Space of holomorphic functions multiplied by smooth functions taking real values 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
 A: 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 $\{f = (0,0)\}$. 


*

*If (INT) fails to hold, then there exists no solution to your problem. 

*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, necessarily that every direction must be represented in the image of $f$ for there to be a solution $g$.
