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pages/Math416/Math416_L12.md

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+# Math 416 Lecture 12
+
+## Continue on last class
+
+### Cauchy's Theorem on triangles
+
+Let $T$ be a triangle in $\mathbb{C}$ and $f$ be holomorphic on $T$. Then
+
+$$
+\int_T f(\zeta) d\zeta = 0
+$$
+
+### Cauchy's Theorem for Convex Sets
+
+Let's start with a simple case: $f(\zeta)=1$.
+
+For any closed curve $\gamma$ in $U$, we have
+
+$$
+\int_\gamma f(\zeta) d\zeta = \int_a^b f(\gamma(t)) \gamma'(t) dt \approx \sum_{i=1}^n f(\gamma(t_i)) \gamma'(t_i) \Delta t_i
+$$
+
+#### Definition of a convex set
+
+A set $U$ is convex if for any two points $\zeta_1, \zeta_2 \in U$, the line segment $[\zeta_1, \zeta_2] \subset U$.
+
+Let $O(U)$ be the set of all holomorphic functions on $U$.
+
+#### Definition of primitive
+
+Say $f$ has a primitive on $U$. If there exists a holomorphic function $g$ on $U$ such that $g'(\zeta)=f(\zeta)$ for all $\zeta \in U$, then $g$ is called a primitive of $f$ on $U$.
+
+#### Cauchy's Theorem for a Convex region
+
+Cauchy's Theorem holds if $f$ has a primitive on a convex region $U$.
+
+$$
+\int_\gamma f(\zeta) d\zeta = \int_\gamma \left[\frac{d}{d\zeta}g(\zeta)\right] d\zeta = g(\zeta_1)-g(\zeta_2)
+$$
+
+Since the curve is closed, $\zeta_1=\zeta_2$, so $\int_\gamma f(\zeta) d\zeta = 0$.
+
+Proof:
+
+It is sufficient to prove that if $U$ is convex, $f$ is holomorphic on $U$, then $f=g'$ for some $g$ holomorphic on $U$.
+
+We pick a point $z_0\in U$ and define $g(\zeta)=\int_{[\zeta_0,\zeta]}f(\xi)d\xi$.
+
+We claim $g\in O(U)$ and $g'=f$.
+
+Let $\zeta_1$ close to $\zeta$, since $f$ is holomorphic on $U$, using the Goursat's theorem, we can find a triangle $T$ with $\xi\in T$ and $\zeta\in T$ and $T\subset U$.
+
+$$
+\begin{aligned}
+0&=\int_{\zeta_0}^{\zeta}f(\xi)d\xi+\int_{\zeta}^{\zeta_1}f(\xi)d\xi\\
+&=g(\zeta)-g(\zeta_1)+\int_{\zeta}^{\zeta_1}f(\xi)d\xi+\int_{\zeta_1}^{\zeta_0}f(\xi)d\xi\\
+\frac{g(\zeta)-g(\zeta_1)}{\zeta-\zeta_1}&=-\frac{1}{\zeta-\zeta_1}\left(\int_{\zeta}^{\zeta_1}f(\xi)d\xi\right)\\
+\frac{g(\zeta_1)-g(\zeta_0)}{\zeta_1-\zeta_0}-f(\zeta_1)&=-\frac{1}{\zeta_1-\zeta_0}\left(\int_{\zeta}^{\zeta_1}f(\xi)d\xi\right)-f(\zeta_1)\\
+&=-\frac{1}{\zeta_1-\zeta_0}\left(\int_{\zeta}^{\zeta_1}f(\xi)-f(\zeta_1)d\xi\right)\\
+&=I
+\end{aligned}
+$$
+
+Use the fact that $f$ is holomorphic on $U$, then $f$ is continuous on $U$, so $\lim_{\zeta\to\zeta_1}f(\zeta)=f(\zeta_1)$.
+
+There exists a $\delta>0$ such that $|\zeta-\zeta_1|<\delta$ implies $|f(\zeta)-f(\zeta_1)|<\epsilon$.
+
+So
+
+$$
+|I|\leq\frac{1}{\zeta_1-\zeta_0}\int_{\zeta}^{\zeta_1}|f(\xi)-f(\zeta_1)|d\xi<\frac{\epsilon}{\zeta_1-\zeta_0}\int_{\zeta}^{\zeta_1}d\xi=\epsilon
+$$
+
+So $I\to 0$ as $\zeta_1\to\zeta$.
+
+Therefore, $g'(\zeta_1)=f(\zeta_1)$ for all $\zeta_1\in U$.
+
+EOP

pages/Math416/_meta.js

@@ -14,4 +14,5 @@ export default {
     Math416_L9: "Complex Variables (Lecture 9)",
     Math416_L10: "Complex Variables (Lecture 10)",
     Math416_L11: "Complex Variables (Lecture 11)",
+    Math416_L12: "Complex Variables (Lecture 12)",
 }