# Proof using Cauchy's Integral Formula

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 $$z \in D(z_0, \varepsilon)$$

$$\frac{g(z) - g(z_0)}{z-z_0} = \frac{1}{2\pi i}\int_{D(z_0, \varepsilon)}\frac{f(\zeta)}{(\zeta-z)(\zeta-z_0)^2}d\zeta$$

My question is if my proof below, which makes use of the Cauchy integral formula is correct

We can write $$\frac{g(z) - g(z_0)}{z-z_0} = \frac{\frac{f(z)-f(z_0)}{z- z_0}-f'(z_0)}{z-z_0} = \frac{f(z)-f(z_0)}{(z- z_0)^2} - \frac{f'(z_0)}{z-z_0}$$

now as epsilon gets arbitrarily small, $$z \rightarrow z_0$$ and so now making use of the Cauchy integral formula and the fact that $$\lim_{z \rightarrow z_0} \frac{f(z_0)}{z-z_0} = 0$$ we should have

$$\lim_{z \rightarrow z_0} \frac{f(z)-f(z_0)}{(z- z_0)^2} - \frac{f'(z_0)}{z-z_0} = \frac{1}{2 \pi i}\int_{D(z_0, \varepsilon)}\frac{f(\zeta)}{(\zeta-z)(\zeta-z_0)^2}d\zeta - 0$$

which gives us the result