Show that the punctured unit disk \(\mathbb{D}^* = \mathbb{D} \setminus \{0\}\) and the annulus \(A = \{z : 1 < |z| < 2\}\) are not conformally equivalent.
Solution
Suppose \(F:\mathbb D^*\to A\) were a biholomorphism. Since \(F\) is bounded, the isolated singularity at \(0\) is removable, so \(F\) extends holomorphically to \(\widetilde F:\mathbb D\to\mathbb C\).
Take any sequence \(z_n\to0\), \(z_n\ne0\). Compactness of the closed annulus \(1\le|w|\le2\) gives a subsequence for which \(F(z_n)\) converges; continuity of \(\widetilde F\) shows every such limit equals \(\widetilde F(0)\). Thus \(|\widetilde F(0)|\in[1,2]\).
If \(1<|\widetilde F(0)|<2\), then this value lies in \(A\), so surjectivity of \(F\) gives \(w\in\mathbb D^*\) with \(F(w)=\widetilde F(0)\), contradicting injectivity of the nonconstant holomorphic map \(\widetilde F\) in a neighborhood of \(0\) (equivalently, zeros of \(\widetilde F-\widetilde F(0)\) near \(0\) have positive multiplicity and \(0\) is not isolated from the image in that fashion). More directly, the open mapping theorem at \(0\) would put values with modulus \(<1\) or \(>2\) in the image if \(|\widetilde F(0)|\) is on either boundary circle, while an interior value would violate injectivity as above. In every case we obtain a contradiction.
Hence no conformal equivalence exists.