I'm going back over some introductory complex analysis, and I'm trying to find an intuitive notion of what makes the $\frac{1}{z}$ term special when calculating complex closed line integrals of analytic functions. My current understanding follows the chain of reasoning:
- Take $f$ as analytic, so $f$ can be locally understood around $z=0$ as a power series $a_iz^i$. Suppose also that $f$ has an isolated zero at $z=0$, so that $i$ ranges from $n$ to $\infty$, for some negative integer $n$.
- The line integral $\int_C f\,\text{ds}$ can thus be locally understood as $\int_C a_iz^i\,\text{ds}$ (assume $C$ is closed, well-behaved, and avoids $0$).
- For each $z^i$, $i\neq -1$, we have $\int z^i\,\text{dz}=\frac{1}{i+1}z^{i+1}+c$, and these antiderivatives are analytic away from $0$, so that $\int_C z^i\,\text{ds}=0$.
- The function $z^{-1}$ is different in that its antiderivative $\text{Log}(z)$ is multi-valued on $\mathbb{C}-\{0\}$, and so the line integral must be evaluated in a way that is always locally on the same branch. A closed line integral with winding number $n$ will have shifted by $n$ branches by the time it ends, and so will have accumulated an integral of $2\pi in$; one step of $2\pi i$ per rotation around the zero.
- We obtain residues by simply applying linearity of the integral, so we get a residue of $2\pi ina_{-1}$ around the pole at zero.
This all makes sense to me - I've studied topology, so the idea of "lifting" the line $C$ to its image in the multi-sheeted codomain of $\text{Log}$ is fine. However, despite following the technical reasoning, I'm still left without an intuitive grasp of why the term of $z^{-1}$ in the power series expansion of $f$ is "special" in this way. Do I just need to accept this as a fact of life, as being due to the non-holomorphicity of its antiderivative, or is there something deeper that makes it all seem more natural?