## Task B-2.6.

Calculate the area of the figure determined in the Gaussian plane by the set of complex numbers $z$ for which $|z-1-i| \leq \sqrt{2}, \operatorname{Re} z \geq 0, \operatorname{Im} z \geq 0$.
## Solution.

For the given set of points $z=x+y i$, it holds that

$$
\begin{aligned}
& |x+y i-1-i| \leq \sqrt{2} \\
& x \geq 0, \quad y \geq 0
\end{aligned}
$$

Furthermore,

$$
\begin{aligned}
|(x-1)+i(y-1)| & \leq \sqrt{2} \\
\sqrt{(x-1)^{2}+(y-1)^{2}} & \leq \sqrt{2}
\end{aligned}
$$

Finally, the set of points to be plotted is

$$
(x-1)^{2}+(y-1)^{2} \leq 2, \quad x \geq 0, \quad y \geq 0
$$

We need to calculate the area of the part of the circle with center at $S(1,1)$ and $r=\sqrt{2}$, which lies in the first quadrant. This shape consists of half of the circle and an isosceles right triangle $O A B$. (See the figure!)

![](https://cdn.mathpix.com/cropped/2024_05_30_c3f5968a3447b3c0b4b2g-08.jpg?height=618&width=792&top_left_y=1390&top_left_x=642)

A precisely drawn figure . .

(2 points)

$$
\begin{aligned}
P & =\frac{1}{2} r^{2} \pi+\frac{1}{2}|O A|^{2}= \\
& =\frac{1}{2} \cdot 2 \pi+\frac{1}{2} \cdot 4=\pi+2
\end{aligned}
$$

Note: The circle with center at $S(1,1)$ and radius $r=\sqrt{2}$ can also be directly derived from the modulus: as the set of all complex numbers $z$ whose distance from $1+i$ is less than or equal to $\sqrt{2}$.

(2 points)