left side of the first equation can be factored as $-2 x(y-1)=0$.

If $x=0$, from the second equation we get $y=0$ or $y=-2$.

Since $z \neq 0$, it must be $y=-2$, which means one solution of the system is $z_{1}=-2 i$.

If $x \neq 0$, then $y=1$, and from the second equation, we get

$$
x= \pm \sqrt{3}
$$

Finally, the solutions of the equation are $z_{1}=-2 i, z_{2}=\sqrt{3}+i, z_{3}=-\sqrt{3}+i$. Let's calculate the sum of the cubes and the product of these solutions.

$$
\begin{aligned}
& z_{1}^{3}=(-2 i)^{3}=8 i \\
& z_{2}^{3}=3 \sqrt{3}+9 i+3 \sqrt{3} i^{2}+i^{3}=8 i \\
& z_{3}^{3}=-3 \sqrt{3}+9 i-3 \sqrt{3} i^{2}+i^{3}=8 i
\end{aligned}
$$

Then we have

$$
\begin{gathered}
z_{1}^{3}+z_{2}^{3}+z_{3}^{3}=3 \cdot 8 i=24 i \\
z_{1} z_{2} z_{3}=-2 i(\sqrt{3}+i)(-\sqrt{3}+i)=-2 i(-3-1)=8 i
\end{gathered}
$$

The desired quotient is $\frac{24 i}{8 i}=3$.