z1, z2, ... , zn are complex numbers with modulus a > 0. Let f(n, m) denote the sum of all products of m of the numbers. For example, f(3, 2) = z1z2 + z2z3 + z3z1. Show that |f(n, m)| /am= |f(n, n-m)| /an-m.
Solution
It is obviously sufficient to take a = 1. Then if z = eiθ we have 1/z = e-iθ = the complex conjugate of z. Now f(n, m)/(z1z2 ... zn) = the sum of the inverses of the terms in f(n, n-m) = the sum of the complex conjugates of the terms in f(n, n-m) = complex conjugate of f(n, n-m). But a number and its complex conjugate have the same modulus.
© John Scholes
jscholes@kalva.demon.co.uk
25 Feb 2002