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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

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


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

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

Last modified on: 2023-01-05 00:02:30