In a paper of Raphael M. Robinson, which can be found here (given Jstor privileges, but I have included screenshots relevant to my question) the author defines a rational function $h(z) \in \mathbb{C}(z)$ such that $|h| < 1$ on some compact set $E \subset \mathbb{C}$. He then chooses some $\lambda < 1$ for which $|h| < \lambda$ on $E$, and integrates some meromorphic function $T$ over the closed curve $C = |h|^{-1}(\lambda)$. He also claims this curve 'encloses' $E$. I am wondering why $|h|^{-1}(\lambda)$ is one of the usual closed curves we can integrate over in complex analysis (so basically, piecewise $C^1$), and I am also unsure exactly what the meaning of 'enclose' here is. I suppose it means $\eta(C,e) = 1$ for all $e \in E$.
Here are the relevant screenshots:
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Also important is that the $q_k,p_k$ are finite (in fact they are integers, but this part does not seem necessary for my question), and $A_r \neq 0$ (in fact $|A_r| \geq 1$ but again I do not think this part is necessary for my question).
Edit (partial solution): Here is a possible hacky way to circumvent the issue I brought up earlier, $h=\frac{P}{Q}$ is a rational function, i.e. some holomorphic map $\hat{\mathbb{C}} \rightarrow \hat{\mathbb{C}}$ , and in particular it is a branched cover in the sense that if we treat it as a map $\hat{\mathbb{C}} \setminus \{q_1,..., q_k \} \rightarrow \hat{\mathbb{C}} \setminus \{\alpha_1,...,\alpha_j \}$
(where the set we removed from the domain are the branch points and the set we remove from the range are the critical values), then it is a bonafide covering map. Now we can just select $\lambda < 1$ so that the circle of radius $\lambda$ does not contain any of the critical values, and lift the circle parameterized positively via the rational function.
The rational function is a degree $\max(\deg(P),\deg(Q))$ cover, so there will be exactly these many disjoint lifts, all of these lifts will be smooth closed curves.
Now I am guessing because of how close $h$ is to a covering map, there will be a way to modify this approach for arbitrary $\lambda$.
The risk with this hacky way is that it is possible $\lambda<1$ being arbitrary so that $|h|<\lambda$ on $E$ is essential for the rest of the proof, luckily it turns out this is not the case.
So the disjoint union (after removing lifts that are reparameterizations of other lifts) of these lifts will be the entire preimage of $|h|^{-1}(\lambda)$, which is a cycle (not exactly a closed curve, but we can still apply the usual residue theorems for cycles so this isn't an issue).
Finally because $h(\infty) = \infty$ each connected components of $|h|^{-1}[0,\lambda]$ will coincide with the closure of the regions bounded by the lifts, and because these curves have winding number $+1$ with respect to points in their interior, we get the result for all but finitely many $\lambda$. I suppose for the bad $\lambda$ there may be some continuity argument to show this still holds.