Prove that if $\displaystyle f$ is derivable in $\displaystyle z_0$ then $\displaystyle f$ is continuous in $\displaystyle z_0$. $\displaystyle \Omega \in \mathbb{C}, ~f:\Omega \rightarrow \mathbb{C}, ~ z_0 \in \Omega$

I started with the definition.

$\displaystyle \lim_{z \to 0} \frac {f(z_0+\Delta z) - f(z_0)}{\Delta z}$

Using Cauchys integral formula.

$\displaystyle \frac {f(z_0+\Delta z) - f(z_0)}{\Delta z}$ =

$\displaystyle \frac{1}{2\pi i \Delta z} \oint_C \frac{f(z)}{z-(z_0+\Delta z)} \, dz -\frac{1}{2\pi i \Delta z} \oint_C \frac{f(z)}{z-z_0} \, dz $ =

$\displaystyle \frac{1}{2\pi i \Delta z} \oint_C \frac{f(z)\Delta z}{(z-z_0-\Delta z)(z-z_0)} \, dz $ =

$\displaystyle \frac{1}{2\pi i } \oint_C \frac{f(z)}{(z-z_0-\Delta z)(z-z_0)} \, dz $

Now comparing the result with Cauchy's Integral formula, first order.

$\displaystyle \oint_C \frac{f(z)}{(z-z_0-\Delta z)(z-z_0)} \, dz - \oint_C \frac{f(z)}{(z-z_0)^2} \, dz$ = $\displaystyle \oint_C \frac{f(z)\Delta z}{(z-z_0-\Delta z)(z-z_0)^2} \, dz$

What happens with the right side if $\displaystyle \Delta z \rightarrow 0$?

I'm sort of in the dark here... I thought maybe I proved something by showing how Cauchy's and the definition of derivation dance so nicely together.

Help or any pointers would be appreciated!