<p>Let \(z = 1 - t + i\sqrt{t^2 + t + 2}\), where \(t\) is a real parameter. The locus of \(z\) in the Argand plane is</p>
Step-by-Step Solution
Key Concept: Separate the complex number into real and imaginary parts, then eliminate the parameter t to find the relationship between x and y coordinates. The constraint on the expression under the square root determines which portion of the resulting curve is traced.
<p><strong>Step 1:</strong> Let z = x + iy where x = 1 - t and y = √(t² + t + 2)</p><p><strong>Step 2:</strong> From x = 1 - t, we get t = 1 - x</p><p><strong>Step 3:</strong> Substitute into the imaginary part: y = √((1-x)² + (1-x) + 2)</p><p><strong>Step 4:</strong> Simplify: y = √(1 - 2x + x² + 1 - x + 2) = √(x² - 3x + 4)</p><p><strong>Step 5:</strong> Square both sides: y² = x² - 3x + 4</p><p><strong>Step 6:</strong> Rearrange: x² - 3x - y² + 4 = 0 or (x - 3/2)² - y² = 9/4 - 4 = -7/4</p><p><strong>Step 7:</strong> This gives y² - (x - 3/2)² = 7/4, which is a hyperbola</p><p><strong>Step 8:</strong> Since y = √(t² + t + 2) ≥ 0, we only consider the upper branch (y ≥ 0)</p><p><strong>Step 9:</strong> Check domain: t² + t + 2 = (t + 1/2)² + 7/4 ≥ 7/4, so y ≥ √(7/4) = √7/2</p><p>∴ The locus is the upper branch of the hyperbola y² - (x - 3/2)² = 7/4 with y ≥ √7/2</p>
Correct Answer: A