p(x, y) = a x^{2} + b x y + c y^{2} is a homogeneous real polynomial of degree 2 such that b^{2} < 4ac, and q(x, y) is a homogeneous real polynomial of degree 3. Show that we can find k > 0 such that p(x, y) = q(x, y) has no roots in the disk x^{2} + y^{2} < k except (0, 0).

**Solution**

The disk is a strong hint that one should use polar coordinates, so put x = r cos θ, y = r sin θ. Then we get r = p(cos θ, sin θ)/q(cos θ, sin θ). Now |cos θ| and |sin θ| ≤ 1, so |q(cos θ, sin θ)| ≤ h, where h is the sum of the absolute values of the coefficients of q. So we are home if we can establish some inequality |p(cos θ, sin θ)| ≥ h', because we can then take k = h'/h.

Without loss of generality a > 0 and hence also c > 0. Now p(cos θ, sin θ) = 1/2 (a + c)(cos^{2}θ + sin^{2}θ) + 1/2 (a - c)(cos^{2}θ - sin^{2}θ) + 1/2 b sin 2θ = 1/2 (a + c) + 1/2 (a - c) cos 2θ + 1/2 b sin 2θ. Put d = √(b^{2} + (a - c)^{2}) and take φ so that cos φ = (a - c)/d, sin φ = b/d. Then 2p(cos θ, sin θ) = (a + c) + d (cos φ cos 2θ + sin φ sin 2θ) = (a + c) + d cos(φ - 2θ) >= (a + c) - d.

But 4ac > b^{2}, so (a + c)^{2} > (a - c)^{2} + b^{2}, so (a + c) > d, and we are done.

© John Scholes

jscholes@kalva.demon.co.uk

25 Aug 2001