I posted my attempted proof to this question here but I realized that I was wrong in taking the limit, and that the proof did not make sense. So I am still stuck on this problem
let $f: \Omega \rightarrow \mathbb{C}$ be analytic and $z_0 \in \mathbb{C}$.
Define
$$g(z) = \begin{cases}
\frac{f(z)-f(z_0)}{z- z_0} & z \not = z_0 \\
f'(z_0) & z = z_0
\end{cases}$$
now pick $\varepsilon$ small enough so that $\overline{D(z_0, \varepsilon)} \subset \Omega$
Show that whenever $z \in D(z_0, \varepsilon)$
$$\frac{g(z) – g(z_0)}{z-z_0} = \frac{1}{2\pi i}\int_{\partial D(z_0, \varepsilon)}\frac{f(\zeta)}{(\zeta-z)(\zeta-z_0)^2}d\zeta$$
So is Cauchy's integral formula still the right way to go? I end up getting that
$$\frac{g(z) – g(z_0)}{z-z_0} = \frac{f(z) – f(z_0)}{z-z_0} – \frac{1}{z-z_0}\int_{\partial D(z_0, \varepsilon)} \frac{f(\zeta)}{(\zeta-z_0)^2}d\zeta$$ and I am not sure how to proceed from here
Best Answer
Here is an approach that avoids the messy calculation. Fix $z$ and $z_0$ in $D(z_0, \varepsilon)$. Then, the residue theorem gives
$\frac{1}{2\pi i}\int_{\partial D(z_0, \varepsilon)}\frac{f(\zeta)}{(\zeta-z)(\zeta-z_0)^2}d\zeta=\frac{f'(z_0)(z_0-z)-f(z_0)}{(z_0-z)^{2}}+\frac{f(z)}{(z-z_0)^{2}}=\frac{f(z)-f(z_0)}{(z-z_0)^{2}}-\frac{f'(z)}{z-z_0}.$
On the other hand, by direct substitution,
$\frac{g(z)-g(z_0)}{z-z_0}=\frac{f(z)-f(z_0)}{(z- z_0)^2}-\frac{f'(z_0)}{z-z_0}.$