20th Putnam 1959 Problem A3 Let C be the complex numbers. f : C → C satisfies f(z) + z f(1 - z) = 1 + z for all z. Find f.
Putting z = 1 - w we have f(1 - w) + (1 - w) f(w)= 2 - w. Hence w f(1 - w) + (w - w 2 ) f(w) = 2w - w 2 . Hence (1 + w - f(w) ) + (w - w 2 ) f(w) = 2w - w 2 , giving (w 2 - w + 1) f(w) = (w 2 - 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. 20th Putnam 1959 © John Scholes jscholes@kalva.demon.co.uk 15 Feb 2002