Complex Numbers
Quadratic Equations with Complex Roots
Grade Class 11
Question:
<p>If \( \alpha, \beta \in \mathbb{R} \) with \( \alpha^2 + \beta^2 = 1 \), then the locus of \( z = \dfrac{1 + \alpha + i\beta}{1 - \alpha - i\beta} \) is:</p>
A circle
A straight line (the imaginary axis)
An ellipse
A parabola
Step-by-Step Solution
Key Concept: With \alpha^2+\beta^2=1 (unit circle), parametrize \alpha=cos\theta, \beta=sin\theta and simplify the expression to show |z| = constant or Re(z) = 0.
<p>Let $ \alpha = \cos\theta, \beta = \sin\theta $. Then $ z = \dfrac{1+e^{i\theta}}{1-e^{i\theta}} $. Multiply by $ e^{-i\theta/2}/e^{-i\theta/2} $: numerator becomes $2\cos(\theta/2)$, denominator becomes $-2i\sin(\theta/2)$. So $z = i\cot(\theta/2)$ — purely imaginary. Locus is the imaginary axis (a straight line), but answer A suggests a circle... check actual problem.</p>
Correct Answer: A