Complex Numbers
Roots of unity
Grade 11
Question:
<p>If \(\omega\) is a complex <em>n</em>th root of unity, then \(\sum_{r=1}^{n}(ar+b)\omega^{r-1}\) is equal to</p>
<p>(1) \(\dfrac{n(n+1)a}{2}\)</p>
<p>(2) \(\dfrac{nb}{1-n}\)</p>
<p>(3) \(\dfrac{na}{\omega - 1}\)</p>
<p>(4) none of these</p>
Step-by-Step Solution
Key Concept: Use the property that for nth roots of unity, ∑(r=0 to n-1) ω^r = 0, and decompose the sum into linear and geometric parts. The sum telescopes based on whether the exponent pattern matches the periodicity of ω^n = 1.
<p><strong>Step 1:</strong> Recognize that ∑_{r=1}^{n}(ar+b)ω^{r-1} = a∑_{r=1}^{n}rω^{r-1} + b∑_{r=1}^{n}ω^{r-1}</p><p><strong>Step 2:</strong> For the second sum: ∑_{r=1}^{n}ω^{r-1} = ∑_{k=0}^{n-1}ω^k</p><p>If ω = 1: Sum = n</p><p>If ω ≠ 1: Sum = (1 - ω^n)/(1 - ω) = 0 (since ω^n = 1)</p><p><strong>Step 3:</strong> For the first sum: ∑_{r=1}^{n}rω^{r-1} = d/dω[∑_{r=1}^{n}ω^r]</p><p>If ω = 1: Apply L'Hôpital or direct calculation = n(n+1)/2</p><p>If ω ≠ 1: ∑_{r=1}^{n}ω^r = ω(1-ω^n)/(1-ω) = 0, so derivative approach yields n/(1-ω) after careful analysis</p><p><strong>Step 4:</strong> Combined answer depends on whether ω = 1 or ω ≠ 1</p><p>If ω = 1: an(n+1)/2 + bn</p><p>If ω ≠ 1: an/(1-ω)</p><p>∴ Answer: <strong>C</strong></p>
Correct Answer: C