Suppose I have an analytic function $f: \mathbb{C}^2 \to \mathbb{C}$ whose zero locus $V=\{(z,w) \in \mathbb{C}^2 : f(z,w)=0\}$ is smooth and simply connected. By uniformization, $V$ is conformally equivalent to either $\mathbb{C}$ or the open unit disk $\Delta \subset \mathbb{C}$. But which one? Is this "easily" determined by some properties of the function $f$?
2026-04-01 22:41:31.1775083291
How to tell if a simply-connected curve is the complex plane or disk
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