Suppose we have a fixed function $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ (sufficiently regular, say $C^\infty$ ). The question is: for which $f$ there exists a scalar function $g: \mathbb{R}^2 \rightarrow \mathbb{R}$ such that the product $fg = (f\_1g, f\_2g)$ is holomorphic (using the standard identificacion $\mathbb{C} \sim \mathbb{R}^2$)? Are there any sufficient or necessary conditions for $f$ in order to exist such a $g$? 
In other words, is there any "nice" description of the product-set of the set of holomorphic functions on the plane with the set of real (sufficiently regular) functions on the plane? 
(Of course, one can try to solve Cauchy-Riemann but I get a pair of transport PDEs $(f\_1g)\_x = (f\_2g)\_y$ and $(f\_1g)\_y = -(f\_2g)\_x$ and is not clear that there is a common solution to both of them). Any references would be very much appreciated


EDIT: As someone commented I am looking for $g\neq 0$, this value of $g$ gives a trivial solution



Space of holomorphic functions multiplied by smooth functions taking real values
First, if $fg$ were holomorphic and nontrivial, then $\{g = 0\}$ has to be discrete. This implies that $g$ has to be either $\geq 0$ or $\leq 0$. So we can assume WLOG $g \geq 0$. 


Restricting away from its zero set, we can study the function $v = \ln g$. The Cauchy-Riemann relations becomes


$$ \begin{cases} f\_1 v\_x - f\_2 v\_y = (f\_2)\_y - (f\_1)\_x \\
 f\_2 v\_x + f\_1 v\_y = - (f\_1)\_y - (f\_2)\_x \end{cases} $$


Unless $f\_1 = f\_2 = 0$ (note also that the set on which this happens is also discrete), the matrix $A\_f = \begin{pmatrix} f\_1 & -f\_2 \\ f\_2 & f\_1\end{pmatrix}$ is invertible, and hence you can explicitly solve the expression for $v\_x, v\_y$ in terms of $f$ and its first derivatives. 


To solve the "local" problem, all you are missing is the integrability conditions, which is $(v\_x)\_y = (v\_y)\_x$. If you write the vector


$$ V = A\_f^{-1} \begin{pmatrix} (f\_2)\_y - (f\_1)\_x \\ - (f\_1)\_y - (f\_2)\_x \end{pmatrix} \tag{Vdef}$$


the integrability condition is $$ (V\_1)\_y = (V\_2)\_x. \tag{INT}$$


### Summary


Starting from $f$, construct $V$ by (Vdef), which is well-defined away from the set $\{f = (0,0)\}$. 


1. If (INT) fails to hold, then there exists no solution to your problem.
2. If $U$ is a simply connected domain on which (INT) holds, then there exists a function $v:U\to \mathbb{R}$ such that $V = \nabla v$. Then on $U$ you have $g = e^v$ solves your problem.


### Additional necessary conditions


By [Picard's Theorem](https://en.wikipedia.org/wiki/Picard_theorem), necessarily that every direction must be represented in the image of $f$ for there to be a solution $g$.


