From 15cd12e43218d63b874022ae13e87b9a6d98aa31 Mon Sep 17 00:00:00 2001
From: Zheyuan Wu <60459821+Trance-0@users.noreply.github.com>
Date: Thu, 27 Mar 2025 12:58:15 -0500
Subject: [PATCH] Update Math416_L19.md

---
 pages/Math416/Math416_L19.md | 67 +++++++++++++++++++++++++++++++++++-
 1 file changed, 66 insertions(+), 1 deletion(-)

diff --git a/pages/Math416/Math416_L19.md b/pages/Math416/Math416_L19.md
index fa5ad16..2cc76f3 100644
--- a/pages/Math416/Math416_L19.md
+++ b/pages/Math416/Math416_L19.md
@@ -72,6 +72,8 @@ the singularity at $z=3$ is a simple pole with order 1 $f(z)=\frac{z+1}{z-3}=\fr
 
 there are three poles at $z=1,5,8$, the order of the poles are 2, 6, 1 respectively.
 
+#### Corollary: order of poles and zeros
+
 If $f$ has a pole of order $m$ at $z_0$,
 
 $$
@@ -83,8 +85,71 @@ then $(z-z_0)^m f(z)$ has a removable singularity at $z_0$. Value of holomorphic
 - $f$ is given by a power series in $A(z_0;0,R)$
 - $f=(z-z_0)^{-m} g(z)$ where $g$ is holomorphic and $g(z_0)\neq 0$, $\frac{1}{f}=(z-z_0)^m \frac{1}{g(z)}$ has a pole of order $m$ at $z_0$. So $f$ has a pole of order $m$ at $z_0$ if and only if $\frac{1}{f}$ has a zero of order $m$ at $z_0$
 
-$e^{1/z}=1+\frac{1}{z}+\frac{1}{2!z^2}+\frac{1}{3!z^3}+\cdots$ has an essential singularity at $z=0$ since it has infinitely many terms with negative powers of $z$.
+> $e^{1/z}=1+\frac{1}{z}+\frac{1}{2!z^2}+\frac{1}{3!z^3}+\cdots$ has an essential singularity at $z=0$ since it has infinitely many terms with negative powers of $z$.
 
+Suppose $f$ is a holomorphic in a neighborhood of $\infty$: $\exists R>0$ s.t. $f$ is holomorphic on $\{z:|z|>R\}$
 
+We defined $g(z)=f(1/z)$ where $g$ is holomorphic on punctured disk center $0$ radius $1/R$
 
+Say $f(z)$ has a zero of order $\infty$ if any only if $g(z)=f(1/z)$ has a zero of order $m$ at $z=0$
 
+Say $f$ has a pole of order $m$ at $\infty$ if and only if $g(z)=f(1/z)$ has a pole of order $m$ at $z=0$
+
+Example:
+
+1. $f(z)=z^2$, $g(z)=f(1/z)=1/z^2$ has a pole of order 2 at $z=0$
+2. $f(z)=\frac{1}{z^3}$ (vanishes to order 3 at $\infty$), $g(z)=f(1/z)=z^3$ has a zero of order 3 at $z=0$
+
+We say $f$ has an isolated singularity at $\infty$ if and only if $g(z)=f(1/z)$ has an isolated singularity at $z=0$.
+
+$f$ has $\begin{cases}
+  \text{removable}\\
+  \text{pole of order } m\\
+  \text{essential}
+\end{cases}$ singularity at $\infty$ if and only if $g(z)=f(1/z)$ has $\begin{cases}
+  \text{removable}\\
+  \text{pole of order } m\\
+  \text{essential}
+\end{cases}$ singularity at $z=0$
+
+#### Theorem: Criterion for a removable singularity (Riemann removable singularity theorem)
+
+Suppose $f$ has an isolated singularity at $z_0$. Then it is removable if and only if $f$ is bounded on a punctured disk centered at $z_0$.
+
+Proof:
+
+($\Leftarrow$) Suppose $z_0$ is a removable singularity. Then $\exists r>0$ such that $B_r(z_0)\setminus\{z_0\}=A(z_0;0,r)$ and $f(z)=\sum_{n=0}^{\infty} a_n (z-z_0)^n$ for $z\in A(z_0;0,r)$. Then $f$ is bounded in $A(z_0;0,r/2)$
+
+($\Rightarrow$) Suppose $|f(z)|\leq M$ for $z\in A(z_0;0,r/2)$. So $f(z)=\sum_{n=-\infty}^{\infty} a_n (z-z_0)^{n-k-1}=\int_{C_r}f(z)(z-z_0)^{-k-1}dz=a_{k}2\pi i$
+
+$a_k=\frac{1}{2\pi i}\int_{C_r}f(z)(z-z_0)^{-k-1}dz$
+
+And $|a_k|\leq \max_{z\in C_r}\left|2\pi|f(z)|z-z_0|^{-k-1}\right|\leq 2\pi M r^{-k-1}2\pi r$
+
+So $|a_k|\leq (4\pi^2M)r^{-k}$ for all $r<R$
+
+So if $k<0$, $|a_k|\leq \lim_{r\to 0} (4\pi^2M)r^{-k}=0$
+
+QED
+
+Corollary:
+
+If $f$ is holomorphic at $\infty$, then $f$ is bounded for large $|z|$.
+
+#### Theorem: Criterion for a pole
+
+Suppose $f$ has an isolated singularity at $z_0$. Then $z_0$ is a pole of order $m$ if and only if $\lim_{z\to z_0} |f(z)|=\infty$
+
+Proof:
+
+($\Rightarrow$) If $z_0$ is a pole of order $m$, then $f(z)=a_{-m}(z-z_0)^{-m}+O((z-z_0)^{-m+1})$
+
+As $z\to z_0$, $|f(z)|\approx |a_{-m}| |z-z_0|^{-m}\to \infty$
+
+($\Leftarrow$) Let $g(z)=\frac{1}{f(z)}$ near $z_0$. Then $g$ has a singularity at $z_0$ and $|g(z)|$ is bounded near $z_0$.
+
+By Riemann removable singularity theorem, $g(z)=(z-z_0)^m h(z)$ for some holomorphic $h$ and $h(z_0)\neq 0$
+
+So $f(z)=\frac{1}{g(z)}=\frac{1}{(z-z_0)^m h(z)}$ has a pole of order $m$ at $z_0$
+
+QED