update notations and fix typos

pages/Math416/Math416_L12.md

@@ -7,114 +7,114 @@
 Let $T$ be a triangle in $\mathbb{C}$ and $f$ be holomorphic on $T$. Then
 
 $$
-\int_T f(\zeta) d\zeta = 0
+\int_T f(z) dz = 0
 $$
 
 ### Cauchy's Theorem for Convex Sets
 
-Let's start with a simple case: $f(\zeta)=1$.
+Let's start with a simple case: $f(z)=1$.
 
 For any closed curve $\gamma$ in $U$, we have
 
 $$
-\int_\gamma f(\zeta) d\zeta = \int_a^b f(\gamma(t)) \gamma'(t) dt \approx \sum_{i=1}^n f(\gamma(t_i)) \gamma'(t_i) \Delta t_i
+\int_\gamma f(z) dz = \int_a^b f(\gamma(t)) \gamma'(t) dt \approx \sum_{i=1}^n f(\gamma(t_i)) \gamma'(t_i) \Delta t_i
 $$
 
 #### Definition of a convex set
 
-A set $U$ is convex if for any two points $\zeta_1, \zeta_2 \in U$, the line segment $[\zeta_1, \zeta_2] \subset U$.
+A set $U$ is convex if for any two points $z_1, z_2 \in U$, the line segment $[z_1, z_2] \subset U$.
 
 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 $U$.
+Say $f$ has a primitive on $U$. If there exists a holomorphic function $g$ on $U$ such that $g'(z)=f(z)$ for all $z \in U$, 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 $U$.
 
 $$
-\int_\gamma f(\zeta) d\zeta = \int_\gamma \left[\frac{d}{d\zeta}g(\zeta)\right] d\zeta = g(\zeta_1)-g(\zeta_2)
+\int_\gamma f(z) dz = \int_\gamma \left[\frac{d}{dz}g(z)\right] dz = g(z_1)-g(z_2)
 $$
 
-Since the curve is closed, $\zeta_1=\zeta_2$, so $\int_\gamma f(\zeta) d\zeta = 0$.
+Since the curve is closed, $z_1=z_2$, so $\int_\gamma f(z) dz = 0$.
 
 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 pick a point $z_0\in U$ and define $g(z)=\int_{[z_0,z]}f(\xi)d\xi$.
 
 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 $T\subset U$.
+Let $z_1$ close to $z$, since $f$ is holomorphic on $U$, using the Goursat's theorem, we can find a triangle $T$ with $\xi\in T$ and $z\in T$ and $T\subset U$.
 
 $$
 \begin{aligned}
-0&=\int_{\zeta_0}^{\zeta}f(\xi)d\xi+\int_{\zeta}^{\zeta_1}f(\xi)d\xi\\
-&=g(\zeta)-g(\zeta_1)+\int_{\zeta}^{\zeta_1}f(\xi)d\xi+\int_{\zeta_1}^{\zeta_0}f(\xi)d\xi\\
-\frac{g(\zeta)-g(\zeta_1)}{\zeta-\zeta_1}&=-\frac{1}{\zeta-\zeta_1}\left(\int_{\zeta}^{\zeta_1}f(\xi)d\xi\right)\\
-\frac{g(\zeta_1)-g(\zeta_0)}{\zeta_1-\zeta_0}-f(\zeta_1)&=-\frac{1}{\zeta_1-\zeta_0}\left(\int_{\zeta}^{\zeta_1}f(\xi)d\xi\right)-f(\zeta_1)\\
-&=-\frac{1}{\zeta_1-\zeta_0}\left(\int_{\zeta}^{\zeta_1}f(\xi)-f(\zeta_1)d\xi\right)\\
+0&=\int_{z_0}^{z}f(\xi)d\xi+\int_{z}^{z_1}f(\xi)d\xi\\
+&=g(z)-g(z_1)+\int_{z}^{z_1}f(\xi)d\xi+\int_{z_1}^{z_0}f(\xi)d\xi\\
+\frac{g(z)-g(z_1)}{z-z_1}&=-\frac{1}{z-z_1}\left(\int_{z}^{z_1}f(\xi)d\xi\right)\\
+\frac{g(z_1)-g(z_0)}{z_1-z_0}-f(z_1)&=-\frac{1}{z_1-z_0}\left(\int_{z}^{z_1}f(\xi)d\xi\right)-f(z_1)\\
+&=-\frac{1}{z_1-z_0}\left(\int_{z}^{z_1}f(\xi)-f(z_1)d\xi\right)\\
 &=I
 \end{aligned}
 $$
 
-Use the fact that $f$ is holomorphic on $U$, then $f$ is continuous on $U$, so $\lim_{\zeta\to\zeta_1}f(\zeta)=f(\zeta_1)$.
+Use the fact that $f$ is holomorphic on $U$, then $f$ is continuous on $U$, so $\lim_{z\toz_1}f(z)=f(z_1)$.
 
-There exists a $\delta>0$ such that $|\zeta-\zeta_1|<\delta$ implies $|f(\zeta)-f(\zeta_1)|<\epsilon$.
+There exists a $\delta>0$ such that $|z-z_1|<\delta$ implies $|f(z)-f(z_1)|<\epsilon$.
 
 So
 
 $$
-|I|\leq\frac{1}{\zeta_1-\zeta_0}\int_{\zeta}^{\zeta_1}|f(\xi)-f(\zeta_1)|d\xi<\frac{\epsilon}{\zeta_1-\zeta_0}\int_{\zeta}^{\zeta_1}d\xi=\epsilon
+|I|\leq\frac{1}{z_1-z_0}\int_{z}^{z_1}|f(\xi)-f(z_1)|d\xi<\frac{\epsilon}{z_1-z_0}\int_{z}^{z_1}d\xi=\epsilon
 $$
 
-So $I\to 0$ as $\zeta_1\to\zeta$.
+So $I\to 0$ as $z_1\toz$.
 
-Therefore, $g'(\zeta_1)=f(\zeta_1)$ for all $\zeta_1\in U$.
+Therefore, $g'(z_1)=f(z_1)$ for all $z_1\in U$.
 
-EOP
+QED
 
 ### 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 $C$.
+Let $U$ be the open set, $f\in O(U)$. Let $C$ be a circle inside $U$ and $z$ be a point inside $C$.
 
 Then
 
 $$
-f(\zeta)=\frac{1}{2\pi i}\int_C\frac{f(\xi)d\xi}{\xi-\zeta} d\xi
+f(z)=\frac{1}{2\pi i}\int_C\frac{f(\xi)d\xi}{\xi-z} d\xi
 $$
 
 Proof:
 
-Let $C_\epsilon$ be a circle with center $\zeta$ and radius $\epsilon$ inside $C$.
+Let $C_\epsilon$ be a circle with center $z$ and radius $\epsilon$ inside $C$.
 
 Claim:
 
 $$
-\int_{C_\epsilon}\frac{f(\xi)d\xi}{\xi-\zeta}=\int_{C}\frac{f(\xi)d\xi}{\xi-\zeta}
+\int_{C_\epsilon}\frac{f(\xi)d\xi}{\xi-z}=\int_{C}\frac{f(\xi)d\xi}{\xi-z}
 $$
 
 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 $\xi\in \mathbb{C}\setminus\{\zeta\}$.
+Notice that $\frac{f(\xi)}{\xi-z}$ is holomorphic whenever $f(\xi)\in U$ and $\xi\in \mathbb{C}\setminus\{z\}$.
 
 So we can apply Cauchy's theorem to the integral on the inside square.
 
 $$
-\int_{C_\epsilon}\frac{f(\xi)d\xi}{\xi-\zeta}=0
+\int_{C_\epsilon}\frac{f(\xi)d\xi}{\xi-z}=0
 $$
 
-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
+Since $\frac{1}{2\pi i}\int_{C_\epsilon}\frac{1}{\xi-z}d\xi=1$, $\sigma=\epsilon e^{it}+z_0$ and $\sigma'=\epsilon e^{it}$, we have
 
 /* TRACK LOST*/
 
 $$
-\int_{C_\epsilon}\frac{f(\xi)d\xi}{\xi-\zeta}=\int_0^{2\pi}\frac{f(\sigma)d\sigma}{\sigma-\zeta}=2\pi i f(\zeta)
+\int_{C_\epsilon}\frac{f(\xi)d\xi}{\xi-z}=\int_0^{2\pi}\frac{f(\sigma)d\sigma}{\sigma-z}=2\pi i f(z)
 $$
 
-EOP
+QED