The polynomial p(x) has all coefficients 0 or 1, and p(0) = 1. Show that if the complex number z is a root, then |z| ≥ (√5 - 1)/2.

**Solution**

Let h = (√5 - 1)/2. h is approx 0. 618, so certainly if |z| ≥ 1, then |z| > h. So it is sufficient to show that if z is a root with |z| < 1, then |z| ≥ h.

For such values of z, 1 + z + z^{2} + ... converges, so we have 2 + z + z^{2} + z^{3} + ... - 2 p(z) = ± z ± z^{2} ± z^{3} ± ... . Now |rhs| ≤ |z| + |z^{2}| + ... = |z|/(1 - |z| ). If z is a root, then p(z) = 0, so lhs = 2 + z + z^{2} + ... = 2 + z/(1 - z) . If we could show that |lhs| ≥ (2 + |z|)/(1 + |z|), then we would have |z|/(1 - |z|) ≥ (2 + |z|)/(1 + |z|) and hence |z|^{2} + |z| - 1 ≥ 0. But |z| ≥ 0, so |z| ≥ h.

We require |2 - z| (1 + |z|) ≥ |1 - z| (2 + |z|). Squaring and using the polar form z = r e^{iθ}, this is equivalent to (4 - 4r cos θ + r^{2})(1 + r)^{2} ≥ (1 - 2r cos θ + r^{2}) (2 + r)^{2}, or after some simplification, 2r (1 - r^{2})(1 + cos θ) ≥ 0, which is true.

*Many thanks to Thomas Linhart for correcting an error in the original proof.*

© John Scholes

jscholes@kalva.demon.co.uk

27 Jan 2001

Last corrected/updated 19 Aug 2003