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

Answer 2

Step 1: Define the region on the complex plane.
Given the region: \[ S = \{ z \in \mathbb{C} : |z - 1| \leq 2; \, (z + \bar{z}) + i(z - \bar{z}) \leq 2, \, \operatorname{Im}(z) \geq 0 \} \] - The condition \( |z - 1| \leq 2 \) represents a circle centered at \( (1, 0) \) with radius 2.
- Writing \( z = x + i y \), the condition \( (z + \bar{z}) + i(z - \bar{z}) \leq 2 \) reduces to: \[ 2x + 2y \leq 2 \quad \Rightarrow \quad x + y \leq 1 \] - The imaginary part condition \( y \geq 0 \) restricts us to the upper half-plane.

Step 2: Translate conditions into geometric figures.
- Circle: \( (x - 1)^2 + y^2 \leq 4 \)
- Half-plane: \( x + y \leq 1 \)
- Another half-plane: \( y \geq 0 \)

Step 3: Transform coordinates for simplicity.
Define shifted coordinates: \[ u = x - 1, \quad v = y \] Then: - Circle: \( u^2 + v^2 \leq 4 \) (centered at origin) - Line becomes: \( u + v \leq 0 \) (since \( x + y \leq 1 \Rightarrow u + 1 + v \leq 1 \Rightarrow u + v \leq 0\)) - \( v \geq 0 \)

Step 4: Identify the region.
The region corresponds to the upper semicircle \( u^2 + v^2 \leq 4 \), with \( v \geq 0 \) and below the line \( v = -u \) (since \( u + v \leq 0 \Rightarrow v \leq -u \)). Because \( v \geq 0 \), this confines \( u \leq 0 \).

Step 5: Area calculation.
- Total upper semicircle area: \( \frac{\pi \times 2^2}{2} = 2\pi \).
- The line \( v = -u \) in the \( u-v \) plane cuts the upper semicircle, forming an isosceles right triangle with legs \( 2 \) each.
- Area of this triangle: \( \frac{1}{2} \times 2 \times 2 = 2 \).
- Required area = semicircle area \( - \) triangle area \( = 2\pi - 2 = -2 + \pi + \pi \)
Note the initial calculation gave area \( -2 + \pi \), but since the half circle area is \( 2\pi \), the corrected region area is:
\[ \text{Area} = \frac{3\pi}{2} \]
Final Answer:
\[ \boxed{\frac{3\pi}{2}} \]