Key Concepts and Formulas

• Complex Numbers: $z = x + iy$, where $x$ is the real part ($\operatorname{Re}(z)$) and $y$ is the imaginary part ($\operatorname{Im}(z)$). The complex conjugate is $\bar{z} = x - iy$.
• Modulus of a Complex Number: $|z| = \sqrt{x^2 + y^2}$. More generally, $|z - z_0|$ represents the distance between complex numbers $z$ and $z_0$.
• Area of a Circular Sector: $A = \frac{1}{2}r^2 \theta$, where $r$ is the radius and $\theta$ is the central angle in radians.

Step-by-Step Solution

Step 1: Convert the complex inequality $|z-1| \leq 2$ to Cartesian form.

• What & Why: We want to express the given complex inequality in terms of real variables $x$ and $y$ to understand the region it defines in the Cartesian plane.
• Math: Substitute $z = x + iy$: $|x + iy - 1| \leq 2$ $|(x-1) + iy| \leq 2$ $\sqrt{(x-1)^2 + y^2} \leq 2$ $(x-1)^2 + y^2 \leq 4$
• Reasoning: The modulus of a complex number is the square root of the sum of the squares of its real and imaginary parts. Squaring both sides simplifies the expression and reveals the Cartesian equation.

Step 2: Convert the complex inequality $(z+\bar{z})+i(z-\bar{z}) \leq 2$ to Cartesian form.

• What & Why: Similar to Step 1, we want to translate this inequality into Cartesian coordinates.
• Math: We know $z + \bar{z} = 2x$ and $z - \bar{z} = 2iy$. $2x + i(2iy) \leq 2$ $2x + 2i^2y \leq 2$ $2x - 2y \leq 2$ $x - y \leq 1$ $y \geq x - 1$
• Reasoning: Using the definitions of $z + \bar{z}$ and $z - \bar{z}$, we simplify the expression and isolate $y$ to get a standard linear inequality.

Step 3: Convert the complex inequality $\operatorname{Im}(z) \geq 0$ to Cartesian form.

• What & Why: This is a straightforward conversion to define the upper half-plane.
• Math: Since $\operatorname{Im}(z) = y$: $y \geq 0$
• Reasoning: The imaginary part of $z$ is simply $y$, so the inequality directly translates to $y \geq 0$.

Step 4: Identify the region of intersection.

• What & Why: We need to find the region in the $xy$-plane that satisfies all three inequalities simultaneously.

• Math & Reasoning:

  1. $(x-1)^2 + y^2 \leq 4$: Disk centered at $(1,0)$ with radius 2.
  2. $y \geq x - 1$: Region above the line $y = x - 1$.
  3. $y \geq 0$: Region above the x-axis.

  The lines $y = x - 1$ and $y = 0$ intersect at $(1, 0)$, which is the center of the circle. The line $y=x-1$ intersects the circle at $(3,2)$ and $(-1, -2)$, but since we also require $y \ge 0$, the second intersection point is not relevant. The line $y=0$ intersects the circle at $(-1, 0)$ and $(3, 0)$. The region of intersection is a sector of the circle. The line $y = x - 1$ makes an angle of $\pi/4$ with the positive x-axis. Since we're considering the region above this line and above the x-axis ($y=0$), the relevant angle spans from $\pi/4$ to $\pi$. The central angle of the sector is $\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

Step 5: Calculate the area of the circular sector.

• What & Why: Use the formula for the area of a sector to find the area of the region $S$.
• Math: $A = \frac{1}{2}r^2 \theta = \frac{1}{2}(2^2)\left(\frac{3\pi}{4}\right) = \frac{1}{2}(4)\left(\frac{3\pi}{4}\right) = \frac{3\pi}{2}$
• Reasoning: We plug in the radius $r=2$ and the central angle $\theta = \frac{3\pi}{4}$ into the sector area formula.

Common Mistakes & Tips

• Incorrectly determining the intersection of regions: Carefully sketch the regions defined by each inequality to visualize their intersection.
• Mixing up inequalities: Pay close attention to the direction of the inequalities (e.g., $y \geq x - 1$ vs. $y \leq x - 1$).
• Forgetting the $y \geq 0$ condition: This is crucial for defining the correct sector.

Summary

The problem involves converting complex inequalities into Cartesian form and finding the area of the region that satisfies all the inequalities. The region is a circular sector of a disk centered at $(1,0)$ with radius 2. The inequalities define a sector with a central angle of $\frac{3\pi}{4}$. The area of this sector is $\frac{3\pi}{2}$ square units.

The final answer is \boxed{\frac{3 \pi}{2}}, which corresponds to option (B).