Task B-4.3. Determine all numbers $z \in \mathbb{C}$ that are solutions to the equation $z^{3}+i^{2013}=0$. Calculate the area of the figure whose vertices in the Gaussian plane are determined by the solutions to this equation.
## Solution.

$$
\begin{gathered}
z^{3}+i^{2013}=0 \\
z^{3}=-i^{2013} \\
z^{3}=-i
\end{gathered}
$$

Let's represent $-i$ in trigonometric form.

$$
-i=\cos \frac{3 \pi}{2}+i \sin \frac{3 \pi}{2}
$$

Then,

$$
z=\sqrt[3]{-i}=\cos \frac{\frac{3 \pi}{2}+2 k \pi}{3}+i \sin \frac{\frac{3 \pi}{2}+2 k \pi}{3}, \quad k=0,1,2
$$

Thus, the solutions to the equation are the following numbers

$$
\begin{aligned}
& z_{1}=\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}=i \\
& z_{2}=\cos \frac{7 \pi}{6}+i \sin \frac{7 \pi}{6}=-\frac{\sqrt{3}}{2}-\frac{1}{2} i \\
& z_{3}=\cos \frac{11 \pi}{6}+i \sin \frac{11 \pi}{6}=\frac{\sqrt{3}}{2}-\frac{1}{2} i
\end{aligned}
$$

They determine the vertices of an equilateral triangle

$$
Z_{1}(0,1), Z_{2}\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right), Z_{3}\left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right)
$$

![](https://cdn.mathpix.com/cropped/2024_06_03_ade005c4283eff439485g-21.jpg?height=403&width=460&top_left_y=1105&top_left_x=819)

The length of the side of this triangle is

$$
a=\left|\overline{Z_{2} Z_{3}}\right|=2 \cdot \frac{\sqrt{3}}{2}=\sqrt{3}
$$

Then its area is

$$
P=\frac{a^{2} \sqrt{3}}{4}=\frac{3 \sqrt{3}}{4}
$$