<p>Let \(z = x + iy\) be a complex number, where \(x\) and \(y\) are real numbers. Let \(A\) and \(B\) be the sets defined by \(A = \{z : |z| \le 2\}\) and \(B = \{z : (1-i)z + (1+i)\bar{z} \ge 4\}\). Find the area of region \(A \cap B\).</p>
Step-by-Step Solution
Key Concept: Convert the complex inequality to a real constraint by substituting z = x + iy and its conjugate, then find the intersection of a disk with a half-plane using geometric integration.
<p><strong>Step 1: Convert set A to Cartesian form</strong></p><p>A = {z : |z| ≤ 2} is a disk centered at origin with radius 2, so x² + y² ≤ 4</p><p><strong>Step 2: Convert set B to Cartesian form</strong></p><p>Let z = x + iy, then z̄ = x - iy</p><p>(1-i)(x+iy) + (1+i)(x-iy) ≥ 4</p><p>= (x+y) + i(y-x) + (x-y) + i(x+y) ≥ 4</p><p>= (x+y + x-y) + i(y-x + x+y) ≥ 4</p><p>= 2x + i(2y) ... Recalculating:</p><p>(1-i)(x+iy) = x + iy - ix + y = (x+y) + i(y-x)</p><p>(1+i)(x-iy) = x - iy + ix + y = (x+y) + i(x-y)</p><p>Sum = 2x + 2y + i(y-x+x-y) = 2x + 2y ≥ 4</p><p>∴ x + y ≥ 2, which is the region above the line x + y = 2</p><p><strong>Step 3: Find intersection geometry</strong></p><p>The line x + y = 2 intersects circle x² + y² = 4. Distance from origin to line x + y = 2 is d = 2/√2 = √2</p><p>Since √2 < 2, the line cuts through the disk creating a circular segment.</p><p><strong>Step 4: Calculate the area</strong></p><p>The line passes through points where x² + y² = 4 and x + y = 2. Solving: x² + (2-x)² = 4 gives 2x² - 4x = 0, so x = 0 or x = 2</p><p>Intersection points: (0, 2) and (2, 0)</p><p>The angle subtended at origin: vectors to (0,2) and (2,0) make angle π/2</p><p>Area of circular sector = (1/2)r²θ = (1/2)(4)(π/2) = π</p><p>Area of triangle with vertices (0,0), (0,2), (2,0) = (1/2)(2)(2) = 2</p><p>Area of segment (region closer to origin) = π - 2</p><p>Area of A ∩ B = Area of disk - Area of near segment = 4π - (4π - (π - 2)) = π - 2</p><p><strong>∴ Answer: π - 2</strong></p>
Correct Answer: π - 2