Suppose \(D\) is a domain and \(f\) and \(g\) are analytic functions on \(D\). Prove that if the product \(fg = 0\) throughout \(D\), then either \(f\) or \(g\) must vanish identically on \(D\).
Solution
1. If \(f\) vanishes identically on \(D\), then the required conclusionholds.
This is one of the two alternatives in the conclusion.
2. Suppose \(f\) does not vanish identically on \(D\). Then there is anonempty open set \(U\subseteq D\) on which \(g\) vanishes.
Choose \(z_0\in D\) with \(f(z_0)\neq0\). Since \(f\) is continuous, there is an open neighborhood \(U\subseteq D\) of \(z_0\) such that \[ f(z)\neq0 \] for every \(z\in U\). The hypothesis \[ f(z)g(z)=0 \] throughout \(D\) therefore implies \[ g(z)=0 \] for every \(z\in U\).
3. Under the hypothesis of step 2, \(g\) vanishes identically on \(D\).
The analytic function \(g\) vanishes on the nonempty open subset \(U\) of the domain \(D\). Since a domain is connected, the identity theorem implies that \(g\) vanishes identically on \(D\).
4. Either \(f\) or \(g\) vanishes identically on \(D\).
If \(f\) vanishes identically, step 1 applies. Otherwise step 3 shows that \(g\) vanishes identically.
5. Q.E.D.
Step 4 is the required conclusion.