In the proof of Cauchy's Integral Formula: $$\oint_C{f(z)\over{z-a}}dz$$ The video said that it would be equals to: $$\oint_\Gamma{f(z)\over{z-a}}dz$$ where $\Gamma$ is a circle centred at $a$.
My question is: Why can the integral with a unknown $C$(closed curve) can be rewritten with form $\Gamma$ which is not same as $C$? Then if it is true, does the path affect the final result?
From the picture, $\int _{C_n \cup \Gamma_n} \frac{f(z)}{z-a} = 0$ by Cauchy Theorem since there is no poles within $C_n \cup \Gamma_n$.
$\lim_{n \rightarrow \infty} \int _{C_n \cup \Gamma_n} \frac{f(z)}{z-a} = \int _{C \cup -\Gamma} \frac{f(z)}{z-a} = \int _{C} \frac{f(z)}{z-a} - \int _{\Gamma} \frac{f(z)}{z-a} = 0$
Hence both integrals are equal since there are no poles in the region between $C$ and $\Gamma$.
Now the same argument extends for arbitrary $C$ and $\Gamma$ other than the nice circles drawn for illustration.
The integrals are same assuming the region between $C$ and $\Gamma$ has no poles and winding number of both $C$ and $\Gamma$ is $1$.