update 416

pages/Math416/Math416_L3.md

@@ -71,7 +71,7 @@ And $u$ and $v$ have continuous partial derivatives at $(x_0,y_0)$.
 
 And let $c=\frac{\partial u}{\partial x}(x_0,y_0)$ and $d=\frac{\partial v}{\partial x}(x_0,y_0)$.
 
-Then $f'(\zeta_0)=c+id$.
+**Then $f'(\zeta_0)=c+id$, is holomorphic at $\zeta_0$.**
 
 ### Holomorphic Functions
 

pages/Math416/Math416_L4.md

@@ -0,0 +1,246 @@
+# Lecture 4
+
+## Review
+
+### Derivative of a complex function
+
+$$
+\frac{\partial f}{\partial z}=\frac{1}{2}\left(\frac{\partial f}{\partial x}-\frac{\partial f}{\partial y}\right)
+$$
+
+$$
+\frac{\partial f}{\partial \bar{z}}=\frac{1}{2}\left(\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\right)
+$$
+
+### Angle between two curves
+
+Let $\gamma_1,\gamma_2$ be two curves in $G\subset \mathbb{C}$ with $\gamma_1(t_0)=\gamma_2(t_0)=\zeta_0$ for some $t_0\in I_1\cap I_2$.
+
+The angle between $\gamma_1$ and $\gamma_2$ at $\zeta_0$ is the angle between the vectors $\gamma_1'(t_0)$ and $\gamma_2'(t_0)$. Denote as $\arg(\gamma_2'(t_0))-\arg(\gamma_1'(t_0))=\arg(\gamma_2'(t_0)\gamma_1'(t_0))$.
+
+### Cauchy-Riemann equations
+
+$$
+\frac{\partial f}{\partial z}=\frac{1}{2}\left(\frac{\partial f}{\partial x}-\frac{\partial f}{\partial y}\right)
+$$
+
+## Continue on last lecture
+
+### Theorem of conformality
+
+Suppose $f:G\to \mathbb{C}$ is holomorphic function on open set $G\subset \mathbb{C}$ and $\gamma_1,\gamma_2$ are regular curves in $G$ with $\gamma_1(t_0)=\gamma_2(t_0)=\zeta_0$ for some $t_0\in I_1\cap I_2$.
+
+If $f'(\zeta_0)\neq 0$, then the angle between $\gamma_1$ and $\gamma_2$ at $\zeta_0$ is the same as the angle between the vectors $f'(\zeta_0)\gamma_1'(t_0)$ and $f'(\zeta_0)\gamma_2'(t_0)$.
+
+### Lemma of function of a curve and angle
+
+If $f:G\to \mathbb{C}$ is holomorphic function on open set $G\subset \mathbb{C}$ and $\gamma$ is differentiable curve in $G$ with $\gamma(t_0)=\zeta_0$ for some $t_0\in I$.
+
+Then,
+
+$$
+(f\circ \gamma)'(t_0)=f'(\gamma(t_0))\gamma'(t_0).
+$$
+
+> Looks like the chain rule.
+
+Proof:
+
+We want to show that
+
+$$
+\lim_{t\to t_0}\frac{(f\circ \gamma)(t)-(f\circ \gamma)(t_0)}{t-t_0}=f'(\gamma(t_0))\gamma'(t_0).
+$$
+
+> Notation:
+>
+> A function $g(h)$ is $O(h)$ if $\exists C>0$ such that $|g(h)|\leq C|h|$ for all $h$ in a neighborhood of $0$.
+>
+> A function $g(h)$ is $o(h)$ if $\lim_{h\to 0}\frac{g(h)}{h}=0$.
+> 
+> <!---TODO: check after lecture-->
+> $f$ is differentiable if and only if $f(z+h)=f(z)+f'(z)h+\frac{1}{2}h^2f''(z)+o(h^3)$ as $h\to 0$. 
+
+
+
+Since $f$ is holomorphic at $\gamma(t_0)=\zeta_0$, we have
+
+$$
+f(\zeta_0)=f(\zeta_0)+(\zeta-\zeta_0)f'(\zeta_0)+o(\zeta-\zeta_0)
+$$
+
+and
+
+$$
+f(\gamma(t_0))=f(\gamma(t_0))+f'(\gamma(t_0))(\gamma(t)-\gamma(t_0))+o(\gamma(t)-\gamma(t_0))
+$$
+
+So,
+
+$$
+\begin{aligned}
+\lim_{t\to t_0}\frac{(f\circ \gamma)(t)-(f\circ \gamma)(t_0)}{t-t_0}
+&=\lim_{t\to t_0}\frac{\left[f(\gamma(t_0))+f'(\gamma(t_0))(\gamma(t)-\gamma(t_0))+o(\gamma(t)-\gamma(t_0))\right]-f(\gamma(t_0))}{t-t_0} \\
+&=\lim_{t\to t_0}\frac{f'(\gamma(t_0))(\gamma(t)-\gamma(t_0))+o(\gamma(t)-\gamma(t_0))}{t-t_0} \\
+&=\lim_{t\to t_0}\frac{f'(\gamma(t_0))(\gamma(t)-\gamma(t_0))}{t-t_0} +\lim_{t\to t_0}\frac{o(\gamma(t)-\gamma(t_0))}{t-t_0} \\
+&=f'(\gamma(t_0))\lim_{t\to t_0}\frac{\gamma(t)-\gamma(t_0)}{t-t_0} +0\\
+&=f'(\gamma(t_0))\gamma'(t_0)
+\end{aligned}
+$$
+
+EOP
+
+#### Definition of conformal function
+
+A function $f:G\to \mathbb{C}$ is called conformal if it preserves the angle between two curves.
+
+#### Theorem of conformal function
+
+If $f:G\to \mathbb{C}$ is holomorphic function on open set $G\subset \mathbb{C}$ and $\gamma_1,\gamma_2$ are regular curves in $G$ with $\gamma_1(t_0)=\gamma_2(t_0)=\zeta_0$ for some $t_0\in I_1\cap I_2$, and $f'(\zeta_0)\neq 0$, then $f$ is conformal at $\zeta_0$.
+
+Example:
+
+$$
+f(z)=z^2
+$$
+
+is not conformal at $z=0$ because $f'(0)=0$.
+
+
+
+#### Lemma of conformal function
+
+Suppose $f$ is real differentiable, let $a=\frac{\partial f}{\partial \zeta}(\zeta_0)$, $b=\frac{\partial f}{\partial \overline{\zeta}}(\zeta_0)$.
+
+Let $\gamma(t_0)=\zeta_0$. Then $(f\circ \gamma)'(t_0)=a\gamma'(t_0)+b\overline{\gamma'(t_0)}$.
+
+Proof:
+
+$f=u+iv$, $u,v$ are real differentiable.
+
+$$
+a=\frac{\partial f}{\partial \zeta}=\frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)+i\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)
+$$
+
+$$
+b=\frac{\partial f}{\partial \overline{\zeta}}=\frac{1}{2}\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)+i\frac{1}{2}\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)
+$$
+
+$$
+\gamma'(t_0)=\frac{d\alpha}{dt}+i\frac{d\beta}{dt}
+$$
+
+$$
+\overline{\gamma'(t_0)}=\frac{d\beta}{dt}-i\frac{d\alpha}{dt}
+$$
+
+$$
+\begin{aligned}
+(f\circ \gamma)'(t_0)&=\frac{\partial f}{\partial \zeta}(\gamma(t_0))\gamma'(t_0)+\frac{\partial f}{\partial \overline{\zeta}}(\gamma(t_0))\overline{\gamma'(t_0)} \\
+&=\left[\frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)+i\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)\right]\left(\frac{d\alpha}{dt}+i\frac{d\beta}{dt}\right)\\
+&+\left[\frac{1}{2}\left(\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}\right)+i\frac{1}{2}\left(\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}\right)\right]\left(\frac{d\beta}{dt}-i\frac{d\alpha}{dt}\right) \\
+&=\left[\frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)\frac{d\alpha}{dt}-\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)\frac{d\beta}{dt}\right]\\
+&+i\left[\frac{1}{2}\left(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right)\frac{d\alpha}{dt}+\frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)\frac{d\beta}{dt}\right] \\
+&=\left[a+b\right]\frac{d\alpha}{dt}+i\left[a-b\right]\frac{d\beta}{dt} \\
+&=\left[u_x+iv_x\right]\frac{d\alpha}{dt}+i\left[v_y-iu_y\right]\frac{d\beta}{dt} \\
+&=a\gamma'(t_0)+b\overline{\gamma'(t_0)}
+\end{aligned}
+$$
+
+EOP
+
+#### Theorem of differentiability
+
+Let $f:G\to \mathbb{C}$ be holomorphic function on open set $G\subset \mathbb{C}$ and real differentiable. $f=u+iv$ where $u,v$ are real differentiable functions.
+
+Then, $f$ is conformal if and only if $f$ is holomorphic at $\zeta_0$ and $f'(\zeta_0)\neq 0,\forall \zeta_0\in G$.
+
+Proof:
+
+<!---TODO: check after lecture-->
+
+Case 1: Suppose $f(\zeta)=a\zeta+b\overline{\zeta}$, Let $b=\frac{\partial f}{\partial \overline{z}}(\zeta)$. We need to prove $a+b\neq 0$. So we want $b=0$ and $a\neq 0$, other wise $f(\mathbb{R})=0$.
+
+$f:\mathbb{R}\to \{(a+b)t\}$ is not conformal.
+
+...
+
+Case 2: Immediate consequence of the lemma of conformal function.
+
+EOP
+
+### Harmonic function
+
+Let $\Omega$ be a domain in $\mathbb{C}$. A function $u:\Omega\to \mathbb{R}$ 
+
+> A domain is a connected open set.
+
+Say $g:\Omega\to \mathbb{R} \text{ or } \mathbb{C}$ is harmonic if it satisfies the Laplace equation
+
+$$
+\Delta g=\frac{\partial^2 g}{\partial x^2}+\frac{\partial^2 g}{\partial y^2}=0.
+$$
+
+#### Theorem of harmonic conjugate
+
+Let $f=u+iv$ be holomorphic function on domain $\Omega\subset \mathbb{C}$. Then $u$ and $v$ are harmonic functions on $\Omega$.
+
+Proof:
+
+$$
+\Delta u=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0.
+$$
+
+Using the Cauchy-Riemann equations, we have
+
+$$
+\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 v}{\partial x\partial y}, \quad \frac{\partial^2 u}{\partial y^2}=-\frac{\partial^2 v}{\partial y\partial x}.
+$$
+
+So,
+
+$$
+\Delta u=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=\frac{\partial^2 v}{\partial x\partial y}-\frac{\partial^2 v}{\partial y\partial x}=0.
+$$
+
+EOP
+
+If $v$ is such that $f=u+iv$ is holomorphic on $\Omega$, then $v$ is called harmonic conjugate of $u$ on $\Omega$.
+
+Example:
+
+$$
+u(x,y)=x^2-y^2
+$$
+
+is harmonic on $\mathbb{C}$.
+
+To find a harmonic conjugate of $u$ on $\mathbb{C}$, we need to find a function $v$ such that
+
+$$
+\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}=2y, \quad \frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}=2x.
+$$
+
+Integrating, we get
+
+$$
+v(x,y)=2xy+G(y)
+$$
+
+$$
+\frac{\partial v}{\partial y}=2x+G'(y)=2x
+$$
+
+So,
+
+$$
+G'(y)=0 \implies G(y)=C
+$$
+
+$$
+v(x,y)=2xy+C
+$$
+
+is a harmonic conjugate of $u$ on $\mathbb{C}$.
+
+Combine $u$ and $v$ to get $f(x,y)=x^2-y^2+2xyi+C=(x+iy)^2+C=z^2+C$, which is holomorphic on $\mathbb{C}$.

pages/Math416/_meta.js

@@ -6,4 +6,5 @@ export default {
     Math416_L1: "Complex Variables (Lecture 1)",
     Math416_L2: "Complex Variables (Lecture 2)",
     Math416_L3: "Complex Variables (Lecture 3)",
+    Math416_L4: "Complex Variables (Lecture 4)",
 }