Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 4 / 3G (Complex Analysis)

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.

The domains are not conformally equivalent. If $F:\mathbb{C}\to \mathbb{H}$ were a conformal equivalence, composing it with a Cayley transform from the upper half-plane $\mathbb{H}$ to the unit disc would produce a bounded nonconstant entire function. This contradicts the Liouville theorem, proving that the complex plane is not conformally equivalent to the upper half-plane.

The domains are conformally equivalent. The Möbius transformationmaps the upper half of the unit disc onto the first quadrant: its diameter $(-1,1)$ maps to the positive real axis and its upper semicircle maps to the positive imaginary axis. Squaring maps the first quadrant bijectively and conformally onto the upper half-plane. Thereforeis the required equivalence, as recorded by the conformal equivalence between the upper half-disc and upper half-plane.