Cauchy Integral Formula

Theorem
Let $f \left({z}\right)$ be an analytic function on and within a closed contour $C$.

Let $z_0$ be any point within $C$.

Then:


 * $\displaystyle \int_C \frac {f \left({z}\right)} {z - z_0} \ \mathrm d z = 2 \pi i f \left({z_0}\right)$

Proof
Let $C$ be any arbitrary closed curve which defines a region $R$ where the function $f \left({z}\right)$ is analytic.

Let $z_0$ be any point in the region $R$ such that:


 * $\dfrac {f \left({z}\right)}{z - z_0}$ is analytic everywhere except at $z_0$.

We draw a circle $C_1$ with center at $z_0$ and radius $r$ such that $r \to 0$. Thus making $C$ and $C_1$ multiply connected region.

According to Cauchy's Integral Theorem for multiply connected region,

$\displaystyle \oint_C { \frac { f \left({z}\right) }{z - z_0} dz } =\oint_{C_1}{ \frac { f \left({z}\right) }{z - z_0} dz } = I $ (say)

$\displaystyle I=\oint_{C_1}{ \frac { f \left({z_0}\right)+\left[ f \left({z}\right)-f \left({z_0}\right) \right] }{z - z_0} dz }

\\ \displaystyle =f \left({z_0}\right)\oint_{C_1}{ \frac { dz }{z - z_0}} + \oint_{C_1}{ \frac {f \left({z}\right) - f \left({z_0}\right)}{z - z_0} dz }$

Let $z - z_0 = r e^{i \theta}$.

And, $dz = i r e^{i\theta}$

So, $\displaystyle \oint_{C_1}{ \frac { dz }{z - z_0} } =\int _{ 0 }^{ 2\pi  }{ \frac { ir{ e }^{ i\theta  } }{ r{ e }^{ i\theta  } } d\theta  } =i\int _{ 0 }^{ 2\pi  }{ d\theta =2\pi i } $

Now, $\displaystyle I=2\pi i+\oint_{C_1}{ \frac { f \left({z}\right)-f \left({z_0}\right)}{z - z_0} dz } $

According to Epsilon-Delta definition of a function, for every $\left| z - z_0 \right| <\delta $ there exists a $\epsilon $ such that $\left| f \left({z}\right)-f \left({z_0}\right) \right| <\epsilon$.

$\displaystyle\left| \oint_{C_1}{ \frac { f \left({z}\right)-f \left({z_0}\right)}{z - z_0} dz } \right| \le \oint_{C_1}{ \frac { \left| f \left({z}\right) - f \left({z_0}\right) \right|  }{z - z_0} \left| dz \right|  } \le \frac {\epsilon}{\delta} \oint_{C_1}{ \left| dz \right|} =2\pi \epsilon $

As $\epsilon \to 0$, $\displaystyle \oint_{C_1}{ \frac { f \left({z}\right)-f \left({z_0}\right)}{z - z_0} dx } =0$.

$\displaystyle I=f \left({z_0}\right)\oint_{C_1}{ \frac { dz }{z - z_0} } } +\oint_{C_1}{ \frac { f \left({z}\right)-f \left({z_0}\right)}{z - z_0} } dx } \\\displaystyle\Rightarrow I=2\pi i f \left({z_0}\right) + 0 \\\displaystyle\Rightarrow \oint_{ C }{ \frac { f \left({z}\right) }{z - z_0} dz } = 2\pi i f \left({z_0}\right)$