C is the complex numbers. f : C → R is defined by f(z) = |z3 - z + 2|. What is the maximum value of f on the unit circle |z| = 1?
Solution
Answer: √13.
Put z = eiθ, and use cos 2θ = 2 c2 - 1, cos 3θ = 4 c3 - 3c, where cos θ = c, to get: |f(z)|2 = 6 - 2 cos(3θ - θ) + 4 cos3θ - 4 cos θ = 4(4c3 - c2 - 4c + 2). The cubic has stationary points where 12c2 - 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