## Task B-4.6.

Determine the area of the set of all points associated with complex numbers $z$ for which

$$
|z| \leqslant\left|\frac{1}{z}\right| \quad \text{and} \quad \frac{28277 \pi}{12} \leqslant \arg \left(\frac{z^{2020}}{1+i}\right) \leqslant \frac{40397 \pi}{12}
$$
## Solution.

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

From $|z| \leqslant\left|\frac{1}{z}\right|$ it follows that $|z|^{2} \leqslant 1$, so $r \leqslant 1$.

We have

$$
\begin{aligned}
\arg \left(\frac{z^{2020}}{1+i}\right) & =\arg \frac{r^{2020}(\cos 2020 \varphi+i \sin 2020 \varphi)}{\sqrt{2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)} \\
& =\arg \left[\frac{r^{2020}}{\sqrt{2}}\left(\cos \left(2020 \varphi-\frac{\pi}{4}\right)+i \sin \left(2020 \varphi-\frac{\pi}{4}\right)\right)\right] \\
& =2020 \varphi-\frac{\pi}{4}
\end{aligned}
$$

Therefore, from the given inequality it follows that

$$
\frac{28277 \pi}{12} \leqslant 2020 \varphi-\frac{\pi}{4} \leqslant \frac{40397 \pi}{12}
$$

or $\frac{7 \pi}{6} \leqslant \varphi \leqslant \frac{5 \pi}{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_0c173970003c85556d5eg-25.jpg?height=605&width=628&top_left_y=1562&top_left_x=634)

The given set of points is a circular sector with radius 1 and central angle $\frac{5 \pi}{3}-\frac{7 \pi}{6}=\frac{\pi}{2}$.

The area is $P=\frac{1}{4} r^{2} \pi=\frac{\pi}{4}$.