Past exam of the mathematics course of the University of Cambridge 2023 ib Paper 4 3G i Solution Created 2026-09-24 Updated 2026-09-25
The domains are not conformally equivalent. Suppose that were a conformal equivalence. Since 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.