Complex Numbers
Purely Imaginary Complex Numbers
Grade 11
Question:
<p>If <span>\(\frac{z - a}{z + a}\)</span> (<i>a</i> ∈ ℝ) is a purely imaginary number and |<i>z</i>| = 2, then a value of <i>a</i> is:</p>
<p>(a) <span>\(\frac{1}{2}\)</span></p>
<p>(b) <span>\(\frac{1}{\sqrt{2}}\)</span></p>
<p>(c) 1</p>
<p>(d) 2</p>
Step-by-Step Solution
Key Concept: A complex number is purely imaginary when its real part is zero. Combined with the modulus constraint, this determines the value of the real parameter.
<p><strong>Note:</strong> The solution text is incomplete in the provided page. Based on the problem setup:</p><p>Since <span>$\frac{z - a}{z + a}$</span> is purely imaginary, its real part must be zero. Let <i>z</i> = 2<i>e</i><sup><i>iθ</i></sup> with |<i>z</i>| = 2. For the quotient to be purely imaginary, <i>z</i> must satisfy the condition that makes the real part of the numerator and denominator relationship zero.</p><p>Using the property that a complex number is purely imaginary if and only if it equals the negative of its conjugate, and given |<i>z</i>| = 2, we find <i>a</i> = 2.</p>
Correct Answer: D