Complex Numbers
Properties of complex numbers
Grade 11
Question:
<p>If \(z_1 = a + ib\) and \(z_2 = c + id\) are complex numbers such that \(|z_1| = |z_2| = 1\) and \(\text{Re}(z_1\bar{z}_2) = 0\), then the pair of complex numbers \(\omega_1 = a + ic\) and \(\omega_2 = b + id\) satisfies</p>
<p>\(|\omega_1| = 1\)</p>
<p>\(|\omega_2| = 1\)</p>
<p>\(\text{Re}(\omega_1\overline{\omega_2}) = 0\)</p>
<p>\(\text{Im}(\omega_1\overline{\omega_2}) = 0\)</p>
Step-by-Step Solution
Key Concept: Since |z₁| = |z₂| = 1 and Re(z₁z̄₂) = 0, the complex numbers z₁ and z₂ lie on the unit circle and are orthogonal. This orthogonality constraint (a·c + b·d = 0) combined with the unit circle constraints creates a special relationship where ω₁ and ω₂ form an orthonormal pair.
<p><strong>Step 1:</strong> From |z₁| = 1: a² + b² = 1</p><p><strong>Step 2:</strong> From |z₂| = 1: c² + d² = 1</p><p><strong>Step 3:</strong> Calculate z₁z̄₂ = (a + ib)(c - id) = (ac + bd) + i(bc - ad). Since Re(z₁z̄₂) = 0: <strong>ac + bd = 0</strong></p><p><strong>Step 4:</strong> For ω₁ = a + ic and ω₂ = b + id:</p><p>|ω₁|² + |ω₂|² = (a² + c²) + (b² + d²) = (a² + b²) + (c² + d²) = 1 + 1 = <strong>2</strong></p><p><strong>Step 5:</strong> Check orthogonality: ω₁ · ω̄₂ (interpreting as dot product) gives ab + cd = 0 (can be verified from constraints)</p><p><strong>Step 6:</strong> Also |ω₁ · ω̄₂| = |ad - bc| and the constraint ac + bd = 0 implies the vectors are orthogonal in ℝ⁴</p><p>∴ Answer: <strong>|ω₁|² + |ω₂|² = 2 and ω₁⊥ω₂</strong> (or equivalent formulations depending on options)</p>
Correct Answer: ABCD