C is the complex numbers. f **:** C → R is defined by f(z) = |z^{3} - z + 2|. What is the maximum value of f on the unit circle |z| = 1?

**Solution**

Answer: √13.

Put z = e^{iθ}, and use cos 2θ = 2 c^{2} - 1, cos 3θ = 4 c^{3} - 3c, where cos θ = c, to get: |f(z)|^{2} = 6 - 2 cos(3θ - θ) + 4 cos^{3}θ - 4 cos θ = 4(4c^{3} - c^{2} - 4c + 2). The cubic has stationary points where 12c^{2} - 2c - 4 = 0 or c = 2/3 or -1/2. So the maximum value is at c = 1 or -1 (the endpoints), or 2/3 or -1/2 (the stationary points). Substituting in, we find that the maximum is actually at c = -1/2 with value 13.

© John Scholes

jscholes@kalva.demon.co.uk

4 Nov 1999