Complex Numbers
Roots of Unity
Grade None
Question:
<p>If \(\alpha\) is a complex \(n^{\text{th}}\) root of unity and if \(z_1\) and \(z_2\) are two complex numbers, then \(\displaystyle\sum_{r=0}^{n-1}\left|z_1+\alpha^r z_2\right|^2 =\)</p>
<p>\(n^2|z_1+z_2|^2\)</p>
<p>\(\left(\dfrac{z_1}{n}+\dfrac{z_2}{n}\right)^2\)</p>
<p>\(n\left(|z_1|^2+|z_2|^2\right)\)</p>
<p>\(n^2\left(|z_1|^2+|z_2|^2\right)\)</p>
Step-by-Step Solution
Key Concept: Expand |z₁ + α^r z₂|² and use the fact that Σ(α^r) = 0 for nth roots of unity (except r=0 contributes n). The sum telescopes because cross terms involving powers of α vanish due to the orthogonality property of roots of unity.
<p><strong>Step 1:</strong> Expand |z₁ + α^r z₂|²</p><p>|z₁ + α^r z₂|² = (z₁ + α^r z₂)(z̄₁ + ᾱ^r z̄₂)</p><p>= |z₁|² + |z₂|² + z₁ᾱ^r z̄₂ + z̄₁α^r z₂</p><p><strong>Step 2:</strong> Sum over r = 0 to n-1</p><p>Σ|z₁ + α^r z₂|² = Σ|z₁|² + Σ|z₂|² + Σz₁ᾱ^r z̄₂ + Σz̄₁α^r z₂</p><p>= n|z₁|² + n|z₂|² + z₁z̄₂Σᾱ^r + z̄₁z₂Σα^r</p><p><strong>Step 3:</strong> Apply the root of unity property</p><p>Since α is an nth root of unity, Σ(r=0 to n-1) α^r = 0 (sum of all nth roots of unity)</p><p>Therefore: Σᾱ^r = Σα^(-r) = 0 and Σα^r = 0</p><p><strong>Step 4:</strong> Final result</p><p>Σ|z₁ + α^r z₂|² = n|z₁|² + n|z₂|² + 0 + 0</p><p>∴ Answer: <strong>n(|z₁|² + |z₂|²)</strong></p>
Correct Answer: C