## Task B-4.6.

Calculate the area of the figure whose vertices are the solutions of the following equation in the set $\mathbb{C}$:

$$
\left(\frac{z}{z+1}\right)^{2}+\left(\frac{z}{z-1}\right)^{2}=-1
$$
## First Solution.

Multiplying the equation by $\left(z^{2}-1\right)^{2}$, we get

$$
z^{2}(z-1)^{2}+z^{2}(z+1)^{2}=-\left(z^{2}-1\right)^{2}
$$

1 point

or, after simplification,

$$
z^{4}=-\frac{1}{3}
$$

. The solutions are all complex numbers $z$ for which

$$
\begin{aligned}
& z^{4}=\frac{1}{3}(\cos \pi+i \sin \pi) \\
& \Rightarrow z_{k}=\frac{\sqrt[4]{27}}{3}\left(\cos \frac{\pi+2 k \pi}{4}+i \sin \frac{\pi+2 k \pi}{4}\right), k=0,1,2,3
\end{aligned}
$$

The numbers $z_{0}, z_{1}, z_{2}, z_{3}$ are the vertices of a square in the complex plane that lie on a circle of radius $r=\frac{\sqrt[4]{27}}{3}$.

The area of the resulting square is

$$
P=\frac{d^{2}}{2}=\frac{(2 r)^{2}}{2}=2 r^{2}=2\left(\frac{\sqrt[4]{27}}{3}\right)^{2}=\frac{2 \sqrt{3}}{2}
$$