Let α1, α2, ... , αn be the nth roots of unity. Find ∏i<j (αi - αj)2.
Solution
Let αk = exp(2pi(k-1)/n). So α1 = 1, and the other αk are roots of xn-1 + xn-2 + ... + x + 1 = 0. Hence (1 - α2), ... , (1 - αn) are roots of (1 - x)n-1 + ... + (1 - x) + 1 = 0. The coefficient of x0 is n and the coefficient of xn-1 is (-1)n-1. Hence ∏2n(1 - αk) = n.
Hence n/αk = (αk - α1)(αk - α2) ... (αk - αn), where the product excludes (αk - αk). Hence nn/(α2...αn) = ∏i not equ j(αi - αj) = (-1)n(n-1)/2 ∏i<j(αi - αj)2. But (α2...αn) = (-1)n-1. Hence ∏i<j(αi - αj)2 = (-1)(n-1)(n-2)/2nn.
© John Scholes
jscholes@kalva.demon.co.uk
11 Mar 2002