p(z) and q(z) are complex polynomials with the same set of roots (but possibly different multiplicities). p(z) + 1 and q(z) + 1 also have the same set of roots. Show that p(z) ≡ q(z).
Solution
This is much easier than it looks. We just have to consider p - q and p' - q'. Suppose that p has A roots, of which B are distinct, and that p(z) + 1 has A roots, of which C are distinct. Without loss of generality we may assume the degree of q is at most A, so that p - q has at most A roots. Clearly the roots of p(z) and p(z) + 1 do not overlap. The B distinct roots of p(z) and the C distinct roots of p(z) + 1 must all be roots of p - q. So B + C ≤ A. On the other hand, p' has at least A - B roots in common with p and at least A - C in common with p(z) + 1, so it has at least 2A - (B + C) in total. But its degree is A - 1, so (B + C) ≥ A + 1. Contradiction.
© John Scholes
jscholes@kalva.demon.co.uk
4 Dec 1999