8th Putnam 1948 Problem A1 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
: √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. 8th Putnam 1948 © John Scholes jscholes@kalva.demon.co.uk 4 Nov 1999