50th Putnam 1989

Prove that all the roots of 11z¹⁰ + 10i z⁹ + 10i z - 11 = 0 have unit modulus.

Solution

Note that z⁹ = (11 - 10iz)/(11z + 10i). But |11 - 10iz|² = (121 + 100|z|²) + 220 Im z, and |11z - 10iz|² = (100 + 121|z|²) + 220 Im z. In other words, if |z⁹| = N/D, then N - D = 21(1 - |z|²). So if |z| > 1, then N < D and hence |z⁹| < 1. Similarly if |z| < 1, then |z⁹| > 1. A contradiction in both cases. Hence |z| = 1.

50th Putnam 1989

© John Scholes
jscholes@kalva.demon.co.uk
1 Jan 2001