I have asked the following question in math.stackexchange, but I could not receive the answer. See [here](https://math.stackexchange.com/questions/2063463/solving-a-system-of-equations-involving-smooth-functions).


Suppose $h\_{i\overline{j}}$, where $1\leq i, j\leq n$, are functions defined on $\mathbb{C}^n$ such that $h\_{i\overline{j}}$ are smooth when viewed as functions defined on $\mathbb{R}^{2n}$
and $H=(h\_{i\overline{j}})$ is a Hermitian positive definite matrix. I wonder if it is always possible to find smooth functions $p\_1,..., p\_n$ defined on $\mathbb{R}^{2n}$ such that 
$$\frac{\partial p\_i}{\partial z\_j}+\overline{\frac{\partial p\_j}{\partial z\_i}}=h\_{i\overline{j}}\mbox{ and }
\frac{\partial p\_i}{\partial \overline{z}\_j}
-\frac{\partial p\_j}{\partial \overline{z}\_i}=0.$$
In a related question, I require $p\_i$ to be holomorphic, which seems to be too strong. 



Solving a system of equations involving smooth functions
I find a possible way to solve it, assuming some properties of polynomials which I will state in the end: 
Write
$p\_j(z)=a\_j+\sqrt{-1}b\_j$ for $1\leq j\leq n$. Equating the real and imaginary parts,
we see that solving $(1)$ is equivalent to solving
$$\tag{2}
\frac{\partial a\_j}{\partial x\_i}+\frac{\partial b\_j}{\partial y\_i}
+\frac{\partial a\_i}{\partial x\_j}+\frac{\partial b\_i}{\partial y\_j}=2h\_{i\overline{j}},\\
\frac{\partial b\_j}{\partial x\_i}-\frac{\partial a\_j}{\partial y\_i}
-\frac{\partial b\_i}{\partial x\_j}+\frac{\partial a\_i}{\partial y\_j}=0,\\
\frac{\partial a\_j}{\partial x\_i}+\frac{\partial b\_j}{\partial y\_i}
-\frac{\partial a\_i}{\partial x\_j}-\frac{\partial b\_i}{\partial y\_j}=0
$$
where $1\leq i, j\leq n$.
Note that $(2)$ is a system of linear PDEs with constant coefficients. Let
\begin{equation}
P\_{ij}^{k}(X\_1,X\_2,...,X\_{2n-1},X\_{2n})=\left\{
 \begin{array}{ll}
 X\_{2j-1}, & \hbox{ if $k=2i-1$;} \\
 X\_{2j}, & \hbox{ if $k=2i$;}\\
 X\_{2i-1}, & \hbox{ if $k=2j-1$;} \\
 X\_{2i}, & \hbox{ if $k=2j$;}\\
 0, & \hbox{ otherwise.}
 \end{array}
 \right.
\end{equation}
and
$$(u\_1,u\_2,...,u\_{2n-1},u\_{2n})=(a\_1,b\_1,...,a\_n,b\_n).$$
Then
the system of equations in the first line of $(2)$ can be written as
$$\sum\_{1\leq k\leq l\leq n}P\_{ij}^{k}\Big(\frac{\partial}{\partial x\_1},\frac{\partial}{\partial y\_1},...,\frac{\partial}{\partial x\_n},\frac{\partial}{\partial y\_n}\Big)u\_k=0$$
where $1\leq i\leq j\leq n$.


I guess that if $\{Q\_{i,j}\}\_{1\leq i\leq j\leq n}$ are polynomials
in $X\_1$, $X\_2$,..., $X\_{2n-1}$, $X\_{2n}$ such that
$$
\sum\_{1\leq i\leq j\leq n}Q\_{i,j}P\_{ij}^k=0\mbox{ for all }1\leq k\leq 2n,
$$
then we must $Q\_{i,j}=0$ for all $1\leq i\leq j\leq n$, which I have posted it as a question [here](https://math.stackexchange.com/questions/2086279/still-another-fact-about-polynomials).
If this is true, we can then apply Theorem 7.6.13 in the book of Hormander [
An introduction to complex analysis in several variables, Third edition, North-Holland, Amsterdam, 1990]
to conclude that the system $(2)$ is solvable. 


