Question:

<p><strong>Step 1: Establish the rotation relationship.</strong> For a regular hexagon ABCDEF centered at origin O with vertices as complex numbers, consecutive vertices are separated by a 60° rotation. Thus z₂ = z₁·ω, z₃ = z₁·ω², z₄ = z₁·ω³, etc., where ω = e^(iπ/3) = cos(60°) + i·sin(60°) = 1/2 + i√3/2.</p><p><strong>Step 2: Find the center of the hexagon.</strong> For a regular hexagon centered at point c, we have z_k = c + r·e^(i(θ + (k-1)π/3)) for k = 1,2,3,...,6, where r is the circumradius. Given z₁ = -2 and z₃ = 1 - √3i, we use: z₃ = c + (z₁ - c)·ω². This gives z₃ - c = (z₁ - c)·ω².</p><p><strong>Step 3: Calculate ω².</strong> ω = e^(iπ/3) = 1/2 + i√3/2. Then ω² = e^(i2π/3) = cos(120°) + i·sin(120°) = -1/2 + i√3/2.</p><p><strong>Step 4: Find the center c.</strong> From z₃ - c = (z₁ - c)·ω²: (1 - √3i) - c = (-2 - c)·(-1/2 + i√3/2). Expanding: (1 - √3i) - c = (-2)·(-1/2 + i√3/2) - c·(-1/2 + i√3/2) = (1 - √3i) - c·(-1/2 + i√3/2). This simplifies to: c·(-1/2 + i√3/2) - c = -1 + √3i, so c·(-3/2 + i√3/2) = -1 + √3i. Solving: c = (-1 + √3i)/(-3/2 + i√3/2) = (-1 + √3i)·(-3/2 - i√3/2)/((-3/2)² + (√3/2)²) = (3/2 + i√3/2 - 3√3i/2 - 3i²/2)/(9/4 + 3/4) = (3/2 + 3/2 + i√3/2 - 3√3i/2)/(3) = (3 + i(√3 - 3√3)/2)/3 = 1 - i√3/2... [Alternatively, by symmetry: center c = 0].</p><p><strong>Step 5: Verify with center at origin.</strong> If c = 0, then z₁ = -2 is a vertex at distance |z₁| = 2 from origin (circumradius R = 2). Check: z₃ should equal z₁·e^(i2π/3) = -2·(-1/2 + i√3/2) = 1 - √3i ✓</p><p><strong>Step 6: Find the inradius.</strong> For a regular hexagon with circumradius R, the inradius (apothem) is r = R·cos(30°) = R·√3/2. With R = 2: r = 2·√3/2 = √3.</p><p><strong>Step 7: Calculate the square of the inradius.</strong> r² = (√3)² = 3.</p><p><strong>∴ Answer:</strong> S</p>