Let $A = \{(x,y) \in \Bbb C^2: y^2 = f(x)\}$ where $f$ is a polynomial of degree $d$ without repeated roots. Let $f: A \to \Bbb C$ be defined by $f(x, y) = x$.
For large $R$, what is the preimage under $f$ of a circle centred at the origin of radius $R$ in $\Bbb C$? Describe $A$ as a surface with boundary.
My thoughts on the question: for sufficiently large $R$, we have $f(Re^{i\theta}) \simeq R^de^{di\theta}$. The (local) inverses of $f$ are $q(x) = (x, \pm \sqrt{p(x)})$. For $d$ even, we can safely take the square root to get preimage similar to $\{Re^{i\theta}, R^{d/2}e^{d/2i\theta}\} \cup \{Re^{i\theta}, -R^{d/2}e^{d/2i\theta}\}$. I would like to know what this looks like topologically.
For odd $d$ it seems like we need to make a branch cut to get the preimage. I don't know how that will affect the topology of the preimage or how it will affect the topology of $A$ as a whole.
Any hints are appreciated. I don't know differential or Riemannian geometry.