Cauchy integral formula

Theorem: Let $\Gamma$ be a simple closed positively oriented contour. If $f$ is analytic in some simply connected domain $D$ contain $\Gamma$, and $z_0$ is any point inside $\Gamma$, then \(f(z_0)= \frac{ 1 }{2\pi i }\int _\Gamma \frac{ f(z) }{ z-z_0 }dz\)

Proof: For sufficiently small $r>0$, $\gamma :\vert z-z_0 \vert =r,$ positively oriented lies inside $\Gamma,$ then by deformation theorem, we have

Now,

\begin{align} \int \gamma \frac{ f(z) }{ z-z_0 } dz &=\int _\gamma \frac{f(z)-f(z_0) + f(z_0)}{ z-z_0 } dz
&=\int _\gamma \frac{ f(z) }{ z-z_0 }dz+ \int\gamma \frac{f(z)-f(z_0) }{ z-z_0 } dz
&= f(z_0) \cdot 2\pi i + \int _\gamma \frac{ f(z)-f(z_0) }{ z-z_0 }dz \end{align}

In order to find the second integral, consider the following expression,

Using the above inequality, the integral will be

\begin{align} \left \vert \int_\gamma \frac{ f(z)-f(z_0 ) }{ z-z_0 } \right \vert &\leq \frac{ M_r }{ r } \int_\gamma \vert dz \vert
&= \frac{ M_r }{ r } \ell(\gamma)
&= \frac{ M_r }{ \cancel{r} }\cdot 2\pi \cancel{r}
&= 2\pi \cdot M_r \end{align}

But $f$ is continuous at $z_0$, hence taking the limit $r\to 0$, we get

By deformation we can make this $r$ arbitralily small, hence taking the limit $r\to 0$, we have

Hence the given integral will be,

Credits: Pinku Kumar