# $\int_{\mathbb{C}} \frac{\partial f }{ \partial z} dxdy = \int_{\mathbb{C}} \frac{\partial f }{ \partial \bar{z}} dxdy = 0$

f is a complex function, $C^{1}$ on $\mathbb{C} \$, $f = 0$ on $K^{c} \$, where $K$ is a compact. Then

$\int_{\mathbb{C}} \frac{\partial f }{ \partial z} dxdy = \int_{\mathbb{C}} \frac{\partial f }{ \partial \bar{z}} dxdy = 0$

I can solve the second integral but not the first.

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I'm assuming $C^1$ here means the partial derivatives of $f$ exist? In that case, if you can prove one of these integrals is zero the exact same proof should work in the other case (from the point of view of continuity, $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial \bar z}$ are indistinguishable. –  Eric O. Korman May 4 '12 at 18:45
@Eric: Sorry but I don't understand this comment –  WLOG May 4 '12 at 21:06

The easiest way is probably to use the formalism of complex differential forms. If $dz = dx + i\,dy$ and $d\bar z = dx - i\,dy$, then $$d\bar z \wedge dz = (dx - i\,dy) \wedge (dx + i\,dy) = 2i\,dx \wedge dy.$$
For the first integral, put $\omega = f \,d\bar z$. Choose a smoothly bounded domain $\Omega$ containing $K$. By Stokes' theorem, $$\int_{\partial \Omega} \omega = \int_{\Omega} d\omega = \int_\Omega \partial \omega + \bar\partial\omega = \int_{\Omega} \frac{\partial f}{\partial z}\,dz\wedge d\bar z = -2i \int_\Omega \frac{\partial f}{\partial z}\,dx\wedge dy .$$
But, the integral on the lefthand side is clearly $0$, since $\omega = 0$ outside of $K$ (in particular, $\omega = 0$ on $\partial\Omega$). For the second integral, do the same with $\omega = f\,dz$.