Solution

ID: 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 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.

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