<p>If complex number \(z = x + iy\) satisfies the equation \(\text{Re}(z + 1) = |z - 1|\), then \(z\) lies on:</p>
Step-by-Step Solution
Key Concept: Separate the condition Re(z+1) = |z-1| into real and imaginary parts by substituting z = x+iy, then simplify to identify the geometric locus (curve/line).
<p><strong>Step 1:</strong> Let z = x + iy where x, y ∈ ℝ.</p><p><strong>Step 2:</strong> Find Re(z+1):<br/>z + 1 = (x+1) + iy<br/>∴ Re(z+1) = x+1</p><p><strong>Step 3:</strong> Find |z-1|:<br/>z - 1 = (x-1) + iy<br/>∴ |z-1| = √[(x-1)² + y²]</p><p><strong>Step 4:</strong> Apply the given condition Re(z+1) = |z-1|:<br/>x + 1 = √[(x-1)² + y²]</p><p><strong>Step 5:</strong> Square both sides (note: x+1 ≥ 0 for validity):<br/>(x+1)² = (x-1)² + y²<br/>x² + 2x + 1 = x² - 2x + 1 + y²<br/>4x = y²</p><p><strong>Step 6:</strong> This is the equation of a parabola with vertex at origin, opening rightward, with parameter a = 1.<br/>∴ Answer: Parabola y² = 4x (or the locus is a parabola)</p>
Correct Answer: A