## Task B-2.7.

Determine the area of the set of points defined by the solutions of the inequalities $\operatorname{Im} z \geqslant|\operatorname{Re} z-1|,|z-1| \leqslant 1$ in the complex plane.
## Solution.

The solution to the inequality $|z-1| \leqslant 1$ is the set of all points $z$ that lie within the circle and on the circumference with center at point $S(1,0)$ and radius 1.

If $z=x+i y, x, y \in \mathbb{R}$, then the inequality $\operatorname{Im} z \geqslant|\operatorname{Re} z-1|$ translates into a system of inequalities

$$
\begin{gathered}
y \geqslant|x-1| \\
-y \leqslant x-1 \leqslant y, \quad y \geqslant 0 \\
y \geqslant x-1 \quad \text { and } \quad y \geqslant-x+1
\end{gathered}
$$

The set of solutions to this system and the solution to the inequality $|z-1| \leqslant 1$ is graphically represented as follows.

![](https://cdn.mathpix.com/cropped/2024_05_30_d3836afd55956f7fff9cg-12.jpg?height=620&width=968&top_left_y=284&top_left_x=453)

(1 point for the circle, 1 point for the lines, 1 point for the marked intersection)

The intersection of these sets of points is a circular segment with a central angle of $90^{\circ}$ (since the lines $y=x-1$ and $y=-x+1$ are perpendicular to each other).

Then, the area of the drawn circular segment is one-fourth of the area of the entire circle, i.e., $P=\frac{1}{4} \cdot 1^{2} \pi=\frac{\pi}{4}$.

## SCHOOL/CITY COMPETITION IN MATHEMATICS

3rd grade - high school - B variant
January 4, 2019.

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