## Task A-4.4.

A quadrilateral with vertices $0, z, \frac{1}{z} \text{ and } z+\frac{1}{z}$ in the complex plane has an area of $\frac{35}{37}$. Determine the smallest possible value of the expression $\left|z+\frac{1}{z}\right|^{2}$.
## Solution.

The observed quadrilateral is a parallelogram, because $z+\frac{1}{z}$ is the sum of the complex numbers $z$ and $\frac{1}{z}$.

Let $z=r(\cos \varphi+i \sin \varphi)$ be the trigonometric form of the complex number $z$.

Then $\frac{1}{z}=\frac{1}{r}(\cos \varphi-i \sin \varphi)$.

![](https://cdn.mathpix.com/cropped/2024_05_30_44b5caa35f0d2fab7d89g-18.jpg?height=502&width=697&top_left_y=894&top_left_x=611)

The angle between the two sides of the parallelogram is $2 \varphi$, so its area is equal to $|z| \cdot \frac{1}{|z|} \cdot \sin 2 \varphi=\frac{35}{37}$, i.e., $\sin 2 \varphi=\frac{35}{37}$.

The desired expression can be transformed as follows:

$$
\begin{aligned}
\left|z+\frac{1}{z}\right|^{2} & =\left(z+\frac{1}{z}\right) \cdot\left(\bar{z}+\frac{1}{\bar{z}}\right) \\
& =z \bar{z}+\frac{z}{\bar{z}}+\frac{\bar{z}}{z}+\frac{1}{z \bar{z}}=r^{2}+\frac{1}{r^{2}}+2 \frac{z^{2}+\bar{z}^{2}}{r^{2}} \\
& =r^{2}+\frac{1}{r^{2}}+2 \cos 2 \varphi
\end{aligned}
$$

From $\sin 2 \varphi=\frac{35}{37}$, it follows that $\cos 2 \varphi= \pm \sqrt{1-\frac{35^{2}}{37^{2}}}= \pm \frac{12}{37}$.

Since $r^{2}+\frac{1}{r^{2}} \geqslant 2$,

the desired minimum is equal to $2-2 \cdot \frac{12}{37}=\frac{50}{37}$.

Note: The area of the parallelogram does not depend on $r$ but only on $\varphi$. There are two values of the angle $\varphi$ for which the area of the parallelogram is equal to the given value. The expression whose minimum is sought represents the square of the length of the diagonal of the observed parallelogram, so most conclusions can be reached through geometric considerations.