Complex analysis review 2(Cauchy Integral Theory)

Theorem 1(Cauchy-Green formula, Pompeiu formula)

Suppose that $U\subset C$ is a bounded domain, having $C^1$ boundary. $f(z) = u(x,y)+iv(x,y)\in C^1(\bar{U})$, then

$$f(z) = \frac{1}{2\pi i}\int_{\partial U}\frac{f(\xi)}{\xi - z}d \xi-\frac{1}{2\pi i}\iint_U \frac{\partial f(\xi)}{\partial \bar{\xi}}\frac{d\bar{\xi}\wedge d\xi}{\xi -z}.$$
We can prove this by first consider the domain $U_{z,\epsilon} = U\backslash D(z,\epsilon)$. Then use Green’s formula in complex form for
$$\frac{f(\xi)}{\xi -z}d\xi$$
in $U_{z,\epsilon}$. And note that
$$d = \partial + \bar{\partial}$$
Finally, let $\epsilon \rightarrow 0$.

Theorem 2 (Cauchy Integral Formula)

Suppose that $U\subset C$ is a bounded domain, having $C^1$ boundary, $f(z)$ is analytic on $U$, $f(z)\in C^1(\bar{U})$. Then

$$f(z) = \frac{1}{2\pi i}\int_{\partial U}\frac{f(\xi)}{\xi-z}dz.$$
This is because
$$\frac{\partial f(\xi)}{\partial \bar{\xi}}=0.$$

Theorem 3 (Cauchy Integral Theorem)

Suppose that $U\subset C$ is a bounded domain, having $C^1$ boundary, $F(z)$ is analytic on $U$, $F(z)\in C^1(\bar{U})$. Then

$$\int_{\partial U}F(\xi)d\xi=0.$$
Note that Theorem 2 and Theorem 3 are equivalent.

Cauchy-Goursat Theorem

It is often too strong to have smooth boundary. We can refine the condition and get the following theorems.

Theorem 2’

Suppose that $U\subset C$ is a bounded domain, and $\partial U$ is rectifiable. $f(z)$ is analytic on $U$, and continuous on $\bar{U}$. Then

$$f(z) = \frac{1}{2\pi i}\int_{\partial U}\frac{f(\xi)}{\xi-z}dz.$$

Theorem 3’

Suppose that $U\subset C$ is a bounded domain, and $\partial U$ is rectifiable. $f(z)$ is analytic on $U$, and continuous on $\bar{U}$. Then

$$\int_{\partial U}f(\xi)d\xi=0.$$
Note that Theorem 2’ and Theorem 3’ are also equivalent. And the theorems are true for multi-connected fields.