Let $U$ be an open set in $\Bbb C$ and $f\in H(U)$.
Fix a point $z\in U$. Consider the line integral $$\displaystyle \oint_{\partial D(z,\varepsilon)} \frac{f(\zeta)-f(z)}{\zeta-z}$$
Since $\frac{f(\zeta)-f(z)}{\zeta-z}$ is continuous on $U-\{z\}$ and $[0,2\pi]$ is compact,
$$\sup_{\theta\in [0,2\pi]} \lvert\frac{f(\zeta)-f(z)}{\zeta-z} \rvert$$ exists and is equal to $$\max_{\theta\in [0,2\pi]} \lvert\frac{f(\zeta)-f(z)}{\zeta-z} \rvert$$
Hence, $$\lvert\displaystyle \oint_{\partial D(z,\varepsilon)} \frac{f(\zeta)-f(z)}{\zeta-z}\rvert \le \sup_{\theta\in [0,2\pi]} \lvert\frac{f(\zeta)-f(z)}{\zeta-z} \rvert 2\pi\varepsilon \to 0 \;\;\text{as } \varepsilon \to 0$$
We get $\displaystyle \oint_{\partial D(z,\varepsilon)} \frac{f(\zeta)-f(z)}{\zeta-z}=0$.
Is my argument correct?
Thanks for your helping.
You have demonstrated that $$ \lim_{\varepsilon \to 0}\oint_{\partial D(z,\varepsilon)} \frac{f(\zeta)-f(z)}{\zeta-z} \, d\zeta = 0 \, . $$ In order to show that the integral itself it zero you would have to show that it is independent of $\varepsilon$. That is correct but requires some justification.
It may be easier to use Cauchy's integral theorem: The function $$ g(\zeta) = \frac{f(\zeta)-f(z)}{\zeta-z} $$ is holomorphic in $U \setminus \{ z \}$ with a removable singularity at $\zeta =z$ (because it has a limit there). It follows that $g$ can be extended homomorphically over $z$, and then Cauchy's integral theorem states that $$ \oint_{\partial D(z,\varepsilon)} \frac{f(\zeta)-f(z)}{\zeta-z} \, d\zeta = \oint_{\partial D(z,\varepsilon)} g(\zeta) \, d\zeta = 0 \, . $$