Complex Numbers
Purely imaginary complex numbers
Grade 11
Question:
<p>If \(\dfrac{z-i}{z+i}\) is a purely imaginary number (where \(z \neq -i\)), then which of the following is true?</p>
<p>\(z + \dfrac{1}{z} = z + \bar{z} = 1\)</p>
<p>\(z + \dfrac{1}{z} = z + \bar{z} = 2\)</p>
<p>\(|z| = 2\)</p>
<p>\(\text{Re}(z) = 0\)</p>
Step-by-Step Solution
Key Concept: A complex number is purely imaginary if and only if its real part is zero and imaginary part is non-zero. For a fraction to be purely imaginary, the numerator and denominator must be complex conjugates (up to a real scalar multiple), making the real part vanish.
<p><strong>Step 1:</strong> Let z = x + iy where x, y ∈ ℝ and y ≠ -1 (since z ≠ -i).</p><p><strong>Step 2:</strong> Calculate the fraction:</p><p>$$\frac{z-i}{z+i} = \frac{(x+iy)-i}{(x+iy)+i} = \frac{x+i(y-1)}{x+i(y+1)}$$</p><p><strong>Step 3:</strong> Multiply numerator and denominator by the conjugate of the denominator:</p><p>$$= \frac{[x+i(y-1)][x-i(y+1)]}{[x+i(y+1)][x-i(y+1)]}$$</p><p><strong>Step 4:</strong> Expand numerator: $$x^2 - ix(y+1) + ix(y-1) + (y-1)(y+1)$$</p><p>$$= x^2 + (y^2-1) + i[x(y-1) - x(y+1)]$$</p><p>$$= x^2 + y^2 - 1 - 2ix$$</p><p><strong>Step 5:</strong> Expand denominator: $$x^2 + (y+1)^2 = x^2 + y^2 + 2y + 1$$</p><p><strong>Step 6:</strong> For the fraction to be purely imaginary, the real part must be zero:</p><p>$$x^2 + y^2 - 1 = 0$$</p><p>$$x^2 + y^2 = 1$$</p><p><strong>Step 7:</strong> Since |z|² = x² + y², we have |z|² = 1, so |z| = 1.</p><p><strong>Conclusion:</strong> The condition is <strong>|z| = 1</strong> (z lies on the unit circle).</p><p>∴ Answer: B</p>
Correct Answer: B