Question

The area (in square units) of the region $S = \\{ z \in \mathbb{C} : |z - 1| \leq 2; (z + \bar{z}) + i (z - \bar{z}) \leq 2, \, \operatorname{Im}(z) \geq 0 \\}$ is:

• A. $\frac{7\pi}{3}$
• B. $\frac{3\pi}{2}$
• C. $\frac{17\pi}{8}$
• D. $\frac{7\pi}{4}$

Answer 1

Answer: (B)

$\frac{3\pi}{2}$

Answer 2

Let $z = x + iy$, where $x, y \in \mathbb{R}$. Rewrite the given inequalities:

From $|z - 1|^2 \leq 4$:
$|z - 1|^2 = (x - 1)^2 + y^2 \leq 4 \implies (x - 1)^2 + y^2 \leq 4.$ This represents a circle with center $(1, 0)$ and radius $2$.

1. From $z + \overline{z} \geq 2$:
   $z + \overline{z} = 2x \implies x \geq 1.$ This represents the half-plane to the right of the line $x = 1$.
2. From $\operatorname{Im}(z) \geq 0$:
   $\operatorname{Im}(z) = y \implies y \geq 0.$ This represents the upper half-plane.

Step 1: Identify the region of intersection.
The region of intersection is the upper semicircular region of the circle $(x - 1)^2 + y^2 \leq 4$ to the right of $x = 1$.

Step 2: Compute the area.
The area of the semicircle is: $\text{Area of semicircle} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (2^2) = 2\pi.$ The area excluded by the sector to the left of $x = 1$ (sector A) is: $\text{Area of sector A} = \frac{\pi r^2}{4} = \frac{1}{4} \pi (2^2) = \pi.$

Step 3: Subtract the areas.
The required area is: $\text{Area} = \text{Area of semicircle} - \text{Area of sector A} = 2\pi - \pi = \frac{3\pi}{2}.$