Definition. Whenever $X$ is a set, define that a run in $X$ is a function $\mathbb{Z} \rightarrow X$ modulo translation on the domain side. So in particular, two functions $f , g : \mathbb{Z} \rightarrow X$ represent the same run iff there exists $n \in \mathbb{Z}$ such that $f(n+i) = g(i)$ for all $i \in \mathbb{Z}$.
Let $\mathrm{Run}(X)$ denote the collection of all runs in $X$.
A "complex run" is a run in $\mathbb{C}$.
In complex analysis, it seems to be the case that if we're given a mapping $f:\mathbb{C} \rightarrow \mathbb{C},$ it's sometimes possible to get a corresponding function $f^{-1} : \mathbb{C} \rightarrow \mathrm{Run}(\mathbb{C})$ that indexes the preimage sets in a potentially useful way. For example, define $$f_n : \mathbb{C} \rightarrow \mathbb{C}, \qquad f_n(z) = z^n.$$
Then we get corresponding functions $$f^{-1}_n : \mathbb{C} \rightarrow \mathrm{Run}(\mathbb{C})$$ by collecting together $n$th roots appropriately. For instance, we might write $$f^{-1}_2(1) = \sqrt{1} = (\ldots,1,-1,1,-1,1\ldots),$$ where the rightmost expression is notation for a complex run.
Something similar happens with $\exp : \mathbb{C} \rightarrow \mathbb{C}$, we get an inverse $\log : \mathbb{C}_{\neq 0} \rightarrow \mathrm{Run}(\mathbb{C}),$ for example: $$\log(1) = (\ldots,-4\pi i,-2\pi i,0,2\pi i, 4\pi i,\ldots).$$
Question. Are there any general principles that explain why "complex runs" seem to arise like this? In particular, for which $f : \mathbb{C} \rightarrow \mathbb{C}$ can we expect to get a corresponding map $f^{-1} : \mathrm{img}(f) \rightarrow \mathrm{Run}(\mathbb{C})$? Is this typical of holomorphic functions? Is it atypical? What's really going on here?
I'd like to know if there's any general principles out there that explain why this occurs and for which functions we can expect it to occur?