$f\in C^1(A)$ implies $\int_{\mathbb{C}}\frac{\partial f}{\partial z}\,dx\,dy=\int_{\mathbb C}\frac{\partial f}{\partial \bar z}\,dx\,dy=0$. Let $A \subset \mathbb{C}$ be an open set and $f\in C^1(A)$. For hypothesis I know that exists a compact set $K$ such that $f=0$ in $\mathbb{C}\setminus K$ and I have to prove that
\begin{equation}
\int_{\mathbb{C}} \frac{\partial f}{\partial z} \, dx \, dy =\int_{\mathbb{C}} \frac{\partial f}{\partial \bar z} dx \,dy=0.
\end{equation}
Can someone give me an hint to resolve this problem?
Since $f=0$ in  $\mathbb{C}\setminus K$ I have to show that
\begin{equation}
\int_K \frac{\partial f}{\partial z} \, dx \, dy =\int_K \frac{\partial f}{\partial \bar z} \, dx \, dy =0.
\end{equation}
Should I use now  the definition of Wirtinger derivates?
Thanks for the help!

First of all, let us be reminded that
$$
\frac{\partial}{\partial z}=\frac{1}{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)
\quad\text{and}\quad
\frac{\partial}{\partial \overline{z}}=\frac{1}{2}\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right).
$$
Hence it suffices to show that
$$
\int f_x\,dx\,dy=\int f_y \,dx\,dy=0.
$$
Let $K$ be a support of $f$ and $R$ a rectangle, $K\subset R$. Say
$$
R=[x_1,x_2]\times [y_1,y_2].
$$
Then
$$
\int f_x\,dx\,dy=\int_R f_x\,dx\,dy=\int_{y_1}^{y_2}\Big(\int_{x_1}^{x_2} f_x(x,y)\,dx\Big)\,dy=\int_{y_1}^{y_2}\Big(f(x_2,y)-f(x_1,y)\Big)\,dy=0,
$$
since $f$ vanishes on the boundary of $R$.
