Let C be the complex numbers. f : C → C satisfies f(z) + z f(1 - z) = 1 + z for all z. Find f.
Solution
Putting z = 1 - w we have f(1 - w) + (1 - w) f(w)= 2 - w. Hence w f(1 - w) + (w - w2) f(w) = 2w - w2. Hence (1 + w - f(w) ) + (w - w2) f(w) = 2w - w2, giving (w2 - w + 1) f(w) = (w2 - w + 1). So provided w is not (1 ± i √3)/2, we have f(w) = 1.
But at these two values f can be different. We can take one of them to be arbitrary. For example, take f( (1 + i √3)/2 ) = k, any complex number. Then f( (1 - i √3)/2 ) = 1 + (1 - k)(1 - i √3)/2.
© John Scholes
jscholes@kalva.demon.co.uk
15 Feb 2002