Cauchy Theorem¶
1. Simply and Multiply Connected Domains¶
Simply Connected Domain

A domain \(D\subseteq \mathbb{ C }\) is simply connected if \(D\) has no holes \(\Leftrightarrow C \setminus D\) is connected.
Multiply Connected Domain

A domain that is not simply connected is called multiply connected.
2. Cauchy Theorem¶
Theorem: Let \(D \subseteq C\) be simply connected domain and \(f:D\to \mathbb{C}\) is any analytic function. Then for any closed contour \(\gamma\) in \(D\) , We have \begin{align}\int_\gamma f(z) dz=0\end{align}
Proof: Let \(f(z)= u(x,y) + iv(x,y)\) and the contour
Now the integration will be
Now usign green theorem, we get the follwoing integral as
Since \(f\) is analytic, hence it will satisfy the CRequation, \(i.e.,\) \(u_x=v_y\) and \(v_x=u_y\)
Ex: Find the following integration
 \(\displaystyle \int_{\vert z \vert = 2}\frac{ e^z }{ z^29 }dz\),
 \(\displaystyle \int_{\vert z \vert = 2}\frac{ 2z+1 }{ e^z }dz\).
Here we have to evaluate the integral on a circle of radius \(2\), centerd at \(0\).
For (1), the function \(\displaystyle f(z) = \frac{ e^z }{ z^29 }\) is analytic in entire complex plane except \(z=\pm 3\), but these points are outside the contour, hence by Cauchy theorem, the integration will be \(0\).
For (2), the function \(\displaystyle f(z) = \frac{ 2z+1 }{ e^z }\) is analytic in entire complex plane, hence by Cauchy theorem, the integration will be \(0\).
Note: A function of the form of \(f/g\) is analytic everywhere except \(g=0\), provided \(f\) and \(g\) are analytic.
Last modified on: 20230105 00:02:30