Complex Numbers
Conditions on Complex Numbers
Grade 11
Question:
<p>If <i>w</i> = <i>a</i> + <i>i b</i>, where <i>b</i> ≠ 0 and <i>z</i> ≠ 1, satisfies the condition that <span>\(\frac{w - wz}{1 - z}\)</span> is purely real, then the set of values of <i>z</i> is</p>
<p>(a) |<i>z</i>| = 1, <i>z</i> ≠ 2</p>
<p>(b) |<i>z</i>| = 1 and <i>z</i> ≠ 1</p>
<p>(c) <i>z</i> = <span>\(\bar{z}\)</span></p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: For a complex number to be purely real, it must equal its conjugate. Use this condition and properties of conjugates to derive constraints on |z|.
<p><strong>Solution:</strong> Let <span>$z_1 = \frac{w - wz}{1 - z}$</span> be purely real.</p><p>Then <span>$z_1 = \bar{z_1}$</span></p><p><span>$\Rightarrow \frac{w - wz}{1 - z} = \frac{\bar{w} - \bar{w}\bar{z}}{1 - \bar{z}}$</span></p><p><span>$\Rightarrow w - wz - \bar{w}z + w\bar{z} = \bar{w} - \bar{w}\bar{z} - wz + w\bar{z}$</span></p><p><span>$\Rightarrow (w - \bar{w}) + (\bar{w} - w)|z|^2 = 0$</span></p><p><span>$\Rightarrow (w - \bar{w})(1 - |z|^2) = 0$</span></p><p>Since <i>b</i> ≠ 0, we have <i>w</i> ≠ <span>$\bar{w}$</span>, therefore |<i>z</i>| = 1 and <i>z</i> ≠ 1.</p><p>∴ Answer is (b).</p>
Correct Answer: B