Is it possible to prove Cauchy integral formula by parametrizing the circle?

Question

Theorem. Let $f$ be analytic everywhere inside and on a simple closed contour $C$, taken in the positive sense. If $z_0$ is any point interior to $C$, then $$f(z_0) = \frac{1}{2\pi i} \int_C \frac{f(z)dz}{z-z_0}.$$

The proof starts with letting $C_r$ denote a positively oriented circle $|z-z_0|=r$, where $r$ is small enough that $C_r$ is interior to $C$. Since the function $f(z)/(z-z_0)$ is analytic between and on the contours $C$ and $C_r$, we have $$ \int_C \frac{f(z)dz}{z-z_0} = \int_{C_r} \frac{f(z)dz}{z-z_0}.$$

Can we then parametrize the contour $C_r$ and calculate $\int_{C_r} \frac{f(z)dz}{z-z_0}$? How to proceed?

Answer 1

One can proceed using mean value theorem and $r\to 0$: \begin{align} \int_{C_r}\frac{f(z)}{z-z_0}\,dz&=\int_0^{2\pi}\frac{f(z_0+re^{i\theta})}{re^{i\theta}}re^{i\theta}i\,d\theta=i\int_0^{2\pi}f(z_0+re^{i\theta})\,d\theta=\\ &=[\text{Re}\,f(z_0+re^{ic_1})+i\text{Im}\,f(z_0+re^{ic_2})]2\pi i\to f(z_0)2\pi i. \end{align}

Answer 2

$f(z)=\sum a_n (z-z_0)^{n}$ gives $\int_0^{2\pi} f(z+re^{i\theta}) d\theta=\sum a_n \int_0^{2\pi}r^{n} e^{in\theta} d\theta=2 \pi a_0=2 \pi f(z_0)$ (since all but the first term vanish). Now $\int_{C_r} \frac {f(z)} {z-z_0} dz =i \int_0^{2\pi} f(z+re^{i\theta}) d\theta=2\pi i f(z_0)$.