<p>If <em>z</em> = <em>x</em> + <em>iy</em>, then the equation |(2<em>z</em> − <em>i</em>)/(<em>z</em> + 1)| = <em>m</em> represents a circle, then <em>m</em> can be</p>
Step-by-Step Solution
Key Concept: When a complex equation of form |az + b|/|cz + d| = m represents a circle, m must be a positive real number not equal to 1 (m ≠ 1). The constraint comes from the Apollonius circle condition: points whose distances to two fixed points have a constant ratio form a circle only when this ratio is not 1.
<p><strong>Step 1:</strong> Rewrite the given equation |(2z − i)/(z + 1)| = m as |2z − i| = m|z + 1|</p><p><strong>Step 2:</strong> Substitute z = x + iy: |2x + i(2y − 1)| = m|x + 1 + iy|</p><p><strong>Step 3:</strong> Expand both sides: √[4x² + (2y − 1)²] = m√[(x + 1)² + y²]</p><p><strong>Step 4:</strong> Square both sides: 4x² + 4y² − 4y + 1 = m²(x² + 2x + 1 + y²)</p><p><strong>Step 5:</strong> Rearrange: (4 − m²)x² + (4 − m²)y² − 4y − 2m²x + (1 − m²) = 0</p><p><strong>Step 6:</strong> For a circle to exist, the coefficient of x² must equal the coefficient of y², which is satisfied here. Additionally, m ≠ 1 (otherwise we get a straight line). Also m must be positive and m ≠ 2 (to avoid degeneracy).</p><p><strong>Step 7:</strong> Therefore, m can be any positive real number except 1 and 2. The answer options would typically give specific values like m = 3, m = 2, m = 0.5, etc. Any value like m = 3 (m > 1 and m ≠ 2) represents a valid circle.</p><p>∴ Answer: 3</p>
Correct Answer: 3