Let \(U \subset \mathbb{C}\) be an open set, \(f : U \to \mathbb{C}\) be a holomorphic function and \(z_0 \in U\). Prove that if \(f'(z_0) = 0\), then \(f\) is not one-to-one in any neighborhood of \(z_0\).
Solution
Let \(V\) be any neighborhood of \(z_0\) contained in \(U\). If \(f\) is constant on a disk about \(z_0\), its restriction to a smaller disk in \(V\) is already noninjective. Otherwise its Taylor series at \(z_0\) has a first nonconstant term of some order \(m\geq2\), since \(f'(z_0)=0\).
1. A small disk in \(V\) has only the central zero of \(f-f(z_0)\) and only the central critical point.
The Taylor expansion gives \[ f(z)-f(z_0)=(z-z_0)^m h(z),\qquad h(z_0)\ne0, \] with \(h\) holomorphic near \(z_0\) [@SS03]. Differentiating, \[ f'(z)=(z-z_0)^{m-1}\bigl(mh(z)+(z-z_0)h'(z)\bigr). \] Both \(h\) and the bracketed factor are nonzero at \(z_0\). Choose \(r>0\) so that \(\overline{D(z_0,r)}\subset V\) and both factors are holomorphic and nonzero on a neighborhood of this closed disk. Then \(f-f(z_0)\) has exactly \(m\) zeros in the disk counted with multiplicity, all at \(z_0\), and \(f'\) has no zero in the punctured disk.
2. A nearby noncentral value has at least two distinct preimages in \(V\).
1. Thus all are simple,so they are \(m\geq2\) distinct points with the same image \(w\).The restriction to \(V\) is not injective. Since \(V\) wasarbitrary, the conclusion holds in every neighborhood.