Well,I just started to study complex analysis,with the textbook "Function Theory of One Complex Variable".In section 2.4,I encountered the following problem:
Let $\gamma$ be the boundary of a disc $D\left(z_{0}, r\right)$ in the complex plane, equipped with counterclockwise orientation. Let $z$ be a point insic the circle $\partial D\left(z_{0}, r\right) .$ Let $$I(z)=\oint_{\gamma} \frac{1}{\zeta-z} d \zeta$$ Then $$\frac{\partial}{\partial \bar{z}} I(z)=\oint_{\gamma} \frac{\partial}{\partial \bar{z}}\left(\frac{1}{\zeta-z}\right) d \zeta$$ Also,$$\frac{\partial}{\partial z} I(z)=\oint_{\gamma} \frac{\partial}{\partial z}\left(\frac{1}{\zeta-z}\right) d \zeta$$ I don't know why it does work.For real function,We may use Lebesgue's dominated convergence theorem .But for complex case,how to prove that?Can someone give a hint or detailed solution?Thanks in advance.