Let C represent the complex numbers. Let g: C → C be an arbitrary function. Let w be a cube root of 1 other than 1 and let v be any complex number. Find a function f: C → C such that f(z) + f(wz + v) = g(z) for all z and show that it is unique.
Solution
(1) f(z") + f(wz"+v) = g(z"), so taking z" = wz' + v, we get
(2) f(wz'+v) + f(w2z'+wv+v) = g(wz'+v), and taking z' = wz + v, we get
(3) f(w2z+wv+v) + f(z) = g(w2z+wv+v)
Taking (1) - (2) + (3) with (with z" and z' as z) we get:
2 f(z) = g(z) - g(wz + v) + g(w2z + wv + v).
© John Scholes
jscholes@kalva.demon.co.uk
28 Dec 2002
Last corrected/updated 28 Dec 02