From ce4fcd7cd641f12282a370c39e4a5ab3e11359fb Mon Sep 17 00:00:00 2001
From: Trance-0 <60459821+Trance-0@users.noreply.github.com>
Date: Mon, 5 May 2025 00:25:24 -0500
Subject: [PATCH] Update Math416_Final.md

---
 pages/Math416/Exam_reviews/Math416_Final.md | 167 +++++++++++++++++++-
 1 file changed, 160 insertions(+), 7 deletions(-)

diff --git a/pages/Math416/Exam_reviews/Math416_Final.md b/pages/Math416/Exam_reviews/Math416_Final.md
index fa3f5fc..316a0d8 100644
--- a/pages/Math416/Exam_reviews/Math416_Final.md
+++ b/pages/Math416/Exam_reviews/Math416_Final.md
@@ -88,22 +88,175 @@ The series $f(z)=\sum_{n=-\infty}^{-1}a_n(z-z_0)^n$ is called the principle part
 
 If the principle part of Laurent series of $f$ at $z_0$ is zero, then $z_0$ is called a removable singularity of $f$.
 
+Criterion for a removable singularity:
+
+If $f$ is bounded on $0<|z-z_0|<R$, then $z_0$ is a removable singularity.
+
 > Example:
 >
-> $f(z)=\frac{e^z-1}{z^2}$ has a removable singularity at $z=0$.
+> $f(z)=\frac{1}{e^z-1}$ has a removable singularity at $z=0$.
 >
-> The Laurent series of $f$ at $z=0$ can be found using the Taylor series of $e^z-1$ at $z=0$.
+> The Laurent series of $f$ at $z=0$ is
 >
-> $$e^z-1=z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots$$
->
-> So the Laurent series of $f$ at $z=0$ is
->
-> $$f(z)=\frac{1}{z^2}+\frac{1}{z}+\sum_{n=0}^{\infty}\frac{z^n}{n!}$$
+> $$f(z)=\frac{1}{z}+\sum_{n=0}^{\infty}\frac{z^n}{n!}$$
 >
 > The principle part is zero, so $z=0$ is a removable singularity.
 
 #### Poles
 
+If the principle part of Laurent series of $f$ at $z_0$ is a finite sum, then $z_0$ is called a pole of $f$.
+
+Criterion for a pole:
+
+If $f$ has an isolated singularity at $z_0$, and $\lim_{z\to z_0}|f(z)|=\infty$, then $z_0$ is a pole of $f$.
+
+> Example:
+>
+> $f(z)=\frac{1}{z^2}$ has a pole at $z=0$.
+>
+> The Laurent series of $f$ at $z=0$ is
+>
+> $$f(z)=\frac{1}{z^2}$$
+>
+> The principle part is $\frac{1}{z^2}$, so $z=0$ is a pole.
+
+#### Essential singularities
+
+If $f$ has an isolated singularity at $z_0$, and it is neither a removable singularity nor a pole, then $z_0$ is called an essential singularity of $f$.
+
+"Criterion" for an essential singularity:
+
+If the principle part of Laurent series of $f$ at $z_0$ has infinitely many non-zero coefficients corresponding to negative powers of $z-z_0$, then $z_0$ is called an essential singularity of $f$.
+
+> Example:
+>
+> $f(z)=\sin(\frac{1}{z})$ has an essential singularity at $z=0$.
+>
+> The Laurent series of $f$ at $z=0$ is
+>
+> $$f(z)=\frac{1}{z}-\frac{1}{6z^3}+\frac{1}{120z^5}-\cdots$$
+>
+> Since there are infinitely many non-zero coefficients corresponding to negative powers of $z$, $z=0$ is an essential singularity.
+
+#### Singularities at infinity
+
+We say $f$ has a singularity (removable, pole, or essential) at infinity if $f(1/z)$ has an isolated singularity (removable, pole, or essential) at $z=0$.
+
+> Example:
+>
+> $f(z)=\frac{z^4}{(z-1)(z-3)}$ has a pole of order 2 at infinity.
+>
+> Plug in $z=1/w$, we get $f(1/w)=\frac{1}{w^2}\frac{1}{(1/w-1)(1/w-3)}=\frac{1}{w^2}\frac{1}{(1-w)(1-3w)}=\frac{1}{w^2}(1+O(w))$, which has pole of order 2 at $w=0$.
+
+#### Residue
+
+The residue of $f$ at $z_0$ is the coefficient of the term $(z-z_0)^{-1}$ in the Laurent series of $f$ at $z_0$.
+
+> Example:
+>
+> $f(z)=\frac{1}{z^2}$ has a residue of 0 at $z=0$.
+>
+> $f(z)=\frac{z^3}{z-1}$ has a residue of 1 at $z=1$.
+
+Residue for pole with higher order:
+
+If $f$ has a pole of order $n$ at $z_0$, then the residue of $f$ at $z_0$ is
+
+$$
+\operatorname{res}(f,z_0)=\frac{1}{(n-1)!}\lim_{z\to z_0}\frac{d^{n-1}}{dz^{n-1}}((z-z_0)^nf(z))
+$$
+
+## Chapter 9: Generalized Cauchy's Theorem
+
+### Winding number
+
+The winding number of a closed curve $C$ with respect to a point $z_0$ is
+
+$$
+\operatorname{ind}_C(z_0)=\frac{1}{2\pi i}\int_C\frac{1}{z-z_0}dz
+$$
+
+_the winding number is the number of times the curve $C$ winds around the point $z_0$ counterclockwise. (May be negative)_
+
+### Contour integrals
+
+A contour is a piecewise continuous curve $\gamma:[a,b]\to\mathbb{C}$ with integer coefficients.
+
+$$
+\Gamma=\sum_{i=1}^p n_j\gamma_j
+$$
+
+where $\gamma_j:[a_j,b_j]\to\mathbb{C}$ is continuous closed curve and $n_j\in\mathbb{Z}$.
+
+### Interior of a curve
+
+The interior of a curve is the set of points $z\in\mathbb{C}$ such that $\operatorname{ind}_{\Gamma}(z)\neq 0$.
+
+The winding number of contour $\Gamma$ is the sum of the winding numbers of the components of $\Gamma$ around $z_0$.
+
+$$
+\operatorname{ind}_{\Gamma}(z)=\sum_{j=1}^p n_j\operatorname{ind}_{\gamma_j}(z)
+$$
+
+### Separation lemma
+
+Let $\Omega\subseteq\mathbb{C}$ be a domain and $K\subset \Omega$ be compact. Then there exists a simple contour $\Gamma\subset \Omega\setminus K$ such that $K\subset \operatorname{int}_{\Gamma}(\Gamma)\subset \Omega$.
+
+Key idea:
+
+Let $0<\delta<d(K,\partial \Omega)$, then draw the grid lines and trace the contour.
+
+### Residue theorem
+
+Let $\Omega$ be a domain, $\Gamma$ be a contour such that $\Gamma\cap \operatorname{int}(\Gamma)\subset \Omega$. Let $f$ be holomorphic on $\Omega\setminus \{z_1,z_2,\cdots,z_p\}$ and $z_1,z_2,\cdots,z_p$ are finitely many points in $\Omega$, where $z_1,z_2,\cdots,z_p\notin \Gamma$. Then
+
+$$
+\int_{\Gamma}f(z)dz=2\pi i\sum_{j=1}^p \operatorname{res}(f,z_j)
+$$
+
+Key: Prove by circle around each singularity and connect them using two way paths.
+
+
+### Homotopy*
+
+Suppose $\gamma_0, \gamma_1$ are two curves from
+$[0,1]$ to $\Omega$ with same end points $P,Q$.
+
+A homotopy is a continuous function of curves $\gamma_t, 0\leq t\leq 1$, deforming $\gamma_0$ to $\gamma_1$, keeping the end points fixed.
+
+Formally, if $H:[0,1]\times [0,1]\to \Omega$ is a continuous function satsifying
+
+1. $H(s,0)=\gamma_0(s)$, $\forall s\in [0,1]$
+2. $H(s,1)=\gamma_1(s)$, $\forall s\in [0,1]$
+3. $H(0,t)=P$, $\forall t\in [0,1]$
+4. $H(1,t)=Q$, $\forall t\in [0,1]$
+
+Then we say $H$ is a homotopy between $\gamma_0$ and $\gamma_1$. (If $\gamma_0$ and $\gamma_1$ are closed curves, $Q=P$)
+
+#### Lemma 9.12 Technical Lemma
+
+Let $\phi:[0,1]\times [0,1]\to \mathbb{C}\setminus \{0\}$ is continuous. Then there exists a continuous map $\psi:[0,1]\times [0,1]\to \mathbb{C}$ such that $e^\phi=\psi$. Moreover, $\psi$ is unique up to an additive constant in $2\pi i\mathbb{Z}$.
+
+
+### General approach to evaluate definite integrals
+
+Choose a contour so that one side is the desired integral.
+
+Handle the other sides using:
+
+- Symmetry
+- Favorite estimate
+- Bound function by another function whose integral goes to 0
+
+
+
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+