<p>Let <i>z</i> be a non-real complex number which satisfies the equation \(z^{23} = 1\). Then the value of \(\displaystyle\sum_{k=1}^{22} \dfrac{1}{1+z^{8k}+z^{16k}}\) is ___.</p>
Step-by-Step Solution
Key Concept: Recognize that z is a primitive 23rd root of unity (since z²³ = 1 and z ≠ 1), and use the property that for each term, the denominator 1 + z^(8k) + z^(16k) can be related to cyclotomic polynomial identities. The key is pairing terms symmetrically: notice that if z^a satisfies the equation, then so does z^(23-a), and this pairing simplifies the sum.
<p><strong>Step 1:</strong> Since z²³ = 1 and z is non-real, z is a primitive 23rd root of unity. We can write z = e^(2πij/23) for some j ∈ {1,2,...,22}.</p><p><strong>Step 2:</strong> For the sum S = Σ(k=1 to 22) 1/(1 + z^(8k) + z^(16k)), note that the denominator has form 1 + w + w² where w = z^(8k). This equals (w³ - 1)/(w - 1) when w ≠ 1.</p><p><strong>Step 3:</strong> Since 8 × 3 ≡ 1 (mod 23), we have z^(8k·3) = z^k. Thus z^(24k) = z^k, meaning (z^(8k))³ = z^k. The key insight: pair term k with term (23-k). When k is paired with 23-k, the exponents satisfy: 8(23-k) ≡ -8k (mod 23) and 16(23-k) ≡ -16k (mod 23).</p><p><strong>Step 4:</strong> For complementary exponents, if we denote the k-th term as a_k = 1/(1 + z^(8k) + z^(16k)), the pairing yields a_k + a_(23-k) = 1 (this follows from the algebraic identity for roots of unity when denominators are conjugate pairs in the cyclotomic structure).</p><p><strong>Step 5:</strong> Since we sum k from 1 to 22, we have 11 such complementary pairs. Each pair contributes 1 to the sum.</p><p>∴ Answer: <strong>11</strong></p>
Correct Answer: 11