Solution
= Solution
The equation factors as
$$
zw-2iw-iz-2=(z-2i)(w-i).
$$
Thus $Y$ 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.