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

The Correct Option is B

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: \[ \text{Area} = \text{Area of semicircle} - \text{Area of sector A} = 2\pi - \pi = \frac{3\pi}{2}. \]