<p>The number of complex numbers \(z\) such that \(|z - 1| = |z + 1| = |z - i|\) is</p>
Step-by-Step Solution
Key Concept: A point equidistant from two fixed points lies on their perpendicular bisector. Find the intersection of perpendicular bisectors of segments joining pairs of the three given points (1, -1, i).
<p><strong>Step 1:</strong> From |z - 1| = |z + 1|, the point z is equidistant from 1 and -1, so z lies on the perpendicular bisector of the segment joining 1 and -1. This is the imaginary axis: z = yi where y ∈ ℝ.</p><p><strong>Step 2:</strong> From |z + 1| = |z - i|, the point z is equidistant from -1 and i. For z = yi: |yi + 1| = |yi - i|</p><p>√(1 + y²) = |y - 1|</p><p>Squaring: 1 + y² = (y - 1)²</p><p>1 + y² = y² - 2y + 1</p><p>0 = -2y, so y = 0</p><p><strong>Step 3:</strong> Therefore z = 0. Verify: |0 - 1| = 1, |0 + 1| = 1, |0 - i| = 1 ✓</p><p><strong>Step 4:</strong> Check if there are other solutions by testing the other perpendicular bisector equations (|z - 1| = |z - i|), but the constraint from |z - 1| = |z + 1| forces the imaginary axis, leaving only z = 0.</p><p>∴ The number of complex numbers is <strong>1</strong></p>
Correct Answer: C