# 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

$\int _\Gamma \frac{ f(z) }{ z-z_0 }=\int_\gamma \frac{ f(z) }{ z-z_0 }dz$

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,

$\left \vert \frac{ f(z)-f(z_0 ) }{ z-z_0 } \right \vert = \frac{ \vert f(z)-f(z_0 )\vert}{ r }\leq \frac{ M_r }{ r }$

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

$\lim_{r \to 0} \vert f(z) - f(z_0) \vert = \lim_{r \to 0} M_r = 0,$

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

\require{cancel} \begin{align} \lim_{r\to 0}\left \vert \int_\gamma \frac{ f(z)-f(z_0 ) }{ z-z_0 } \right \vert &\leq \lim_{r\to 0} 2\pi M_r = 0 \end{align}

Hence the given integral will be,

\begin{align} \int _\Gamma \frac{ f(z) }{ z-z_0 } dz &= f(z_0) \cdot 2\pi i \\ f(z_0) &= \frac{ 1 }{ 2\pi i } \int _\Gamma \frac{ f(z) }{ z-z_0 } dz \end{align}

Credits: Pinku Kumar