Unless it vanishes identically, a holomorphic function has isolated zeros.Suggest a slogan
Let \(f\colon\CC\to \CC\) be holomorphic and not identically zero. Then the zeros of \(f\) are isolated: every \(z_0\) with \(f(z_0)=0\) has a punctured neighborhood on which \(f\) has no zeros.
By PR-5A64G, near a zero \(z_0\) we have \(f(z)=(z-z_0)^ng(z)\) with \(n\ge1\) and \(g\) holomorphic and nonvanishing on a neighborhood \(U\) of \(z_0\). For \(z\in U\setminus\theset{z_0}\) both factors are nonzero, so \(f(z)\neq0\).