Complex Numbers
Purely Imaginary Complex Numbers
Grade 11
Question:
<p>Given \(\dfrac{z - \alpha}{z + \alpha}\) (\(\alpha \in \mathbb{R}\)) is a purely imaginary number. Its real part is equal to zero. Then which of the following holds?</p>
<p>\(|z| = \alpha\)</p>
<p>\(|z| = 1\)</p>
<p>\(|z|^2 = \alpha\)</p>
<p>\(\text{Re}(z) = \alpha\)</p>
Step-by-Step Solution
Key Concept: For a complex number to be purely imaginary, its real part must equal zero AND it cannot be zero itself. This means the numerator and denominator must have equal real parts but opposite imaginary parts when rationalized.
<p><strong>Step 1:</strong> Let z = x + iy where x, y ∈ ℝ and α ∈ ℝ.</p><p><strong>Step 2:</strong> Express the given expression:</p><p>$$\frac{z - \alpha}{z + \alpha} = \frac{(x-\alpha) + iy}{(x+\alpha) + iy}$$</p><p><strong>Step 3:</strong> Rationalize by multiplying by the conjugate of the denominator:</p><p>$$= \frac{[(x-\alpha) + iy][(x+\alpha) - iy]}{[(x+\alpha) + iy][(x+\alpha) - iy]}$$</p><p><strong>Step 4:</strong> Expand numerator:</p><p>$$[(x-\alpha)(x+\alpha) + y^2] + i[y(x+\alpha) - y(x-\alpha)]$$</p><p>$$= [(x^2 - \alpha^2 + y^2)] + i[2y\alpha]$$</p><p><strong>Step 5:</strong> Expand denominator:</p><p>$$(x+\alpha)^2 + y^2$$</p><p><strong>Step 6:</strong> For purely imaginary, real part = 0:</p><p>$$\frac{x^2 - \alpha^2 + y^2}{(x+\alpha)^2 + y^2} = 0$$</p><p><strong>Step 7:</strong> This gives: x² - α² + y² = 0</p><p>$$\therefore |z|^2 = \alpha^2$$</p><p>$$\therefore |z| = |\alpha|$$</p><p>∴ Answer: B</p>
Correct Answer: B