update notations and fix typos

pages/Math416/Math416_L11.md

@@ -4,23 +4,23 @@
 
 ### Continue on last example
 
-Last lecture we have:Let $R$ be a rectangular start from the $-a$ to $a$, $a+ib$ to $-a+ib$, $\int_{R} e^{-\zeta^2}d\zeta=0$, however, the integral consists of four parts:
+Last lecture we have:Let $R$ be a rectangular start from the $-a$ to $a$, $a+ib$ to $-a+ib$, $\int_{R} e^{-z^2}dz=0$, however, the integral consists of four parts:
 
 Path 1: $-a\to a$
 
-$\int_{I_1}e^{-\zeta^2}d\zeta=\int_{-a}^{a}e^{-\zeta^2}d\zeta=\int_{-a}^{a}e^{-x^2}dx$
+$\int_{I_1}e^{-z^2}dz=\int_{-a}^{a}e^{-z^2}dz=\int_{-a}^{a}e^{-x^2}dx$
 
 Path 2: $a+ib\to -a+ib$
 
-$\int_{I_2}e^{-\zeta^2}d\zeta=\int_{a+ib}^{-a+ib}e^{-\zeta^2}d\zeta=\int_{0}^{b}e^{-(a+iy)^2}dy$
+$\int_{I_2}e^{-z^2}dz=\int_{a+ib}^{-a+ib}e^{-z^2}dz=\int_{0}^{b}e^{-(a+iy)^2}dy$
 
 Path 3: $-a+ib\to -a-ib$
 
-$-\int_{I_3}e^{-\zeta^2}d\zeta=-\int_{-a+ib}^{-a-ib}e^{-\zeta^2}d\zeta=-\int_{a}^{-a}e^{-(x-ib)^2}dx$
+$-\int_{I_3}e^{-z^2}dz=-\int_{-a+ib}^{-a-ib}e^{-z^2}dz=-\int_{a}^{-a}e^{-(x-ib)^2}dx$
 
 Path 4: $-a-ib\to a-ib$
 
-$-\int_{I_4}e^{-\zeta^2}d\zeta=-\int_{-a-ib}^{a-ib}e^{-\zeta^2}d\zeta=-\int_{b}^{0}e^{-(-a+iy)^2}dy$
+$-\int_{I_4}e^{-z^2}dz=-\int_{-a-ib}^{a-ib}e^{-z^2}dz=-\int_{b}^{0}e^{-(-a+iy)^2}dy$
 
 > #### The reverse of a curve 6.9
 >
@@ -37,12 +37,12 @@ If we keep $b$ fixed, and let $a\to\infty$, then
 >
 > Then
 >
-> $$\left|\int_{\gamma}f(\zeta)d\zeta\right|\leq L(\gamma)M$$
+> $$\left|\int_{\gamma}f(z)dz\right|\leq L(\gamma)M$$
 
 _Continue on previous example, we have:_
 
 $$
-\left|\int_{\gamma}f(\zeta)d\zeta\right|\leq L(\gamma)\max_{\zeta\in\gamma}|f(\zeta)|\to 0
+\left|\int_{\gamma}f(z)dz\right|\leq L(\gamma)\max_{z\in\gamma}|f(z)|\to 0
 $$
 
 Since,
@@ -89,7 +89,7 @@ $$
 J=\sqrt{\pi}
 $$
 
-EOP
+QED
 
 ## Chapter 7 Cauchy's theorem
 
@@ -109,7 +109,7 @@ Cauchy's theorem is true if $\gamma$ is a triangle.
 
 Proof:
 
-We plan to keep shrinking the triangle until $f(\zeta+h)=f(\zeta)+hf'(\zeta)+\epsilon(h)$ where $\epsilon(h)$ is a function of $h$ that goes to $0$ as $h\to 0$.
+We plan to keep shrinking the triangle until $f(z+h)=f(z)+hf'(z)+\epsilon(h)$ where $\epsilon(h)$ is a function of $h$ that goes to $0$ as $h\to 0$.
 
 Let's start with a triangle $T$ with vertices $z_1,z_2,z_3$.
 
@@ -127,21 +127,21 @@ Since $L(T_1)=\frac{1}{2}L(T)$, we iterate after $n$ steps, get a triangle $T_n$
 
 Since $K_n=T_n\cup \text{interior}(T_n)$ is compact, we can find $K_n+1\subset K_n$ and $diam(K_n+1)<\frac{1}{2}diam(K_n)$. $diam(K_n)\to 0$ as $n\to\infty$. (Using completeness theorem)
 
-Since $f$ is holomorphic on $u$, $\lim_{\zeta\to z_0}\frac{f(\zeta)-f(z_0)}{\zeta-z_0}=f'(z_0)$ exists.
+Since $f$ is holomorphic on $u$, $\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}=f'(z_0)$ exists.
 
-So $f(\zeta)=f(z_0)+f'(z_0)(\zeta-z_0)+R(\zeta)$, we have
+So $f(z)=f(z_0)+f'(z_0)(z-z_0)+R(z)$, we have
 
 $$
-\int_{T_n}f(\zeta)d\zeta=\int_{T_n}f(z_0)d\zeta+\int_{T_n}f'(z_0)(\zeta-z_0)d\zeta+\int_{T_n}R(\zeta)d\zeta
+\int_{T_n}f(z)dz=\int_{T_n}f(z_0)dz+\int_{T_n}f'(z_0)(z-z_0)dz+\int_{T_n}R(z)dz
 $$
 
-since $f(z_0)d\zeta+\int_{T_n}f'(z_0)(\zeta-z_0)$ is in form of Cauchy integral formula, we have
+since $f(z_0)dz+\int_{T_n}f'(z_0)(z-z_0)$ is in form of Cauchy integral formula, we have
 
 $$
-\int_{T_n}f(z_0)d\zeta+\int_{T_n}f'(z_0)(\zeta-z_0)d\zeta=0
+\int_{T_n}f(z_0)dz+\int_{T_n}f'(z_0)(z-z_0)dz=0
 $$
 
-Let $e_n=\max\{\frac{R(\zeta)}{\zeta-z_0}:z_0\in T_n\}$
+Let $e_n=\max\{\frac{R(z)}{z-z_0}:z_0\in T_n\}$
 
 Since $diam(K_n)\to 0$ as $n\to\infty$, we have $e_n\to 0$ as $n\to\infty$.
 
@@ -149,9 +149,9 @@ So
 
 $$
 \begin{aligned}
-|I|&\leq 4^n\left|\int_{T_n}f(\zeta)d\zeta\right|\\
-&\leq 4^n\left|\int_{T_n}R_n(\zeta)d\zeta\right|\\
-&\leq 4^n\cdot L(T_n)\cdot \max_{\zeta\in T_n}|R_n(\zeta)|\\
+|I|&\leq 4^n\left|\int_{T_n}f(z)dz\right|\\
+&\leq 4^n\left|\int_{T_n}R_n(z)dz\right|\\
+&\leq 4^n\cdot L(T_n)\cdot \max_{z\in T_n}|R_n(z)|\\
 &\leq 4^n\cdot \frac{L(T_0)}{2^n}\cdot e_n L(T_n)\\
 &\leq 4^n\cdot \frac{L(T_0)}{2^n}\cdot e_n\cdot \frac{L(T_0)}{2^n}\\
 &\leq e_n\cdot L(T_0)^2
@@ -163,7 +163,7 @@ Since $e_n\to 0$ as $n\to\infty$, we have $I\to 0$ as $n\to\infty$.
 So
 
 $$
-\int_{T_n}f(\zeta)d\zeta\to 0
+\int_{T_n}f(z)dz\to 0
 $$
 
-EOP
+QED