Let \(U \subset \mathbb{C}\) be an open set. Assume \(f, g : U \to \mathbb{C}\) are holomorphic functions such that \(\bar{f} g\) is holomorphic. Prove that on each connected component of \(U\), either \(f\) is constant or \(g\) is identically zero. In particular, if \(U\) is connected, prove the global dichotomy.
Connectedness is necessary for the global conclusion. On \(U=D(0,1)\cup D(3,1)\), let \(f\) equal zero on the first disk and one on the second, and let \(g=1\) throughout \(U\). Then \(f\), \(g\), and \(\overline f g\) are holomorphic, although \(f\) is not constant on \(U\) and \(g\) is not identically zero.
Solution
1. The Cauchy–Riemann equations give \(\overline{f'}g=0\) on \(U\).
For a continuously differentiable function \(h\), set \(\partial_{\bar z}h=(h_x+ih_y)/2\). A holomorphic function has \(\partial_{\bar z}h=0\); moreover \(\partial_{\bar z}\overline f=\overline{f'}\) for holomorphic \(f\), directly from its Cauchy–Riemann equations [@SS03]. The product rule therefore gives \[ 0=\partial_{\bar z}(\overline f g) =\overline{f'}g+\overline f\,\partial_{\bar z}g =\overline{f'}g. \] The differentiability needed here follows from holomorphy of \(f\) and \(g\).
2. The dichotomy holds on each connected component.
Fix a component \(V\) of \(U\), which is open. If \(g|_V=0\), one alternative already holds. Otherwise choose \(a\in V\) with \(g(a)\ne0\). Continuity gives a disk about \(a\) on which \(g\) is nonzero. Step 1 then implies \(f'=0\) on that disk. The holomorphic function \(f'\) vanishes on a nonempty open subset of the connected domain \(V\), so the identity theorem gives \(f'|_V=0\) [@SS03]. A holomorphic function with zero derivative is locally constant, by its Taylor series; connectedness makes this constant the same throughout \(V\). Thus \(f|_V\) is constant. When \(U\) is connected, take \(V=U\) to obtain the stated global alternative.