I've just missed a point on complex analysis lectures. Namely, we did an integral representation formula using Green's formula: Suppose we have function $f$ continuously differentiable on open set $D$ in complex plane, with piecewise continuously differentiable boundary. Then $$f(z) = 2i \int_D f_{\bar{z}} dxdy. $$ If we cut out some enough small disk around $a$, we get a result similar that generalizes Cauchy integral formula: $$f(a)= \frac{1}{2\pi i}\int_{\partial D}\frac{f(z)dz}{z-a}-\frac{1}{\pi}\int_{D}\frac{f _{\bar{z}}(\zeta) d\xi d\eta}{\zeta - a}, \textrm{where } \zeta= \xi+i \eta.$$
Now define the operator $T$ such that, $$T_D(u) (z) = -\frac{1}{\pi}\int_{D}\frac{u _{\bar{z}}(\zeta) d\xi d\eta}{\zeta - z}. $$
Finally, the question: There is a theorem, in which we suppose that $u$ satisfies the same conditions as above. And suppose $f = T_D (u)$. Then $f$ is a solution of $f_{\bar{z}} = u$ on $D$. In the middle of proof, however we require another hypothesis: $u$ has a compact support on $D$ and in the end, to finish the proof we somehow say that $$\frac{1}{2\pi i}\int_{\partial D} u(\zeta)d\zeta = 0.$$ Is it because of this additional hypothesis or? I know what is support, but cannot connect it with this at the moment.
The support of a function $u$ defined on $\mathbb{C}$ is $\operatorname{supp}u = \overline{\{z \in \mathbb{C} \mid u(z) \neq 0\}}$.
Note that $D$ is an open set and $u$ has compact support in $D$ (in particular, $\operatorname{supp}u \subset U$). Let $z \in \partial D$. As $D$ is open, $z \not\in D$ and therefore $z \not\in \operatorname{supp}u$. So $u(z) = 0$ by definition of $\operatorname{supp}u$. As $z$ was arbitrary, we see that $u|_{\partial D} = 0$ and so
$$\frac{1}{2\pi i}\int_{\partial D} u(\zeta) d\zeta = \frac{1}{2\pi i}\int_{\partial D} 0d\zeta = 0.$$