Complex Numbers
Locus in Complex Plane
Grade 11
Question:
<p>\(z \neq 1\) and \(\dfrac{z^2}{z - 1}\) is real, then the point represented by the complex number \(z\) lies</p>
<p>either on the real axis or on a circle passing through the origin.</p>
<p>on a circle with centre at the origin.</p>
<p>either on the real axis or on a circle not passing through the origin.</p>
<p>on the imaginary axis.</p>
Step-by-Step Solution
Key Concept: If a complex number expression is real, its imaginary part must be zero. Express z = x + iy and use the condition that Im(z²/(z-1)) = 0 to find the locus.
<p><strong>Step 1:</strong> Let z = x + iy where x, y ∈ ℝ and z ≠ 1.</p><p><strong>Step 2:</strong> Calculate z² = (x + iy)² = x² - y² + 2ixy</p><p><strong>Step 3:</strong> Calculate z - 1 = (x - 1) + iy</p><p><strong>Step 4:</strong> Find z²/(z - 1) by multiplying numerator and denominator by the conjugate (x - 1) - iy:</p><p>z²/(z - 1) = [(x² - y² + 2ixy)·((x - 1) - iy)] / [(x - 1)² + y²]</p><p><strong>Step 5:</strong> Expand the numerator:<br/>= [(x² - y²)(x - 1) + 2xy²] + i[2xy(x - 1) - (x² - y²)y]</p><p><strong>Step 6:</strong> For this to be real, the imaginary part = 0:<br/>2xy(x - 1) - y(x² - y²) = 0<br/>y[2x(x - 1) - (x² - y²)] = 0<br/>y[2x² - 2x - x² + y²] = 0<br/>y[x² - 2x + y²] = 0</p><p><strong>Step 7:</strong> Either y = 0 (the real axis) OR x² - 2x + y² = 0<br/>(x - 1)² + y² = 1 (a circle with center (1, 0) and radius 1)</p><p><strong>Step 8:</strong> Excluding z = 1 (where the denominator is zero), the locus is the circle (x - 1)² + y² = 1 together with the real axis (excluding the point z = 1).</p><p>∴ <strong>Answer: The point z lies on a circle with center (1, 0) and radius 1, or on the real axis (excluding z = 1)</strong></p>
Correct Answer: A