### 34th Putnam 1973

**Problem B2**

The real and imaginary parts of z are rational, and z has unit modulus. Show that |z^{2n} - 1| is rational for any integer n.

**Solution**

We may put z = cos θ + i sin θ. Then z^{2n} - 1 = (cos 2nθ - 1) + i sin 2nθ, so |z^{2n} - 1|^{2} = 2 - 2 cos 2nθ = 4 sin^{2}nθ. Hence |z^{2n} - 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 cos^{r}θ sin^{s}θ 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

jscholes@kalva.demon.co.uk

22 Aug 2001