### 7th Putnam 1947

**Problem B4**

p(z) ≡ z^{2} + az + b has complex coefficients. |p(z)| = 1 on the unit circle |z| = 1. Show that a = b = 0.

**Solution**

In particular, |p(1)| = |p(-1)| = 1, so 1 + a + b and 1 - a + b lie on the unit circle. Hence their midpoint 1 + b lies in the unit disk. Similarly, |p(i)| = |p(-i)| = 1, so -1 + ia + b and -1 - ia + b lie on the unit circle and hence their midpoint -1 + b lies in the unit disk. But 1 + b and -1 + b are a distance 2 apart, so they must lie at either end of a diameter of the unit circle and hence b = 0. Now 1 + a and 1 - a lie on the unit circle, as does their midpoint 1. Hence they must coincide and so a = 0.

7th Putnam 1947

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002