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 \(\text{Re}(z_1\overline{z_2})=0\), then the pair of complex numbers \(w_1 = a+ic\) and \(w_2 = b+id\) satisfies</p>
<p>\(|w_1|=|w_2|=1\)</p>
<p>\(|w_2|\neq|w_2|\)</p>
<p>\(\text{Re}(w_1\overline{w_2})=0\)</p>
<p>\(\text{Re}(w_1\overline{w_2})=-1\)</p>
Step-by-Step Solution
Key Concept: When |z₁|=|z₂|=1 and Re(z₁z̄₂)=0, the real and imaginary parts of z₁ and z₂ form orthogonal vectors. This orthogonality translates to |w₁|=|w₂|=1 and w₁⊥w₂ in the complex plane representation.
<p><strong>Step 1:</strong> From |z₁|=1: a²+b²=1, and from |z₂|=1: c²+d²=1</p><p><strong>Step 2:</strong> Compute z₁z̄₂ = (a+ib)(c-id) = (ac+bd)+i(bc-ad)</p><p><strong>Step 3:</strong> From Re(z₁z̄₂)=0: ac+bd=0 (the vectors (a,b)·(c,d)=0 are orthogonal)</p><p><strong>Step 4:</strong> For w₁=a+ic and w₂=b+id:<br/>|w₁|² = a²+c² and |w₂|² = b²+d²<br/>Since a²+b²=1 and c²+d²=1, and ac+bd=0, we can verify: (a²+c²)+(b²+d²) = (a²+b²)+(c²+d²) = 2</p><p><strong>Step 5:</strong> Also check: Re(w₁w̄₂) = Re[(a+ic)(b-id)] = ab+cd. From ac+bd=0 and the constraint equations, one can show |w₁|=|w₂|=1 and w₁⊥w₂ (as vectors in ℂ)</p><p>∴ Answer: A (typically |w₁|=|w₂|=1 and Re(w₁w̄₂)=0, or w₁w̄₂ is purely imaginary)</p>
Correct Answer: A