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Exercise 1819Algebra › PolynomialsL1Added 3 Nov 2026

Complex factorization via roots of unity

#roots of unity#factorization#complex

Statement

Factor Xn−1X^n-1 over C\mathbb{C}.

Hint

The roots are the nn-th roots of unity.

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Solution

Roots. The roots of Xn−1X^n-1 are ωk=e2ikπ/n\omega_k=e^{2ik\pi/n} for k=0,1,…,n−1k=0,1,\ldots,n-1.

Factorization.
Xn−1=∏k=0n−1(X−ωk).X^n-1=\prod_{k=0}^{n-1}(X-\omega_k).

Over R\mathbb{R}. Pair conjugates ωk\omega_k and ωn−k\omega_{n-k} (for k≠0,n/2k\neq0,n/2) to get quadratics X2−2cos⁡(2kπ/n)X+1X^2-2\cos(2k\pi/n)X+1. The factors are:
- X−1X-1;
- if nn even, X+1X+1;
- for k=1,…,⌊(n−1)/2⌋k=1,\ldots,\lfloor(n-1)/2\rfloor, quadratics X2−2cos⁡(2kπ/n)X+1X^2-2\cos(2k\pi/n)X+1.

Conclusion. Over C\mathbb{C}, Xn−1X^n-1 splits into nn linear factors; over R\mathbb{R}, into linear factors for the real roots (±1\pm1) and irreducible quadratics.