Again, I'm struggling with a proof.
Cauchy integral theorem for convex sets (Preliminary lemma)
Let $\Omega\subseteq \mathbb{C}$ open and convex, $ p\in\Omega,\ f$ : $\Omega\rightarrow \mathbb{C}$ continuous, $f\in H(\Omega\backslash \{p\})$ ($f$ analytic on $\Omega \backslash \{p\}$. Then there is a $F\in H(\Omega)$ (F analytic on $\Omega$) with $F'(z)=f(z)$ for all $ z\in\Omega$.
It also folows that $\displaystyle \int\limits_{\gamma}f(z)\mathrm{d}z=0$ for every piecewise continuously differentiable and closed path $\gamma\in\Omega$.
Cauchy integral formula
Let $\Omega\subseteq \mathbb{C}$ be open and convex, $f\in H(\Omega)$ ($f$ analytic on $\Omega$) and $\gamma$ a piecewise continuously differentiable and closed path in $\Omega$. Then, we have: $ f(z)\cdot \mathrm{ind}_{\gamma}(z)=\frac{1}{2\pi i}\int\frac{f(\xi)}{\xi-z}\mathrm{d}\xi\ , (z\in\Omega\backslash \gamma^{*}) $ where $\mathrm{ind}_\gamma$ is the winding number and $\gamma^* = \gamma([a,b])$ for $\gamma:[a,b]\rightarrow\mathbb{C}$.
Proof
Fix $z\in\Omega\backslash \gamma^{*}$ and define $g:\Omega\rightarrow \mathbb{C}$ by: $ g(\xi):=\left\{\begin{array}{l} \frac{f(\xi)-f(z)}{\xi-z},\ \xi\in\Omega\backslash \{z\}\\ f'(z),\ \xi=z \end{array}\right. $ Then we have $g$ continuous on $\Omega$ and $g \in H(\Omega\backslash \{z\})$.
$\color{red}{\text{(1) Why is g continuous on } \Omega \text{ and analytical on } \Omega\backslash \{z\}}$
From the Cauchy integral theorem for convex sets we have $\displaystyle \frac{1}{2\pi i}\int\limits_{\gamma}g(\xi)\mathrm{d}\xi=0$, so:
$\color{red}{\text{(2) Where does that } \frac{1}{2\pi i} \text{ come from? There is no factor in the mentioned Cauchy integral theorem}}$
$ \frac{1}{2\pi i}\int_{\gamma}\frac{f(\xi)-f(z)}{\xi-z}\mathrm{d}\xi\ =0\ (z\in\Omega\backslash \gamma^{*}) $
This completes the proof.