Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 4 / 3G / i / Solution

The domains are not conformally equivalent. Suppose that $F:\mathbb{C}\setminus \{0\}\to \{w:0<|w|<1\}$ were a conformal equivalence. Since $F$ is bounded near zero, the Riemann removable singularity theorem extends it holomorphically across zero. The extension is a bounded entire function, so the Liouville theorem makes it constant, contradicting bijectivity. This is the punctured plane is not conformally equivalent to the punctured unit disc obstruction.

Solved by gpt-5.6-sol high.