This question was transferred here following Mathoverflow suggestions.
Let us consider two functions $f(z)$ and $g(z)$, both holomorphic on a domain $U$ (a simply connected subset of $\mathbb{C}$). Thus, because of Cauchy's integral theorem, along any closed rectifiable path in $U$: $$ \oint f(z)g(z)dz\ = 0 $$ Let us further assume both $f(z)$ and $g(z)$, and their respective derivatives, to be not identically zero on $U$. Considering that the product $f(z)g(\overline{z})$ is not holomorphic, in case $f(z)g(\overline{z})$ is continuous on $U$, could there still exist particular closed rectifiable paths in $U$ for which the line integral of $f(z)g(\overline{z})$ would be zero? or would such line integrals necessarily differ from zero for all closed rectifiable path in $U$? or are there specific theorems for special cases whereby that would be true?
Because $f(z)g(\overline{z})$ is not holomorphic, Morera's Theorem implies that the contour integral above cannot $= 0$ for all said closed rectifiable paths. I did realise that for not holomorphic integrands said integral would in general depend from the particular path, but I was unable to find theorems, and-or research papers, identifying in a more systematic way the additional conditions which such integrands and paths would need to satisfy in order to guarantee that the above integral $\neq 0$
It appears that starting from around 1930 there was some relevant research about the properties of what were called polygenic functions (is that definition still in use?), to which also said $f(z)g(\overline{z})$ would belong. I was able to find the following articles, but not much after that:
http://www.ams.org/journals/tran/1930-032-02/S0002-9947-1930-1501534-9/S0002-9947-1930-1501534-9.pdf
http://projecteuclid.org/download/pdf_1/euclid.bams/1183496499
NOTE: the additional condition $f(z)g(\overline{z})$ continuous on $U$ is specified as, depending on the domain $U$, it does not necessarily follow from the holomorphicity of $f(z)g(z)$. Just take the example: $$ g(z) = \frac{1}{z+\beta} \;\;\;\;\;\;\; \beta = 1+2i \;\;\;\; z \in U\;\;\;\;\;U=\left\{z \in \mathbb{C}: \; \Im(z)> 0 \right\} $$ is continuous on $U$, but $g(\overline{z})$ is not continuous on $U$ (featuring a pole at $z=-1+2i$)