The real and imaginary parts of z are rational, and z has unit modulus. Show that |z2n - 1| is rational for any integer n.
We may put z = cos θ + i sin θ. Then z2n - 1 = (cos 2nθ - 1) + i sin 2nθ, so |z2n - 1|2 = 2 - 2 cos 2nθ = 4 sin2nθ. Hence |z2n - 1| = 2 |sin nθ |. But sin nθ is the imaginary part of (cos θ + i sin θ)n. If we expand by the binomial theorem we get a series of terms cosrθ sinsθ with integer coefficients. Each term is rational since cos θ and sin θ are rational. Hence sin nθ is rational.
34th Putnam 1973
© John Scholes
22 Aug 2001