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    If $fg=0$ on a domain then $f\equiv 0$ or $g\equiv 0$

    P-AKPY4

    Area
    Complex Analysis
    Topics
    Identity Theorem, Zeros
    Status
    draft

    Suppose \(D\) is a domain and \(f, g\) are analytic on \(D\).

    Prove that if \(fg = 0\) on \(D\), then either \(f \equiv 0\) or \(g\equiv 0\) on \(D\).

    Solution

    Suppose \(fg=0\) on \(D\) but \(f\not\equiv 0\), we’ll show \(g\equiv 0\) on \(D\). Since \(f\not \equiv 0\), \(f(z_0)\neq 0\) at some point \(z_0\). Since \(f\) is holomorphic, in particular \(f\) is continuous, so there is a neighborhood \(U\ni z_0\) where \(f(z)\neq 0\) for any \(z\in U\). But \(f(z)g(z) = 0\) for all \(z\in U\), and since \(\CC\) is an integral domain, this forces \(g(z) = 0\) for every \(z\in U\). So \(g\equiv 0\) on \(U\). Now \(U\) is a set with a limit point, so by the identity principle, \(g\equiv 0\) on \(D\).

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