<p><b>For Problems 12 and 13:</b> The roots of the equation \(z^4 + az^3 + (12+9i)z^2 + bz = 0\) (where \(a\) and \(b\) are complex numbers) are the vertices of a square.</p><p>The area of the square is</p>
Step-by-Step Solution
Key Concept: If four complex numbers form a square centered at origin, they must be of the form re^(iθ), re^(i(θ+π/2)), re^(i(θ+π)), re^(i(θ+3π/2)) where r is the distance from center. The coefficient of z³ (sum of roots) determines the center location, and the coefficient of z² constrains the side length.
<p><strong>Step 1:</strong> One root is z = 0 (evident from the equation having no constant term). So the square has one vertex at the origin.</p><p><strong>Step 2:</strong> Let the square have vertices at 0, w, w+iw, iw where w is a complex number. These form a square with one vertex at origin and adjacent vertices at distances |w| apart.</p><p><strong>Step 3:</strong> The roots are: 0, w, w(1+i), iw. Sum of roots = w + w(1+i) + iw = w(2+2i) = -a, so a = -w(2+2i).</p><p><strong>Step 4:</strong> Sum of products of roots taken two at a time: w·w(1+i) + w·iw + w(1+i)·iw = w²(1+i) + iw² + iw²(1+i) = w²(1+i+i+i²+i) = w²(1+3i-1) = 3iw² = 12+9i.</p><p><strong>Step 5:</strong> From 3iw² = 12+9i, we get w² = (12+9i)/(3i) = (12+9i)·(-i)/(3i·(-i)) = (-12i+9)/3 = 3-4i.</p><p><strong>Step 6:</strong> |w|² = |3-4i| = √(9+16) = 5, so |w| = √5. The side length of the square is |w| = √5.</p><p><strong>Step 7:</strong> Area of square = (√5)² = <strong>5</strong>.</p><p>∴ Answer: A</p>
Correct Answer: A