Cauchy Theorem

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.

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

\begin{align} \gamma(t) &= x(t) + i y(t);& t\in [a,b]
\gamma’(t) &= x’(t) + i y’(t)& \end{align}

Now the integration will be

\begin{align} \int_\gamma f(z) dz &= \int_a^b f(\gamma(t))\gamma’(t)dt
&= \int_a^b (u(x,y) + iv(x,y)) (x’ + iy’) dt
&= \int_a^b (ux’-vy’) dt + i\int_a^b(vx’+uy’) dt
&= \oint_\gamma udx - vdy + i \oint_\gamma vdx +udy \end{align}

Now usign green theorem, we get the follwoing integral as

\begin{align} \phantom{\int_\gamma f(z) dz} &= \iint\limits_{R}(-v_x - u_y)dy dx +\iint\limits_{R}(u_x - v_y) dy dx \end{align}

Since $f$ is analytic, hence it will satisfy the CR-equation, $i.e.,$ $u_x=v_y$ and $v_x=-u_y$

\begin{align} \phantom{\int_\gamma f(z) dz} &= \iint\limits_{R}(-v_x - u_y)dy dx +\iint\limits_{R}(u_x - v_y) dy dx
&= 0. \end{align}

Ex: Find the following integration

$\displaystyle \int_{\vert z \vert = 2}\frac{ e^z }{ z^2-9 }dz$,
$\displaystyle \int_{\vert z \vert = 2}\frac{ 2z+1 }{ e^z }dz$.

Here we have to evaluate the integral on a circle of radius $2$, centerd at $0$.

For (1), the function $\displaystyle f(z) = \frac{ e^z }{ z^2-9 }$ 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.