How can i prove this problem about Cauchy integral? Let $V$ be an open connected subset of $\mathbb{C}$.
Let $f:\rightarrow \mathbb{C}$ be a function continuous on a contour $C$.
Define $F(z)=\int_C \frac{f(\zeta)}{\zeta - z} d\zeta$ on $V$.
My question is:


*

*How can i show that $F$ is analytic at every point except $C$?

*How can I show that  $F'(z)=\int_C \frac{f(\zeta)}{(\zeta - z)^2} d\zeta$ where $z$ not in $C$?
 A: 1) Let $D$ be a small disc avoiding $C$. Let $\gamma$ be a simple closed curve in $D$. Then
$$
\int_\gamma \left( \int_C \frac{f(\zeta)}{\zeta-z}\,d\zeta \right)\,dz =
\int_C \left( \int_\gamma \frac{f(\zeta)}{\zeta-z}\,dz\right)\,d\zeta = 0,
$$
by Cauchy's integral theorem (and Fubini), since $z\mapsto \dfrac{f(\zeta)}{\zeta-z}$ is holomorphic on $D$ for every $\zeta \in C$. Hence Morera's theorem implies that $F$ is holomorphic on $D$, but $D$ was arbitrary and $F$ is thus holomorphic everwhere outside $C$.
2) If you don't know strong enough theorems that allow you to differentiate under the integral sign, write
\begin{align}
\frac{F(z+h)-F(z)}{h} &= \frac1h \int_C \left( \frac{f(\zeta)}{\zeta-(z+h)} - \frac{f(\zeta)}{\zeta-z)} \right)\,d\zeta \\
&= \int_C \frac{f(\zeta)}{(\zeta-z)(\zeta-z-h)}\,d\zeta.
\end{align}
Fix $\delta < \operatorname{dist}(z,C)$ and take $h$ so small that $|h| < \delta/2$. Then
\begin{align}
\left|\frac{F(z+h)-F(z)}{h} -  \int_C \frac{f(\zeta)}{(\zeta-z)^2}\,d\zeta \right|
&= \left| \int_C \frac{hf(\zeta)}{(\zeta-z)^2(\zeta-z-h)}\, d\zeta \right| \\
&\le \int_C \left|\frac{hf(\zeta)}{(\zeta-z)^2(\zeta-z-h)}\right|\, d\zeta \\
&\le \frac{2|h|\sup_C |f|}{\delta^3}
\end{align}
which tends to $0$ as $h \to 0$. 
