My question is similar to the questions
For $g(x) = 1/x$, determine the antiderivative, and determine its definite integral.
Except I want to use complex-valued variables and functions.
First, I hypothesize the most parsimonious auxilary function possible that might work. My intuition says to use overlay a Dirac delta distribution on top of the reciprocal function. Maybe the following?
By $A \subset \mathbb{C}$, I denote a set defined by $\left\{x + i 0 \in \mathbb{C} \mid x \leq 0 \right\}$. By $g : ? \to ?$ I denote an auxiliary function that I define as $$g(x,y) = \begin{cases} i\delta_{x+iy}(A) &~\text{for}~ x+iy \in A; \text{and} \\ \frac{1}{x+iy} &~\text{otherwise}. \end{cases}$$
Questions:
(1) Does $g(x,y)$ have an antiderivative?
(2) What is its antiderivative?
(3) Is the definite integral zero around any closed path?
In the pticture below, I give a closed path. The path integral should equal zero taken along the path.
DISCLAIMER: Since OP edited his original question, changing it completely, the following post doesn’t seem to answer the question anymore.
A complex function $g:\Omega \to \mathbb{C}$ ($\Omega \subseteq \mathbb{C}$ open) has an antiderivative iff $\oint_\gamma g(z)\ \text{d} z = 0$ for any closed path $\gamma$ contained in $\Omega$. Your $g$ does not satisfy this assumption in $\Omega = \mathbb{C} \setminus \{ 0\}$, therefore it fails to have an antiderivative.
On the other hand, your $g$ possesses a multi-valued antiderivative, namely $G(z) = \log z = \ln |z| + \imath\ \operatorname{arg} z$ (here $\ln$ stands for the real natural logarithm): in fact, each single-valued branch $\phi$ of $G$ is holomorphic and satisfies $\phi^\prime (z) = \frac{1}{z} = g(z)$ for each $z$ in its domain.