Abstract. Suppose that h, k ∈ L(H) are two selfadjoint bounded operators on a Hilbert space H. It is elementary to show that hk is selfadjoint precisely when hk = kh. We answer the following question: Under what circumstances must hk be selfadjoint given that it is normal?Let h, k ∈ L(H) be two selfadjoint bounded linear operators on a Hilbert space, shows that hk need not be selfadjoint; note that hereHowever, if either h or k is positive, then hk is normal if and only if it is selfadjoint; for example, when k ≥ 0,and a normal operator with real spectrum is selfadjoint. In fact,2 ) when h is selfadjoint and k is positive (see [6]; we thank Prof. J. Zemánek for this reference).We note further that, because σ(hk) \ {0} = {w : 0 = w ∈ σ(hk)} = σ((hk) * ) \ {0} = σ(kh) \ {0} = σ(hk) \ {0}, the spectrum σ(hk) is symmetric with respect to the real axis. Conversely, any compact subset K ⊂ C of the complex plane which is symmetric with respect to the real axis can occur as the spectrum σ(hk) for such a pair of operators. To see this, choose a countable sequence (λ n ) n∈N dense in K. The question naturally arises of finding conditions that will guarantee that hk is selfadjoint given that it is normal; we shall answer it in the wider context of Banach algebras.