From 305cdc283e1ef2e94bbcfe89e635bdc0bab529ca Mon Sep 17 00:00:00 2001
From: Zheyuan Wu <60459821+Trance-0@users.noreply.github.com>
Date: Tue, 28 Jan 2025 12:49:38 -0600
Subject: [PATCH] update

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 pages/Math416/_meta.js      |   1 +
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+# Lecture 5
+
+## Review
+
+Let $f$ be a complex function. that maps $\mathbb{R}^2$ to $\mathbb{R}^2$. $f(x+iy)=u(x,y)+iv(x,y)$.
+
+$Df(x+iy)=\begin{pmatrix}
+\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\
+\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}
+\end{pmatrix}=\begin{pmatrix}
+\alpha & \beta\\
+\sigma & \delta
+\end{pmatrix}$
+
+So 
+
+$$\begin{aligned}
+\frac{\partial f}{\partial \zeta}&=\frac{1}{2}\left(u_x+v_y\right)-i\frac{1}{2}\left(v_x+u_y\right)\\
+&=\frac{1}{2}\left(\alpha+\delta\right)-i\frac{1}{2}\left(\beta-\sigma\right)\\
+\end{aligned}$$
+
+$$
+\begin{aligned}
+\frac{\partial f}{\partial \overline{\zeta}}&=\frac{1}{2}\left(u_x+v_y\right)+i\frac{1}{2}\left(v_x+u_y\right)\\
+&=\frac{1}{2}\left(\alpha-\delta\right)+i\frac{1}{2}\left(\beta+\sigma\right)\\
+\end{aligned}
+$$
+
+When $f$ is conformal, $Df(x+iy)=\begin{pmatrix}
+\alpha & \beta\\
+-\beta & \alpha
+\end{pmatrix}$.
+
+So $\frac{\partial f}{\partial \zeta}=\frac{1}{2}(\alpha+\alpha)+i\frac{1}{2}(\beta+\beta)=a$
+
+$\frac{\partial f}{\partial \overline{\zeta}}=\frac{1}{2}(\alpha-\alpha)+i\frac{1}{2}(\beta-\beta)=0$
+
+> Less pain to represent a complex function using four real numbers.
+
+## Chapter 3: Linear fractional Transformations
+
+Let $a,b,c,d$ be complex numbers. such that $ad-bc\neq 0$.
+
+The linear fractional transformation is defined as
+
+$$
+\phi(\zeta)=\frac{a\zeta+b}{c\zeta+d}
+$$
+
+If we let $\psi(\zeta)=\frac{e\zeta-f}{-g\zeta+h}$ also be a linear fractional transformation, then $\phi\circ\psi$ is also a linear fractional transformation.
+
+New coefficients can be solved by
+
+$$
+\begin{pmatrix}
+a & b\\
+c & d
+\end{pmatrix}
+\begin{pmatrix}
+e & f\\
+g & h
+\end{pmatrix}
+=
+\begin{pmatrix}
+k&l\\
+m&n
+\end{pmatrix}
+$$
+
+So $\phi\circ\psi(\zeta)=\frac{k\zeta+l}{m\zeta+n}$
+
+### Complex projective space
+
+$\mathbb{R}P^1$ is the set of lines through the origin in $\mathbb{R}^2$.
+
+We defined $(a,b)\sim(c,d),(a,b),(c,d)\in\mathbb{R}^2\setminus\{(0,0)\}$ if $\exists t\neq 0,t\in\mathbb{R}\setminus\{0\}$ such that $(a,b)=t(c,d)$.
+
+$R\mathbb{P}^1=S^1\setminus\{\pm x\}\cong S^1$
+
+Equivalently,
+
+$\mathbb{C}P^1$ is the set of lines through the origin in $\mathbb{C}$.
+
+We defined $(a,b)\sim(c,d),(a,b),(c,d)\in\mathbb{C}\setminus\{(0,0)\}$ if $\exists t\neq 0,t\in\mathbb{C}\setminus\{0\}$ such that $(a,b)=(tc,td)$.
+
+So, $\forall \zeta\in\mathbb{C}\setminus\{0\}$:
+
+If $a\neq 0$, then $(a,b)\sim(1,\frac{b}{a})$.
+
+If $a=0$, then $(0,b)\sim(0,-b)$.
+
+So, $\mathbb{C}P^1$ is the set of lines through the origin in $\mathbb{C}$.
+
+### Linear fractional transformations
+
+Let $M=\begin{pmatrix}
+a & b\\
+c & d
+\end{pmatrix}$ be a $2\times 2$ matrix with complex entries. That maps $\mathbb{C}^2$ to $\mathbb{C}^2$.
+
+Suppose $M$ is non-singular. Then $ad-bc\neq 0$.
+
+If $M\begin{pmatrix}
+\zeta_1\\
+\zeta_2
+\end{pmatrix}=\begin{pmatrix}
+\omega_1\\
+\omega_2
+\end{pmatrix}$, then $M\begin{pmatrix}
+t\zeta_1\\
+t\zeta_2
+\end{pmatrix}=\begin{pmatrix}
+t\omega_1\\
+t\omega_2
+\end{pmatrix}$.
+
+So, $M$ induces a map $\phi_M:\mathbb{C}P^1\to\mathbb{C}P^1$ defined by $M\begin{pmatrix}
+\zeta\\
+1
+\end{pmatrix}=\begin{pmatrix}
+\frac{a\zeta+b}{c\zeta+d}\\
+1
+\end{pmatrix}$.
+
+$\phi_M(\zeta)=\frac{a\zeta+b}{c\zeta+d}$.
+
+If we let $M_2=\begin{pmatrix}
+e &f\\
+g &h
+\end{pmatrix}$, where $ad-bc\neq 0$ and $eh-fg\neq 0$, then $\phi_{M_2}(\zeta)=\frac{e\zeta+f}{g\zeta+h}$.
+
+So, $M_2M_1=\begin{pmatrix}
+a&b\\
+c&d
+\end{pmatrix}\begin{pmatrix}
+e&f\\
+g&h
+\end{pmatrix}=\begin{pmatrix}
+\zeta\\
+1
+\end{pmatrix}$.
+
+This also gives $\begin{pmatrix}
+k\zeta+l\\
+m\zeta+n
+\end{pmatrix}\sim\begin{pmatrix}
+\frac{k\zeta+l}{m\zeta+n}\\
+1
+\end{pmatrix}$.
+
+So, if $ab-cd\neq 0$, then $\exists M^{-1}$ such that $M_2M_1=I$.
+
+So non-constant linear fractional transformations form a group under composition.
+
+When do two matrices gives the $t_0$ same linear fractional transformation?
+
+$M_2^{-1}M_1=\alpha I$
+
+We defined $GL(2,\mathbb{C})$ to be the group of general linear transformations of order 2 over $\mathbb{C}$.
+
+This is equivalent to the group of invertible $2\times 2$ matrices over $\mathbb{C}$ under matrix multiplication.
+
+Let $F$ be the function that maps $M$ to $\phi_M$.
+
+$F:GL(2,\mathbb{C})\to\text{Homeo}(\mathbb{C}P^1)$
+
+So the kernel of $F$ is the set of matrices that represent the identity transformation. $\ker F=\left\{\alpha I\right\},\alpha\in\mathbb{C}\setminus\{0\}$.
+
+#### Corollary of conformality
+
+If $\phi$ is a non-constant linear fractional transformation, then $\phi$ is conformal.
+
+Proof:
+
+Know that $\phi_0\circ\phi(\zeta)=\zeta$,
+
+Then $\phi(\zeta)=\phi_0^{-1}\circ\phi\circ\phi_0(\zeta)$.
+
+So $\phi(\zeta)=\frac{a\zeta+b}{c\zeta+d}$.
+
+$\phi:\mathbb{C}\cup\{\infty\}\to\mathbb{C}\cup\{\infty\}$ which gives $\phi(\infty)=\frac{a}{c}$ and $\phi(-\frac{d}{c})=\infty$.
+
+So, $\phi$ is conformal.
+
+EOP
+
+#### Proposition 3.4 of Fixed points
+
+Any non-constant linear fractional transformation except the identity transformation has 1 or 2 fixed points.
+
+Proof:
+
+Let $\phi(\zeta)=\frac{a\zeta+b}{c\zeta+d}$.
+
+Case 1: $c=0$
+
+Then $\infty$ is a fixed point.
+
+Case 2: $c\neq 0$
+
+Then $\phi(\zeta)=\frac{a\zeta+b}{c\zeta+d}$.
+
+The solution of $\phi(\zeta)=\zeta$ is $c\zeta^2+(d-a)\zeta-b=0$.
+
+Such solutions are $\zeta=\frac{-(d-a)\pm\sqrt{(d-a)^2+4bc}}{2c}$.
+
+So, $\phi$ has 1 or 2 fixed points.
+
+EOP
+
+#### Proposition 3.5 of triple transitivity
+
+If $\zeta_1,\zeta_2,\zeta_3\in\mathbb{C}P^1$ are distinct, then there exists a non-constant linear fractional transformation $\phi$ such that $\phi(\zeta_1)=\zeta_2$ and $\phi(\zeta_3)=\infty$.
+
+Proof as homework.
+
+#### Theorem 3.8 Preservation of clircles
+
+We defined clircle to be a circle or a line.
+
+If $\phi$ is a non-constant linear fractional transformation, then $\phi$ maps clircles to clircles.
+
+Proof:
+
+Continue on next lecture.
diff --git a/pages/Math416/_meta.js b/pages/Math416/_meta.js
index a1fc1b3..1fd32f5 100644
--- a/pages/Math416/_meta.js
+++ b/pages/Math416/_meta.js
@@ -7,4 +7,5 @@ export default {
     Math416_L2: "Complex Variables (Lecture 2)",
     Math416_L3: "Complex Variables (Lecture 3)",
     Math416_L4: "Complex Variables (Lecture 4)",
+    Math416_L5: "Complex Variables (Lecture 5)",
 }