Complex Numbers
Locus of Complex Numbers
Grade 11
Question:
<p>Let \(a, b \in \mathbb{R}\) and \(a^2 + b^2 \neq 0\). Suppose \(S = \left\{z \in C : z = \dfrac{1}{a + ibt},\, t \in \mathbb{R},\, t \neq 0\right\}\), where \(i = \sqrt{-1}\). If \(z = x + iy\) and \(z \in S\), then \((x, y)\) lies on</p>
<p>(1) the circle with radius \(\dfrac{1}{2a}\) and centre \(\left(\dfrac{1}{2a}, 0\right)\) for \(a > 0\), \(b \neq 0\)</p>
<p>(2) the circle with radius \(-\dfrac{1}{2a}\) and centre \(\left(-\dfrac{1}{2a}, 0\right)\) for \(a < 0\), \(b \neq 0\)</p>
<p>(3) the \(x\)-axis for \(a \neq 0\), \(b = 0\)</p>
<p>(4) the \(y\)-axis for \(a = 0\), \(b \neq 0\)</p>
Step-by-Step Solution
Key Concept: Rationalize z = 1/(a + ibt) to get x + iy form, then eliminate the parameter t to find the locus by recognizing that x² + y² = x/a represents a circle.
<p><strong>Step 1: Rationalize the complex number</strong></p><p>Let z = 1/(a + ibt). Multiply numerator and denominator by the conjugate:</p><p>z = (a - ibt)/(a² + b²t²) = a/(a² + b²t²) - i·bt/(a² + b²t²)</p><p><strong>Step 2: Identify x and y coordinates</strong></p><p>If z = x + iy, then:</p><p>x = a/(a² + b²t²)</p><p>y = -bt/(a² + b²t²)</p><p><strong>Step 3: Eliminate the parameter t</strong></p><p>From the expressions above: x(a² + b²t²) = a and y(a² + b²t²) = -bt</p><p>Dividing: y/x = -bt/a, so t = -ay/(bx)</p><p>Also note: x² + y² = [a² + b²t²]/(a² + b²t²)² = 1/(a² + b²t²)</p><p>Since x = a/(a² + b²t²), we have: (a² + b²t²) = a/x</p><p><strong>Step 4: Derive the locus equation</strong></p><p>From x² + y² = 1/(a² + b²t²) and a/(a² + b²t²) = x:</p><p>x² + y² = x/a</p><p>Rearranging: x² - x/a + y² = 0, or (x - 1/(2a))² + y² = 1/(4a²)</p><p><strong>Step 5: Conclusion</strong></p><p>This is a circle with center (1/(2a), 0) and radius 1/(2|a|), excluding the origin (which corresponds to t = 0).</p><p>∴ Answer: AD (Circle passing through origin with specific center and radius properties)</p>
Correct Answer: AD