## Task B-4.7.

In the set of complex numbers, solve the equation $z^{3}+|z|=0$.
## First Solution.

Let's write the number $z$ in trigonometric form, $z = r \cos \varphi + i \sin \varphi, r \geqslant 0, 0 \leqslant \varphi < 2 \pi$.

Then from the given equation, we have

$$
\begin{gathered}
r^{3}(\cos 3 \varphi + i \sin 3 \varphi) + r = 0 \\
r\left(r^{2} \cos 3 \varphi + r^{2} i \sin 3 \varphi + 1\right) = 0
\end{gathered}
$$

Then either $r = 0$ or $r^{2} \cos 3 \varphi + r^{2} i \sin 3 \varphi + 1 = 0$. 2 points

If $r = 0$, then $z = 0$ is a solution to the given equation.

If $r^{2} \cos 3 \varphi + r^{2} i \sin 3 \varphi + 1 = 0$, then $r^{2} \sin 3 \varphi = 0$ and $r^{2} \cos 3 \varphi + 1 = 0$. 2 points From the first equation, we get

$$
\sin 3 \varphi = 0 \Rightarrow \varphi \in \left\{0, \frac{\pi}{3}, \frac{2 \pi}{3}, \pi, \frac{4 \pi}{3}, \frac{5 \pi}{3}\right\}
$$

From the second equation, we have

$$
r^{2} \sin 3 \varphi = -1 \Rightarrow \cos 3 \varphi < 0 \Rightarrow \varphi \in \left\{\frac{\pi}{3}, \pi, \frac{5 \pi}{3}\right\}, r = 1
$$

2 points

The solutions to the equation are

$z_{1} = 0, z_{2} = 1\left(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}\right) = \frac{1}{2} + i \frac{\sqrt{3}}{2}, z_{3} = \cos \pi + i \sin \pi = -1, z_{4} = \cos \frac{5 \pi}{3} + i \sin \frac{5 \pi}{3} = \frac{1}{2} - i \frac{\sqrt{3}}{2}$.

2 points