Task B-2.7. Calculate the area of that part of the complex plane in which all complex numbers $z$ for which $\operatorname{Im}\left(z^{2}\right) \leq \operatorname{Re}(z) \text{ and } |z| \leq 1$ are located.
## Solution.

Let $z=x+y i$, where $x, y \in \mathbb{R}$. Then,

$$
z^{2}=\left(x^{2}-y^{2}\right)+2 x y i
$$

The inequality $\operatorname{Im}\left(z^{2}\right) \leq \operatorname{Re}(z)$ is equivalent to the inequality

$$
\begin{aligned}
& 2 x y \leq x \\
\Leftrightarrow & x(2 y-1) \leq 0 \\
\Leftrightarrow & \left\{\begin{array}{l}
x \leq 0, y \geq \frac{1}{2} \\
x \geq 0, y \leq \frac{1}{2}
\end{array}\right.
\end{aligned}
$$

The shaded set represents the obtained system of inequalities.

![](https://cdn.mathpix.com/cropped/2024_05_30_8c954f607b7f1570b01bg-10.jpg?height=420&width=439&top_left_y=880&top_left_x=819)

The inequality $|z| \leq 1$ is satisfied for all complex numbers that lie within the circle of radius 1, centered at the origin.

(2 points) The part of the shaded set that is inside the unit circle is the desired region of the plane. Due to the symmetry of the unit circle with respect to the y-axis, the area of this set of points is equal to the area of half the circle, i.e.,

$$
P=\frac{r^{2} \pi}{2}=\frac{\pi}{2}
$$

## SCHOOL/CITY COMPETITION IN MATHEMATICS

## 3rd grade - high school - B variant

27. January 2014.

IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO EVALUATE AND GRADE THE APPROACH APPROPRIATELY.