Let \(D=\{z\in\CC:\abs{z}<1\}\) and let \(f\) be holomorphic on \(D\setminus\{0\}\). Suppose there is a positive integer \(n\) such that \(f^{-1}(w)\) contains at most \(n\) points for every \(w\in\CC\). Prove that zero is a removable singularity or a pole.
Solution
Suppose, for a contradiction, that zero is an essential singularity. In particular \(f\) is nonconstant.
1. The image of every punctured disk about zero is dense in \(\CC\).
Fix \(0<r<1\). If \(f(\{0<\abs{z}<r\})\) were not dense, there would be \(w\in\CC\) and \(\delta>0\) such that \(\abs{f(z)-w}\geq\delta\) throughout this punctured disk. Then \(g(z)=1/(f(z)-w)\) is holomorphic and bounded by \(1/\delta\) there. It extends holomorphically across zero by Riemann's removable singularity theorem; denote the extension by \(G\).
If \(G(0)\ne0\), the identity \(f=w+1/G\) extends \(f\) holomorphically across zero. If \(G(0)=0\), then \(G\) is not identically zero, since it is nonzero on the punctured disk. Its Taylor expansion has the form \(G(z)=z^mH(z)\) with \(m\geq1\) and \(H(0)\ne0\). Consequently \(f(z)=w+z^{-m}/H(z)\) has a pole at zero. Both alternatives contradict essentiality. Thus the image must be dense.
2. There are \(n+1\) pairwise disjoint disks whose images have nonempty intersection.
Choose an open disk \(B_1\) with closure contained in \(D\setminus\{0\}\). The restriction of \(f\) to \(B_1\) is nonconstant: otherwise the identity theorem on the connected punctured disk would make \(f\) constant everywhere. Thus \(V_1=f(B_1)\) is nonempty and open by the open mapping theorem.
Inductively, suppose \(B_1,\ldots,B_k\) have pairwise disjoint closures in \(D\setminus\{0\}\) and \(V_k=\bigcap_{j=1}^k f(B_j)\) is nonempty and open. Choose \(r>0\) so small that \(r<1\) and the punctured disk \(0<\abs{z}<r\) misses all those closures. Step 1 gives a point \(a\) in this punctured disk with \(f(a)\in V_k\). Choose an open disk \(B_{k+1}\) about \(a\) whose closure is contained in \(0<\abs{z}<r\). It is disjoint from all the earlier closures. Its image is open by the same nonconstancy and open mapping argument. Hence \[ V_{k+1}=V_k\cap f(B_{k+1}) \] is open and contains \(f(a)\), so it is nonempty. This completes the induction up to \(k=n+1\).
3. The common image value contradicts the fiber bound.
Choose \(w\in\bigcap_{j=1}^{n+1}f(B_j)\). For each \(j\) there is \(z_j\in B_j\) with \(f(z_j)=w\). The disks are disjoint, so these are \(n+1\) distinct points of \(f^{-1}(w)\), contrary to the hypothesis. Zero is therefore not essential. The classification of isolated singularities leaves only a removable singularity or a pole, as required.
4. Q.E.D.
Step 3 proves the required classification of the singularity at zero.