Complex Numbers
Algebra of Complex Numbers
Grade 11
Question:
<p>Given <br/> \(z = 1 + i\alpha \Rightarrow z^2 = 1 - \alpha^2 + 2i\alpha\)<br/>If \(x + iy = z^2\), then which of the following is satisfied?</p><p>(2) \(y^2 + 4x - 4 = 0\)</p>
<p>\(y^2 + 4x + 4 = 0\)</p>
<p>\(y^2 + 4x - 4 = 0\)</p>
<p>\(y^2 - 4x + 4 = 0\)</p>
<p>\(y^2 - 4x - 4 = 0\)</p>
Step-by-Step Solution
Key Concept: When z = 1 + iα, squaring it gives z² = (1 - α²) + 2iα. Equating x + iy = z², we get x = 1 - α² and y = 2α, from which we can eliminate α to find the locus relation between x and y.
<p><strong>Step 1:</strong> From z = 1 + iα, we square to get z² = (1 + iα)² = 1 + 2iα + (iα)² = 1 + 2iα - α²</p><p>Therefore: z² = (1 - α²) + 2iα</p><p><strong>Step 2:</strong> Equating x + iy = z², we obtain:<br/>x = 1 - α² ... (1)<br/>y = 2α ... (2)</p><p><strong>Step 3:</strong> From equation (2): α = y/2</p><p><strong>Step 4:</strong> Substitute α = y/2 into equation (1):<br/>x = 1 - (y/2)²<br/>x = 1 - y²/4<br/>4x = 4 - y²<br/>y² + 4x - 4 = 0</p><p><strong>Verification:</strong> This is a parabola opening left with vertex at (1, 0), which is the locus of all points (x, y) obtained by squaring complex numbers of the form 1 + iα.</p><p>∴ Answer: B</p>
Correct Answer: B