<p>Let \(z = x + iy\). The maximum distance between the two points on the curve \(2|2x| = 32 + \left(\dfrac{2y}{i}\right)^2\) is ___.</p>
Step-by-Step Solution
Key Concept: Simplify the equation by recognizing that (2y/i)² = -4y², then rearrange to identify the curve as a hyperbola. The maximum distance is between the two vertices farthest apart on opposite branches.
<p><strong>Step 1:</strong> Simplify the term (2y/i)². Note that 1/i = -i, so (2y/i)² = 4y²/i² = 4y²/(-1) = -4y²</p><p><strong>Step 2:</strong> Substitute into the equation: 2|2x| = 32 - 4y², which gives |2x| = 16 - 2y²</p><p><strong>Step 3:</strong> This yields 2|x| = 16 - 2y², or |x| = 8 - y². For the right branch: x = 8 - y² (where x ≥ 0 requires y² ≤ 8)</p><p><strong>Step 4:</strong> This is a parabola in the complex plane. Rewrite as y² = 8 - x, a parabola opening leftward with vertex at (8, 0)</p><p><strong>Step 5:</strong> The two branches correspond to |x| = 8 - y²: left branch at x = -(8 - y²) and right branch at x = 8 - y². These are symmetric parabolas with vertices at (-8, 0) and (8, 0) respectively</p><p><strong>Step 6:</strong> The maximum distance between any two points on the curve is between the two vertices: distance = |8 - (-8)| = 16</p><p>∴ Answer: <strong>16</strong></p>
Correct Answer: 16