The paper gives an upper bound for the valence of harmonic polynomials. An example is given to show that this bound is sharp. Interest in harmonic mappings in the complex plane has increased due to the publication of [1] in 1984. Most of this interest has centered on univalent harmonic mappings. It is, however, the purpose of this paper to present the fundamental valence results of complex valued polynomials which are harmonic in the plane. Specifically a harmonic polynomial is a function P(z) = S(z) + T(z) where S(z) and T(z) are analytic polynomials. The degree of P is defined as the larger of the degrees of S and T. Lyzzaik studied the local topology of harmonic mappings in [4]; in particular he put a bound on the local valence of P but did not obtain a global valence bound. In this paper it will be shown that, apart from certain degenerate examples, P(z) is at most n2-valent, where n is the degree of P. An example is exhibited to show this to be the sharp bound. To prove the n2 valence bound we need the following classical result from algebraic geometry; a proof may be found in [2]. Theorem 1 (Bezout's Theorem). Let A and B be relatively prime polynomials in the real variables x and y with real coefficients, and let deg A = n and deg B = m. Then the two algebraic curves A(x, y) = 0 and B(x, y) = 0 have at most mn points in common. To apply Bezout's theorem to the valence problem we restate it in an equivalent form. Theorem 2 (Alternate form of Bezout's Theorem). Let A and B be polynomials in the real variables x and y with real coefficients. If deg A = n and deg B m, then either A and B have at most mn common zeros or have infinitely many common zeros. Proof. If A and B are relatively prime, then, by Bezout's theorem, the polynomial equations A(x, y) = 0 and B(x, y) = 0 have at most mn common zeros. Otherwise A and B are not relatively prime and they have a greatest common factor, say C. Let deg C = q. If C(x, y) = 0 has no solutions, then A and B have at most (n q)(m q) common zeros. But if there exists a point on C(x, y) = 0 with either Oc 74 0 or 4c 04 o, then, by the implicit function theorem, C(x, y) = 0 Received by the editors June 26, 1995 and, in revised form, September 20, 1996 and January 2, 1997. 1991 Mathematics Subject Classification. Primary 30C55. (?)1998 American Mathematical Society