Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 24G / iii / Solution

The equation factors asThus $Y$ is the union of the two complex lines $z=2i$ and $w=i$, meeting at $(2i,i)$. A sufficiently small neighborhood of the intersection, with that point removed, has two connected components, whereas a punctured disc is connected. The intersection therefore has no neighborhood homeomorphic to an open subset of $\mathbb{C}$. Hence $Y$ cannot be a Riemann surface.