update

pages/Math416/Math416_L12.md

@@ -75,4 +75,46 @@ So $I\to 0$ as $\zeta_1\to\zeta$.
 
 Therefore, $g'(\zeta_1)=f(\zeta_1)$ for all $\zeta_1\in U$.
 
-EOP
+EOP
+
+### Cauchy's Theorem for a disk
+
+Let $U$ be the open set, $f\in O(U)$. Let $C$ be a circle inside $U$ and $\zeta$ be a point inside $C$.
+
+Then
+
+$$
+f(\zeta)=\frac{1}{2\pi i}\int_C\frac{f(\xi)d\xi}{\xi-\zeta} d\xi
+$$
+
+Proof:
+
+Let $C_\epsilon$ be a circle with center $\zeta$ and radius $\epsilon$ inside $C$.
+
+Claim:
+
+$$
+\int_{C_\epsilon}\frac{f(\xi)d\xi}{\xi-\zeta}=\int_{C}\frac{f(\xi)d\xi}{\xi-\zeta}
+$$
+
+We divide the integral into four parts:
+
+![Integral on a disk](https://notenextra.trance-0.com/Math416/Cauchy_disk.png)
+
+Notice that $\frac{f(\xi)}{\xi-\zeta}$ is holomorphic whenever $f(\xi)\in U$ and $\xi\in \mathbb{C}\setminus\{\zeta\}$.
+
+So we can apply Cauchy's theorem to the integral on the inside square.
+
+$$
+\int_{C_\epsilon}\frac{f(\xi)d\xi}{\xi-\zeta}=0
+$$
+
+Since $\frac{1}{2\pi i}\int_{C_\epsilon}\frac{1}{\xi-\zeta}d\xi=1$, $\sigma=\epsilon e^{it}+\zeta_0$ and $\sigma'=\epsilon e^{it}$, we have
+
+/* TRACK LOST*/
+
+$$
+\int_{C_\epsilon}\frac{f(\xi)d\xi}{\xi-\zeta}=\int_0^{2\pi}\frac{f(\sigma)d\sigma}{\sigma-\zeta}=2\pi i f(\zeta)
+$$
+
+EOP