Review

$Lnz=ln|z|+iargz+3k\pi i (k\in Z)=ln|z|+iArgz$
$lnz=ln|z|+iargz\quad -\pi<argz\leq\pi$
- $arg z=\arctan\cfrac{y}{x}$
$e^z=e^{z+2k\pi i}\quad (k\in Z)$
- $e^{Lnz}=e^{lnz}=z$
- Example: $(1+i)^n=(1-i)^n$
$z^\alpha=e^{\alpha(ln|z|+iargz+2k\pi i)}$
$\sin z=\cfrac{e^{iz}-e^{-iz}}{2i}\qquad \cos z=\cfrac{e^{iz}+e^{-iz}}{2}$
- 复数计算展开到$a+bi$
$f(z)=zImz$在复平面处处连续
可导的充分条件：
- $u_x,u_y,v_x,v_y$在$(x,y)$连续，且在$(x,y)$满足C-R方程，则$f(z)$在$z=x+iy$可导，且$f’(z)=u_x+iv_x=v_y+iv_x=u_x-iu_y=v_y-iu_y$
解析函数
- 所有解析点的合集必为开集，不可能仅在一个点或一条曲线上解析
- 平面上处处解析的函数
  - $f(z)=e^z,z^n,\sin z,\cos z$
- 处处不解析的函数
  - $f(z)=Rez,Imz,|z|,\overline{z},z|z|$
- $\oint_{|z|=1}Rezdz$:参数方程求解
调和函数
- $u_{xx}+u_{yy}=0$
- Example $u=x^2-y^2+xy$
  - $u_x=2x+y,\quad u_y=-2y+x$
  - $u_{xx}=2,\qquad u_{xy}=1,\qquad u_{yx}=1,\qquad u_{yy}=-2$
  - $u_xx+u_yy=0$, 调和
复积分
- 非闭区间
  - 参数方程法
    - $\int_Cf(z)dz=\int_a^bf(z(t))z’(t)dt$
    - Example $\int_CRezdz$, $C=1-i$ to $3+2i$直线段
      - $C:(2t+1)+i(3t-1)$
      - $I=\int_0^1(2t+1)(2+3i)dt$
    - Example $\int_C|z|dz$, C:上半单位圆周线，起点-1
      - $C:z=e^{i\theta}, \theta:\pi\to 0$
      - $|z|=1$
      - $I=\int_\pi^0 1(ie^{i\theta})d\theta=e^{i\theta}\big|_\pi^0=2$
  - 牛顿-莱布尼兹公式
- 周线积分
  - 参数方程
    - Example $\oint_{|z|=2}(Rez+e^z)dz$
      - $I=\oint_{|z|=2}Rezdz+\oint_{|z|=2}e^zdz=\oint_{|z|=2}Rezdz$
      - $\int_0^{2\pi}2\cos\theta ie^{i\theta}d\theta=\cdots$
  - 推广的柯西积分定理
  - 柯西积分公式
    - $\oint_C\cfrac{f(z)}{(z-z_0)^{n+1}}dz=\cfrac{2\pi i}{n!}f^{(n)}(z_0)$
      - Example $I=\oint_C\cfrac{e^z}{z(z-1)^2}dz, C:x^2+y^2=4x+5$
        $C:(x-2)^2+y^2=3^2$
        $I=\oint_{C_1}\cfrac{\frac{e^z}{(z-1)^2}}{z}dz+\oint_{C_2}\cfrac{\frac{e^z}{z}}{(z-1)^2}$
  - 留数定理
    - $\oint_Cf(z)dz=2\pi i\sum\limits_{k=1}^n Res[f(z),z_k]$
    - $Res[f(z),z_k]:$ f(z)在$z_k$去心邻域的洛朗展开中$\cfrac{1}{z-z_k}$系数$C_{-1}$
    - 本性奇点
      - $I=\oint_{|z|=1}ze^{-\cfrac{1}{z^2}}dz=2\pi i Res[ze^{-\cfrac{1}{z^2}},0]$
    - 可去奇点
    - 极点
      - 洛朗展开求留数
      - $Res[f(z),z_0]=\cfrac{1}{(m-1)!}\lim\limits_{z\to z_0}\cfrac{d^{m-1}}{dz^{m-1}}[(z-z_0)^mf(z)]$
    - 留数定理应用
  - 泰勒展开
    - $e^z=\sum\limits_{n=0}^\infty\cfrac{z^n}{n!},\quad |z|<\infty$
    - $\sin(z)=\sum\limits_{n=0}^\infty (-1)^n\cfrac{z^{2n+1}}{(2n+1)!}$
    - $\cos(z)=\sum\limits_{n=0}^\infty(-1)^n\cfrac{z^2n}{(2n)!}$
    - $\cfrac{1}{1-z}=\sum\limits_{n=0}^\infty z^n,\quad |z|<1$
    - $\cfrac{1}{1+z}=\sum\limits_{n=0}^\infty (-1)^nz^n,\quad |z|<1$
    - $\cfrac{1}{1-z^2}=\sum\limits_{n=0}^\infty z^{2n},\quad |z|<1$
    - $\cfrac{1}{1+z^2}=\sum\limits_{n=0}^\infty (-1)^n z^{2n},\quad |z|<1$
  - 洛朗展开
    - $e^{\frac{1}{z}}=1+\cfrac{1}{z}+\cfrac{1}{2!}(\cfrac{1}{z})^2+\cdots+\cfrac{1}{n!}(\cfrac{1}{z})^n+\cdots,\qquad 0<|z|<+\infty$
    - $\sin\cfrac{1}{z}=\cfrac{1}{z}-\cfrac{1}{3!}\cfrac{1}{z^3}+\cdots+(-1)^n\cfrac{1}{(2n+1)!}(\cfrac{1}{z})^{2n+1}+\cdots,\qquad 0<|z|<+\infty$
    - $\cos\cfrac{1}{z}1-\cfrac{1}{2!}\cfrac{1}{z^2}+\cdots+(-1)^n\cfrac{1}{(2n)!}(\cfrac{1}{z})^{2n}+\cdots,\qquad 0<|z|<+\infty$
  - 傅里叶变换
    - $F(\omega)=\mathscr{F}[f(t)]=\int_{-\infty}^\infty f(t)e^{-i\omega t}dt,\qquad T\to \infty$
    - $f(t)=\mathscr{F}^{-1}[F(\omega)]=\cfrac{1}{2\pi}\int_{-\infty}^\infty F(\omega)e^{i\omega t}d\omega$
    - $\mathscr{F}[\delta(t)]=1$
    - $\mathscr{F}[e^{i\omega_0 t}]=2\pi\delta(\omega-\omega_0)$
    - $\mathscr{F}[e^{i\omega_0 t}f(t)]=F(\omega-\omega_0)$
    - $\mathscr{F}[f^{(n)}(t)]=(i\omega)^n\mathscr{F}[f]$
    - $\cfrac{d^nF(\omega)}{d\omega^n}=(-i)^n\mathscr{F}[t^nf(t)]$
  - 拉普拉斯变换
    - $F(s)=\mathscr{L}[f(t)]=\int_0^{+\infty}f(t)e^{-st}dt$
    - $\mathscr{L}[1]=\cfrac{1}{s}$
    - $\mathscr{L}[e^{at}]=\cfrac{1}{s-a}$
    - $\mathscr{L}[\sin bt]=\cfrac{b}{s^2+b^2}$
    - $\mathscr{L}[\cos bt]=\cfrac{s}{s^2+b^2}$
    - $\mathscr{L}[t]=\cfrac{1}{s^2}$
    - $\mathscr{L}[t^2]=\cfrac{2!}{s^3}$
    - $\mathscr{L}[t^n]=\cfrac{n!}{s^{n+1}}$
    - $\mathscr{L}[\alpha f+\beta g]=\alpha F(s)+\beta G(s)$
    - $\mathscr{L}[e^{at}f(t)]=F(s-a)$
    - $\mathscr{L}[t^nf(t)]=(-1)^n F^{(n)}(s)$
    - $\mathscr{L}[\int_0^tf(t)dt]=\cfrac{F(s)}{s}$
    - $\mathscr{L}[f^{(n)}(t)]=s^nF(s)-s^{n-1}f(0)-s^{n-2}f’(0)-\cdots-f^{(n-1)}(0)$

Class

$f’(z_0)=\lim_{\Delta z\to 0}\cfrac{\Delta w}{\Delta z}$
$\cfrac{\Delta w}{\Delta z}=f’(z_0)+\eta(\Delta z)\Delta z$
$Prove f(z)=z^2 \text{is derivatable and } (z^2)’=2z$ $\cfrac{\Delta w}{\Delta z}=\cfrac{(z+\Delta z)^2-z^2}{\Delta z}=\cfrac{2z\Delta z+\Delta z^2}{\Delta z}$ add lim $=2z$
$f(z)=\overline{z}, Re z, Im z, |z| \text{ continuous on }z$
$u(x,y) \text{ is derivatable at }(x,y)\Leftrightarrow \Delta u=A\Delta x+B\Delta y+o(\sqrt{(\Delta x)^2+(\Delta y)^2})$
$f(z)=u(x,y)+iv(x,y)$
- $u(x,y),v(x,y) \text{is differentiable at }(x,y)$
- $u(x,y), v(x,y) \text{ supports } u_x=v_y, u_y=-v_x\textbf{ C-R Euqation}$
$\text{if f(z) is derivatable at z, }f’(x)=u_x+iv_x=v_y+iv_x=u_x+iu_y=v_y-iu_y$
if $u=u(x,y)$ is partial derivative at $(x,y)$, $u_x$,$u_y$ are continuous at $(x,y)$, and $u(z)$ is differentiable at $(x,y)$.
- Example 1 $f(z)=|z|^2$. 1. $f(z)=|z|^2=x^2+y^2$ 2. $u(x,y)=x^2+y^2, v(x,y)=0$ 3. $u_x=2x,u_y=2y,v_x=v_y=0$ 4. $u_x=v_y,u_y=-v_x$ (C-R Equation) get $z=0$ 5. $f(z)|{z=0}=u_x|{z=0}+iv_x|{z=0}=2x|{z=0}=0$
- Example 2 $f(z)=x^2+2y^2$ 1. $u(x,y)=x^2, v(x,y)-y^2$ 2. $u_x=2x,u_y=0,v_x=0,v_y=2y$ 3. $y=x$ 4. $f’(z)|_{y=x}=2x$
analytic function
- Example $f(z)=\overline{z}$ 1. $f(z)=x-iy$ 2. not analytic
$f(z)=\overline{z},Imz,Rez,|z|$ are not analytic
The relationship between Analytic Function and Harmonic Function
Construct: Given a differentiable function u,(v), if exsist a v,(u) make $u+iv$ analytical
- C-R
  - $u_x=v_y, u_y=-v_x$
  - $u_{xx}+u_{yy}=v_{yx}-v_{xy}!=0$
- C-R
  - $u_{xx}+u_{yy}=0$
Harmonic Function : $u: u_{xx}+u_{yy}=0$ ($u\in C^2(D)$) $u_x$, $u_{xx}$ continuous
- Example: $u(x,y)=x^3-3xy^2$
  - $u_{xx}=6x, u_{xy}=-6y, u_{yx}=-6y, u_{yy}=-6x$
  - $u\in C^2(C)$
  - Harmonic
$f(z)=u+iv$ analytic $\Leftrightarrow$ u,v harmonic in D and support C-R Equation.
If $u(x,y)$ and $v(x,y)$ harmonic in D and satisfying C-R equation $u_x=v_y$, $u_y=-v_x$, then v is the convergent harmonic function of u.
Partial Integrate Method
- $u_x=v_y (1)$
- $\int v_y(x,y)dy=\int u_x(x,y)dy$
- $\Rightarrow v(x,y)+\psi (x)=\int u_x dy$
- $v(x,y)=\int u_x dy+\psi (x)$
- $v_x=-u_y (2)$
- $(\int u_x(x,y) dy)_x+\phi ‘(x)=-u_y$
Example $u(x,y)=x^3-3xy^2$
- $u_x=3x^2-3y^2, u_y=-6xy$
- $v_y=u_x=3x^2-3y^2$
- $v(x,y)=3x^2y-y^3+\phi (x)$
- $v_x=-u_y=6xy$
- $6xy+\phi’(x)=6xy$
- $\Rightarrow \phi(x)=C$
- $f(z)=(3x^2y-y^3+C)i+(x^3-3xy^2)$
- $f(0)=i$
- $Ci=i$
- $C=1$
Primary Analytic Function
- $w=e^z=e^{x(cos y+i sin(y))}$
  - $z=|z|=(cos \theta +i\sin\theta)=|z|e^{i\theta}$
  - $(e^z)’=e^z$
  - $|e^z|=e^x>0$
  - $e^{z_1}\cdot e^{z_2}=e^{z_1+z_2}$
  - $e^{z+2k\pi i}=e^z\qquad (k\in Z)$
  - example $e^z=1-\sqrt{3}i$
    - $e^z=e^{i(-\frac{\pi}{3}+ln2)}$
- $e^w=z\Leftrightarrow w=Ln(z)$
  - $z=re^{i\theta}, w=u+iv,e^{u+iv}=re^{i\theta}$
  - $\Rightarrow u=\ln r, v=\theta+2k\pi =Arg z$
  - $w=Ln(z)=\ln |z|+i Arg z(z\not = 0)=\ln|z|+i argz+2k\pi i (k\in Z)$
  - Example $Ln(-1-i)$
    - $Ln(-1-i)=ln(\sqrt{2})+i(-\cfrac{3\pi}{4})+2k\pi i, k\in Z$
  - Example $Ln(-1)$
    - $Ln(-1)=i\pi+2k\pi i, k\in Z$
  - Example $Ln(2)$
    - $Ln(2)=ln2+2k\pi i, k\in Z$+
  - $Ln(z_1z_2)=Lnz_1+Lnz_2$
  - $Ln(\cfrac{z_1}{z_2})=Ln z_1-Ln z_2$, ($z_1\not ={0}, z_2\not ={0}$)
  - $Ln z^n\not ={nLnz}, Lnz-Lnz\not ={0}$
  - $2Ln z= ln r^2+i (2\theta +4k\pi), k\in Z$
  - *$e^{Lnz}=e^{lnz}=z$
  - Example: solve $lnz=1+\pi i$
    - $z=e^{1+\pi i}$
    - $z=e\cdot(cos \pi + isin \pi)=-e$
- $\begin{cases}e^{iy}=\cos y+i\sin y\\e^{-iy}=\cos y-i\sin y\\\end{cases}$
  - $sin z=\cfrac{1}{2i}(e^{iz}-e^{-iz})$
  - $cos z=\cfrac{1}{2}(e^{iz}+e^{-iz})$
  - zero point of $sin z$ is $z=k\pi, k\in Z$
  - zero point of $cos z$ is $z=(k+1/2)\pi, k\in Z$
  - Example $sinz=0$
    - $\Rightarrow e^{ir}-e^{-ir}=0$
    - $\Rightarrow e^{2iz}=1$
- $w=z^\alpha, \alpha \in C$
  - $z^\alpha=e^{\alpha Lnz}=e^{\alpha (ln|z|+iargz+2k\pi i)}$
  - Example $i^i$
    - $i^i=e^{iLni}=e^{i(i(\pi/2+2k\pi ))}=e^{-(\pi /2+2k\pi)}$
Homework 2.9(1)(4), 2.13(1)(2)(4), 2.14
Integration of Complex Function
- Complex Integral
  - $\oint_Cf(z)dz=-\oint_{C^{-}}f(z)dz$
  - If $f(z)=u(x,y)+iv(x,y)$ continuous on smooth curve $C$ from $a$ to $b$ ($u$,$v$ are continuous), $\int_C f(z)dz=\int_C udx-vdy+i(\int_C vdx+udy)$
    - If $f(z)$ is continuous function and $C$ is a smooth curve, integral $\int_Cf(z)dz$ exist.
    - $\int_Cf(z)dz$ can be simplified into real variable function.
  - Parametric Equation
    - $z(t)=x(t)+iy(t)\qquad t: a\rightarrow b$
  - (Important!) Parametric Equation method: *$\int_Cf(z)dz=\int_a^bf(z(t))z’(t)dt$
  - Example:
    - Prove: $I_n=\oint_C\cfrac{dz}{(z-z_0)^n}=\begin{cases}2\pi i & n=1\\0& n\in Z^+,\quad n\not ={0}\\\end{cases}$ where $C:|z-z_0|=r\quad(r>0)$
      - $z(\theta)=z_0+re^{i\theta}$
      - $\oint_C\cfrac{1}{(z-z_0)^n}dz=\int_0^{2\pi}ir^{1-n}e^{i(1-n)\theta}d\theta$
      - $=ir^{1-n}\int_0^{2\pi}e^{i(1-n)\theta}d\theta=$ then split it into sin and cos.
  - $|\oint_Cf(z)dz|\leqslant\oint_C|f(z)|ds$
Cauchy Integral Theorm
- $f(z)$ integral in D is unrelated to path$\Leftrightarrow$ $\oint_lf(z)dz=0$
- If $F(z)$ is analytic in single connected region $D$, $\oint_Cf(z)dz=0$ for any contour $C$ in $D$.
  - Example1:
    - $f(z)=e^z,z^n,\sin z,\cos z$ analytic on $z$ area, $\oint_Cf(z)dz=0$ where C is a contour in the complex area.
  - Example2: solve $\oint_C\cfrac{1}{z+2}dz, C:|z|=1$
    - singular point: $z=-2$
    - Construct $D$: 1. $-2\notin D$ 2. $C \subset D$
    - $D: |z|\leq 1.5$
    - $\oint_C\cfrac{1}{z+2}dz=0$
- Cauchy integral theorm on complex connected region
  - $f(z)$ is analytic in $D$ bounded in $C_1$ and $C_2$ (contours) and continuous on $\overline{D}=D+C_1+C_2$, then $\oint_{C_1}f(z)dz=\oint_{C_2}f(z)dz$
  - $\oint_Cf(z)dz=\sum_{k=1}^{n}\oint_{C_{k}}f(z)dz$
  - Example: solve $\oint_C\cfrac{1}{(z-1)(z+1)}dz$, $C:$ bound 1 and -1
    - $f(z)=\cfrac{1}{(z-1)(z+1)}=\cfrac{1}{2}\big(\cfrac{1}{z-1}-\cfrac{1}{z+1}\big)$
    - Construct $C_1$, $C_2$ around -1 and 1.
    - $\oint_Cf(z)dz=\oint_{C_1}f(z)dz+\oint_{C_1}f(z)dz$
      - $=\oint_{C_1}(\cfrac{1}{z-1}-\cfrac{1}{z+1})+\oint_{C_2}(\cfrac{1}{z-1}-\cfrac{1}{z+1})$
      - $=0-2\pi i+2\pi i+ 0$
      - $=0$
- $f(z)$ is analytic in $D$, $z_0\in D$, $F(z)$ is analytic in $D$ and $F’(z)=f(z)$
- $f(z)$ is analytic in single connect region $D$, $\int_{z_0}^{z_1}f(z)dz=F(z_1)-F(z_0), z_0,z_1\in D$
Calculate Complex Integration
- $C:$ not close
  - Paramater
  - N-L Formula
- $C: $ contour
  - Parameter
  - Cauchy Theory
  - Important Integral: $\oint_C\cfrac{dz}{(z-z_0)^n}=\begin{cases}2\pi i & n=1\\0& n\in Z^+,\quad n\not ={0}\\\end{cases}(C:|z-z_0|=r>0)$
Cauchy Integral Formula
- If $f(z)$ is analytic in $D$ surrounded by contour $C$, continuous on $C$ and $D$, for all $z_0\in D$,
  - $f(z_0)=\cfrac{1}{2\pi i}\oint_C\cfrac{f(z)}{z-z_0}dz$
- Application:$\oint_C\cfrac{f(z)}{z-z_0}=2\pi if(z_0)$
  - where $g(z)=\cfrac{f(z)}{z-z_0}$, singular point $z_0$ $g(z)$ has only one sigular point $z_0$
- Example: $I_1=\oint_C\cfrac{\sin z}{z} dz, C:|z|=2$
  - $g(z)=\cfrac{\sin z}{z}$, singular point $z=0$
  - $I_1=2\pi i sin(0)$
  - $=0$
- Example2: $I_2=\oint_{|z-i|=1}\cfrac{1}{z^2+i}dz$
  - $g(z)=\cfrac{1}{z^2+i}$, singular point $z_0=\sqrt[n]{|z|}e^{i(\frac{-\pi/2+2k\pi)}{n}}, k \in {0,1}$
  - $z_0=\cfrac{\sqrt{2}}{2}-\cfrac{\sqrt{2}}{2}i$, $z_1=-\cfrac{\sqrt{2}}{2}+\cfrac{\sqrt{2}}{2}i$
  - $z_1$ in C
  - $\oint_C\cfrac{1}{z^2+i}dz=\oint_C\cfrac{\cfrac{1}{z-z_0}}{z-z_1}dz=2\pi i \cfrac{1}{z-z_0}\bigg|_{z=z_1}=\cfrac{2\pi i}{-\sqrt{2}+\sqrt{2}i}=\cfrac{\sqrt{2}}{2}\pi(1-i)$
- Example3: $I_3=\oint_{|z|=2}\cfrac{e^z}{(z^2+1)(2z-1)}$
if $f(z)$ analytic on $D$, $f^{(n)}=\cfrac{n!}{2\pi i}\oint_C\cfrac{f(z)}{(z-z_0)^{n+1}},n\in N$
- $\oint_C\cfrac{f(z)}{(z-z_0)^{n+1}}=\cfrac{2\pi i }{n!}f^{(n)}(z_0)$
- Example: $I_1=\oint_{|z-i|=1}\cfrac{\cos z}{(z-i)^3}$
  - $g(z)=\cfrac{\cos z}{(z-i)^3}$, singular point $z=i$
  - $I_1=2\pi i\cfrac{(\cos z)’’}{2!}\bigg|_{z=i}=\pi i (-\cos i)=-\pi i \cfrac{e^{-1}+e}{2}$
- Example2: $I_2=\oint_{|z|=1}\cfrac{\sin z}{z^{2020}}dz$
  - $g(z)=\cfrac{\sin z}{z^{2020}}$, $z_0=0$
  - $I_2=2\pi i\cfrac{(\sin z)^{(2019)}}{2019!}\bigg|_{z=z_0}=2\pi i\cfrac{-\cos 0}{2019!}=-\cfrac{2\pi i}{2019!}$
- Example3: $I_3=\oint_{|z|=4}\cfrac{e^z}{z^2(z-1)^2}dz$
  - $z_0=0$, $z_1=1$
  - $\oint_{|z|=4}g(z)dz=\oint_{C_1}\cfrac{\frac{e^z}{(z-1)^2}}{z^2}dz+\oint_{C_2}\cfrac{\frac{e^z}{z^2}}{(z-1)^2}=\cfrac{2\pi i }{1!}\cfrac{e^z}{(z-1)^2}\bigg|{z=0}+\cfrac{2\pi i}{1!}\cfrac{e^z}{z^2}\bigg|{z=1}=…$
infinite series
- $\sum_{n=1}^{\infty}z_n$
- $\sum_{n=1}^{\infty}q^n$ convergence on $|q|<1$
- $z_n=x_n+iy_n$ convergent then $\sum_{n=1}^\infty x_n$ and $\sum_{n=1}^\infty y_n$ convergent.
- Example $\sum_{n=1}^{\infty}\cfrac{i^n}{n}$
  - $=i-\cfrac{1}{2}-\cfrac{1}{3}i+\cfrac{1}{4}+\cfrac{1}{5}i$
  - $=(-\cfrac{1}{2}+1/3-1/6+\cdots)+i(1-1/3+1/5-\cdots)$
- $\sum_{n=1}^{\infty}|z_n|$ convergent, then $\sum_{n=1}^{\infty}z_n$ convergent.
- power series
  - $\sum_{n=0}^{\infty}C_n(z-z_0)^n$
  - Abel’s Theorm
    - if $\sum_{n=1}^{\infty}C_n(z-z_0)$ convergent at $z_1$, it definitely convergent on $K:|z-z_0|<|z_1-z_0|$
Taylor Series
- Taylor Expansion Theorem: If $f(z)$ analytic in $D$, for all $z_0\in D$, if $K:|z-z_0|<R$ in $D$, $f(z)$ could be expand into $f(z)=\sum_{n=0}^{\infty}C_n(z-z_0)^n$, where
  - $C_n=\cfrac{1}{2\pi i}\oint_{C_\rho}\cfrac{f(\xi)}{(\xi-z_0)^{n+1}}d\xi=\cfrac{f^{(n)}(z_0)}{n!}$: Taylor Coefficient
  - $z_0$: Expand Center
  - $C_\rho:|\xi-z_0|=\rho$, $0<\rho<R$
  - Trick: $\cfrac{1}{\xi-z}=\cfrac{1}{(\xi-z_0)-(z-z_0)}=\cfrac{1}{\xi-z_0}\cfrac{1}{1-\cfrac{z-z_0}{\xi-z_0}}=\cfrac{1}{\xi-z_0}\sum_{n=0}^{\infty}t^n=\sum_{n=0}^{\infty}\cfrac{(z-z_0)^n}{(\xi-z_0)^{(n+1)}}$
  - Example 1: $f(z)=\cfrac{1}{1+z}$, $z_0=0$
    - Singular Point: $z=-1$, $D:z\not ={-1}$
    - $K: |z-z_0|=|z|<1$
    - $\cdots$
- Taylor Expansion Method 1. Solve Taylor Coefficient Directly: $C_n=\cfrac{f^{(n)}(z_0)}{n!}$ + $e^z=\sum_{n=0}^{\infty}\cfrac{z^n}{n!}, |z|<\infty$ + $\sin(z)=\sum_{n=0}(-1)^n\cfrac{z^{2n+1}}{(2n+1)!}$ + $\cos(z)=\sum_{n=0}(-1)^n\cfrac{z^{2n}}{(2n)!}$ 2. Solve with known expansion like $\cfrac{1}{1-z}=\sum_{n=0}^\infty z^n, |z|<1$ + $\cfrac{1}{1-z^2}$, $z_0=0$ + $R=1$, $|z|<1$ + $\cfrac{1}{1-z^2}=\sum_{n=0}^\infty z^2n$ + $\cfrac{1}{1+z^2}=\sum_{n=0}^\infty (-1)^n z^{2n}$ + $\cfrac{1}{1+z}=\sum_{n=0}^\infty (-1)^n z^n$ + $\cfrac{z-1}{z+1}$, $z_0=1$ + $R=2, |z-1|<2$ + $=1-\cfrac{2}{1+z}=1-\cfrac{1}{2+(z-1)}=1-\cfrac{1}{2}\cfrac{1}{1+\cfrac{z-1}{2}}=1-\cfrac{1}{2}\sum_{n=0}^\infty (-1)^n(\cfrac{z-1}{2})^{2n}$ + $\cfrac{z}{z^2+3z+2}$, $z_0=2$ + $R=3$, $|z-2|<3$ + $=\cfrac{z}{(z+1)(z+2)}=\cfrac{a}{z+1}+\cfrac{b}{z+2}$ + $a(z+2)+b(z+1)=z$ + $a=-1$, $b=2$ + $=-\cfrac{1}{z+1}+\cfrac{2}{z+2}=\cdots$ + $\cfrac{2}{z+2}=\cfrac{2}{(z-2)+4}=\cfrac{1}{2}\cfrac{1}{1+\frac{z-2}{4}}=\cfrac{1}{2}\sum_{n=0}^\infty(-1)^n(\cfrac{z-2}{4})^n$ + $f(z)=\cfrac{z}{z^2-2z+5}$, $z_0=1$ + singular point: $z^2-2z+5=0$, $1\pm 2$ + $R=2$ + $f(z)=\cfrac{(z-1)+1}{(z-1)^2+4}=((z-1)+1)\cfrac{1}{(z-1)^2+4}$ + $=\cfrac{(z-1)+1}{4}\sum_{n=0}^\infty (-1)^n(\cfrac{z-1}{2})^{2n}$ + $=\cfrac{z-1}{4}\sum_{n=0}^\infty (-1)^n(\cfrac{z-1}{2})^{2n}+\cfrac{1}{4}\sum_{n=0}^\infty (-1)^n(\cfrac{z-1}{2})^{2n}$ 3. Derivative within contour $\cfrac{1}{(z-a)^2}=-(\cfrac{1}{z-a})’$ + Example: $f(z)=\cfrac{1}{(z+2)^2}$, $z_0=-1$ + $R=1$, $|z+1|<1$ + $\cfrac{1}{(z+1)^2}=(-\cfrac{f1}{z+2})’$ + $-\cfrac{1}{z+2}=-\cfrac{1}{(z+1)+1}=-\sum_{n=0}^\infty (-1)^{n+1}(z+1)^n$ + $\Rightarrow \cfrac{1}{(z+2)^2}=\sum_{n=1}^\infty (-1)^{n-1}n(z+1)^{n-1}$ + Example2: $f(z)=\cfrac{1}{(z^2+1)^2}$, $z_0=0$ + $|z|<1$ + $\cfrac{1}{(z^2+1)^2}=(-\cfrac{1}{z^2+1})’\cfrac{1}{2z}$ + $(\sum_{n=0}^\infty (-1)^n z^2n)’\cfrac{1}{2z}$ + $=\sum_{n=1}^\infty (-1)^{n+1}z^{2n-1}(2n)\cfrac{1}{2z}$ + $=\sum_{n=1}^\infty (-1)^{n+1}(2n)z^{2n-2}$ + Example3: $f(z)=\cfrac{z^2}{(z+1)^2}$, $z_0=1$ + $|z-1|<2$ + $f(z)=\cfrac{(z+1)^2-2(z+1)+1}{(z+1)^2}=1-\cfrac{}{z+1}+\cfrac{1}{(z+1)^2}$ + $\cdots$
Laurent Series
- Laurent Expansion
  - Definition: $C_0+C_1(z-z_0)+C_2(z-z_0)^2+\cdots$(1) and $\cfrac{C_{-1}}{z-z_0}+\cfrac{C_{-2}}{(z-z_0)^2}+\cdots$(2), (1)+(2) called Laurent Series.
    - Expression: $\sum\limits_{n=-\infty}^\infty C_n(z-z_0)^n$
  - Convergence Region:
    - Let $\xi=\cfrac{1}{z-z_0}$, then $(2)=c_{-1}\xi+c_-2\xi^2+\cdots$
    - $|\xi|<\cfrac{1}{r}(0\leq r\le \infty)$ convergent
    - $\Rightarrow$ (2) convergent in $|z-z_0| \ge r(0 \leq r\le \infty$
    - When $r<R$, $r \le |z-z_0| \le R$
      - Especially when $r=0$, $0<|z-z_0|<R$
- $f(z)=\sum\limits_{n=-\infty}^{+\infty}C_n(z-z_0)^n$ could be derivative by steps within $H$
- $f(z)$ could be integrate along $C$ within $H$
- Laurent Expansion Theorem
  - $C_n\cfrac{1}{2\pi i}\oint_C\cfrac{f(\xi)}{(\xi-z_0)^{n+1}}d\xi$
    - $C_n\not ={\cfrac{f^{(n)}(z_0)}{n!}}$
  - Example1: $f(z)=\cfrac{z}{z^2+z-2}$, $z_0=-2$
    - singular point: $z=-2,1$
    - $\rho_0=0$,$\rho_1=3$
    - $0< |z+2|< 3$
    - $\cdots$
- Laurent Expansion Method
  - Expand through exsisted Taylor Expansion $\cfrac{1}{1-z}=\sum\limits_{n=0}^\infty z^n,|z|<1$
    - $z_0$ is an analyic point on $f(z)$
    - Example $f(z)=\cfrac{z}{z^2+z-2}$, $z_0=-1$
      - $f(z)=\sum_{n=-\infty}^\infty C_n(z+1)^n$
      - $f(z)=\cfrac{z}{(z+2)(z-1)}=\cfrac{a}{z+2}+\cfrac{b}{z-1}=\cfrac{1}{3}(\cfrac{2}{z+2}+\cfrac{1}{z-1})$
        (1) $\cfrac{2}{z+2}=\cfrac{2}{(z+1)+1}=\sum_{n=0}^\infty(-1)^n(z+1)^n$
        (1) $\cfrac{1}{z-1}=\cfrac{1}{(z+1)-2}=\cfrac{1}{2}\sum_{n=0}^\infty(\cfrac{z+1}{2})^n$
        (2) $\cfrac{2}{z+2}=\cfrac{2}{(z+1)+1}=\cfrac{2}{z+1}\sum_{n=0}^\infty(-1)^n(\cfrac{1}{z+1})^n$
        (2) $\cfrac{1}{z-1}=\cfrac{1}{(z+1)-2}=\cfrac{1}{2}\sum_{n=0}^\infty(\cfrac{z+1}{2})^n$
    - Example: $f(z)=\sin \cfrac{1}{z}$, $z_0=0$
      - $0<|z|<+\infty$, $f(z)=\sum C_n z^n$
      - $\sin \frac{1}{z}=\sin t(t=1/z)=\sum_{n=0}^\infty (-1)^n\cfrac{z^{2n+1}}{(2n+1)!}$, $0<|t|<\infty$
      - $\cdots$
  - Integrate/Derivate by steps within domain region
    - Example $f(z)=\cfrac{1}{(z+2)^2}$
      - $f(z)=-(\cfrac{1}{z+2})’$, $z_0=-1$
      - $\cfrac{1}{z+2}=\cfrac{1}{1+(z+1)}=\cfrac{1}{z+1}\cfrac{1}{1+\frac{1}{z+1}}=\cfrac{1}{z+1}\sum_{n=0}^\infty(-1)^n(\cfrac{1}{z+1})^n=\sum_{n=0}^\infty(-1)^{n++1}(\cfrac{1}{z+1})^{n+1}$
      - $(-\cfrac{1}{z+2})’=\sum (-1)^{n+1}(\cfrac{1}{z+1})^n(-\cfrac{1}{(z+1)^2})$
  - Integrate
    - $\oint_Cf(z)dz=\oint_{C_1}f(z)dz+\oint_{C_2}f(z)dz$
    - $\oint_{C_1}f(z)dz ==\oint_{C}\sum_{C_n}(z-z_0)^n$
      - $f(z)=\sum_{n=0}^\infty C_n(z-z_0)^n+\sum_{n=-\infty}^{-1}c_n(z-z_0)^n$
        Analytic Part + Main Part
    - $=\sum_{n=-\infty}^\infty \oint_{C_1}C_n(z-z_1)^ndz$
    - $=\sum_{n=0}^\infty\oint_{C_1}C_n(z-z_0)^n+\sum_{n=-\infty}^{-1}\oint_{C_1}C_n(z-z_0)^ndz$
    - $=\sum_{n=-\infty}^{-1}\oint_{C_1}C_n(z-z_0)^ndz$
    - $=\sum_{n=1}^\infty C_{-n}\oint_{C_1}\cfrac{1}{(z-z_0)^n}$
    - $=C_{-1}2\pi i$
Residue: $\text{Res}[f(z),z_0]=C_{-1}$
- Residue Theroem: $\oint_Cf(z)dz=2\pi i\sum_{k=1}^n\text{Res}[f(z),z_k]$
- Isolated Singularity
  - Classification of Isolated Singularity
    - Removable Singularity
      - Main part of $f(z)$ equals to 0, $f(z)=C_0+C_1(z-z_0)+\cdots$, $0<|z-z_0|<R$
      - Append definition $f(z_0)=\lim\limits_{z\to z_0}f(z)=C_0$ then $f(z)$ analytic on $z_0$.
      - Example 1: Prove that $z=0$ is the removable singularity of $f(z)=\cfrac{\sin z}{z}$
        Singularity: $z_0=0$, $0<|z|<+\infty$
        $f(z)=\cfrac{1}{z}\sum_{n=0}^\infty(-1)^n\cfrac{z^{2n+1}}{(2n+1)!}$
        $\cdots$
      - Th: $z_0$ is the removable singularity of $f(z)$ $\Leftrightarrow$ $\lim\limits_{z\to z_0}f(z)=C_0 (\not ={\infty})$
      - L’Hospital Rule: $\lim\limits_{z\to z_0}\cfrac{f(z)}{g(z)}=\lim\limits_{z\to z_0}\cfrac{f’(z)}{g’(z)}$
      - $f(z)=\cfrac{\sin z}{z}$, $z_0=0$
      - $f(z)=\cfrac{e^z-1}{z}$, $z_0=0$
      - $f(z)=\cfrac{1}{e^z-1}-\cfrac{1}{z}$, $z_0=0$ is removable singularity
    - Pole Singularity
      - $f(z)=\cfrac{C_{-m}}{(z-z_0)^m}+\cdots+ \cfrac{C_{-1}}{z-z_0}+C_0+C_1(z-z_0)+\cdots$ and $C_m\not ={0}$, then $z_0$ is the m-order pole singularity of $f(z)$
      - Example: prove that $z=0$ is the 2018-order polar singularity of $f(z)=\cfrac{e^z-1}{z^{2019}}$
        $z_0=0$
        $f(z)=\cfrac{1}{z^{2019}}(-1+ 1+\cfrac{z^2}{2!}+\cfrac{z^3}{3!}+\cdots$
        $\cdots$
      - methods
        $z=z_0\Leftrightarrow \lim_{z\to z_0}f(z)=\infty$, $\cfrac{1}{0}=\infty, \cfrac{1}{\infty}=0, a\cdot \infty=\infty$
        Example: $f(z)=\cfrac{e^z-1}{z^{2019}}$
        $\lim\limits_{z\to 0}\cfrac{e^z-1}{z^2019}=\lim\limits\cfrac{e^z}{2019z^{2018}}=\infty$
        $z_0$ is the pole singularity of $f(z)\Leftrightarrow f(z)$ can be expressed as $f(z)=\cfrac{\phi(z)}{(z-z_0)^m}$ and $\phi(z)$ analytic in $z_0$ and $\phi(z_0)\not ={0}$
        Requirements
        Polynomial denominator
        Important: $\frac{1}{0}$
        Example: $f(z)=\cfrac{1}{z^2(z-1)}$, $z_0=0$
        $f(z)=\cfrac{\frac{1}{z-1}}{z^2}$
        2-order
        $z_0$ is m-order pole singularity of $f(z)\Leftrightarrow z_0$ is m-order 0-point of $g(z)=\cfrac{1}{f(z)}$
        Example: $f(z)=\cfrac{1}{e^z-1}$, $z_0=2k\pi i$
        $g(z)=e^z-1$
        $g^{(1)}(z)=e^z$
        $g^{(1)}(2k\pi i)=0$
        1-order
        Example 2: $f(z)=\cfrac{z-1}{z(e^z-1)}$
        $z_0=0$ or $2k\pi i$
        $z_0=0$
        $g(z)=\cfrac{z(e^z-1)}{z-1}=\cfrac{1}{z-1}h(z)$
        $h(z)=z(e^z-1)$
        $g’(z)=(\cfrac{1}{z-1})‘h(z)+(\cfrac{1}{z-1})h’(z)$
        $g’(0)=(\cfrac{1}{0-1})h’(0)=0$
        $g’’(0)=(\cfrac{1}{z-1})‘‘h(z)+(\cfrac{1}{z-1})‘h’(z)+\cfrac{1}{z-1}h’’(z)\Leftrightarrow g’’(0)=0\Leftrightarrow h’’(0)=0$
        $\sin z=\cfrac{e^{iz}-e^{-iz}}{2i}$
      - $f(z)=g(z)h(z)$ where $g(z)$ is removable singularity and $f(z)$ is m-order pole singuarity, then $f(z)$ is m-order pole singularity
        $f(z)=\cfrac{\sin z}{z(e^z-1)}=\cfrac{\sin z}{z}\cdot \cfrac{1}{e^z-1}$
    - Essential Singularity
      - $f(z)=\cdots + \cfrac{C_{-m}}{(z-z_0)^m}+ \cdots$
      - $z=0$, $f(z)=e^{\frac{1}{z}}$, $\sin\cfrac{1}{z}$, $\cos \cfrac{1}{z}$
- Residue of removable singularity: $Res[f(z),z_0]=C_{-1}=0$
  - $Res[\cfrac{z}{e^z-1}]=0$, $\oint_{|z|=10}\cfrac{z}{e^z-1}dz=2\pi i*$
- Residue of essential singularity: solve with Laurent Expansion
  - Example: $f(z)=\cos\cfrac{1}{z}$
    - $f(z)=1-\cfrac{1}{2!}(1/2)^2$
    - 0
  - Example: $f(z)=\cfrac{z^3}{1+z}e^{\cfrac{1}{z}}$
    - $z=0$
      - $f(z)=(z^3(1-z+z^2-z^3\cdots))(1+\cfrac{1}{z}+\cfrac{1}{2!z}+\cdots)$
      - $=(\cfrac{1}{4!}-\cfrac{1}{5!}+\cfrac{1}{6!}-\cdots)\cfrac{1}{z}+ \cdots$
      - $=(e^{-z}-\cfrac{1}{2!}+\cfrac{1}{3!})(1/z)+\cdots$
- Residue of m-order pole singularity
  - solve with Laurent Expansion: $C_{-1}$
    - Example: $Res[\cfrac{e^z-1}{z^{2019}},0]$
      - $=\cfrac{1}{z^{2019}}(z+\cfrac{1}{2!}z^2+\cdots+\cfrac{1}{n!}z^n+\cdots)$
      - $C_{-1}=\cfrac{1}{2018!}$
    - $I=\oint_{|z|=1}\cfrac{e^z-1}{z^{2019}}dz=2\pi i*\cfrac{1}{2018!}$
  - if $z_0$ is m-order pole singularity of $f(z)$, $Res[f(z),z_0]=\cfrac{1}{(m-1)!}\lim\limits_{z\to z_0}\cfrac{d^{m-1}}{dz^{m-1}}[(z-z_0)^mf(z)]$
    - Especially when $m=1$, $Res[f(z),z_0]=\lim\limits_{z\to z_0}[(z-z_0)f(z)]$
    - Example1 $Res[\cfrac{e^z}{(z-1)^2(2z+1)}]$
      - $z_0=1$: 2-order pole singularity
        $Res[\cfrac{e^z}{(z-1)^2(2z+1)},1]=\cfrac{1}{1!}(\cfrac{e^z}{2z+1})’=\cfrac{e}{9}$
      - $z_0=-\cfrac{1}{2}$: 1-order
        $Res[\cfrac{e^z}{(z-1)^2(2z+1)},-\cfrac{1}{2}]=\lim\limits_{z\to-1/2}\cfrac{e^z}{(z-1)^2}$
    - Example $Res[\cfrac{z-1}{z(e^z-1)},0]$
      - 2-order
      - $=\lim_{z\to 0}({\cfrac{(z-1)z}{e^z-1}})'$
      - $=\lim_{z\to 0}{\cfrac{(2z-1)(e^z-1)-z(z-1)e^z}{(e^z-1)^2}}$
      - $=\cfrac{3}{2}$
    - Example $Res[\tan\pi z,k+\cfrac{1}{2}], I_3=\oint_{|z|=2}\tan \pi zdz$
      - $z_k=k+\cfrac{1}{2}$: 1-order;
      - $=\lim_{z\to z_k}(z-z_k)\cfrac{\sin \pi z}{\cos\pi z}$
      - $\cdots$
    - Example $f(z)=\cfrac{1}{x^4+1}$
      - $z_k=e^{\cfrac{(2k+1)\pi i}{4}}$
      - 1-order
      - $Res[f(z),z_k]=\lim_{z\to z_k}f(z)(z-z_k)=\cfrac{1}{4z_k^3}=\cfrac{z_k}{4\cdot (-1)}$
  - Process 1. find all singularity 2. check type 3. get residue based on singularity type 4. use residue theroem
- Application of Residue theorem
  - $\int_0^{2\pi} R(\cos\theta,\sin\theta)d\theta$
    - $\int_0^{2\pi} R(\cos\theta,\sin\theta)d\theta=\oint_Cf(z)dz$
    - let $z=e^{i\theta}$
    - $\int_0^{2\pi}R(\cos\theta,\sin\theta)d\theta=\oint_{|z|=1}f(z)dz$
      - $d\theta=\cfrac{1}{iz}dz$
      - $\cos\theta=\cfrac{e^{i\theta}+e^{-i\theta}}{2}=\cfrac{z^2+1}{2z}$
      - $\sin\theta=\cfrac{e^{i\theta-e^{-i\theta}}}{2i}=\cfrac{z^2-1}{2iz}$
    - $I=\oint_{|z|=1}R(\cfrac{z^2+1}{2z},\cfrac{z^2-1}{2iz})\cfrac{1}{2z}dz$
    - Example: $I=\int_0^{2\pi}\cfrac{d\theta}{5-3\cos\theta}$
      - $z=e^{i\theta}$
      - $\cos\theta=\cfrac{z^2+1}{2z}$
      - $d\theta=\cfrac{1}{iz}dz$
      - $I=\oint_{|z|=1}\cfrac{1}{5-3(\cfrac{z^2+1}{2z})}\cfrac{1}{iz}dz$
      - $=-\cfrac{2}{i}\oint_{|z|=1}\cfrac{dz}{(3z-1)(z-3)}=-\cfrac{1}{8}$
  - $\int_{-\infty}^\infty\cfrac{P(x)}{Q(x)}dx=\lim\limits_{R\to\infty}\int_{-R}^R\cfrac{P(x)}{Q(x)}dx$
    - where $P(x)=C_0x^m+C_1x^{m-1}+\cdots+C_m$ and $Q(x)=b_0x^n+b_1x^{n-1}+\cdots+b_n$
      - $\partial Q(x)-\partial P(x)\ge 2$
      - $Q(x)\ne 0$
    - then $\int_{-\infty}^\infty\cfrac{P(x)}{Q(x)}dx=2\pi i\sum_{i=1}^n Res[f(z),z_k]$, where $f(z)=\cfrac{P(z)}{Q(z)}$ and $z_k$ is the singularity on above part of $f(z)$
    - Example: $I=\int_{-\infty}^\infty\cfrac{1}{1+x^4}dx$
      - $f(z)=\cfrac{1}{1+z^4}$
      - $z_k=e^{\cfrac{\pi+2k\pi}{4}}$, $k=0,1$
      - $I=2\pi i(Res(f(z),0)+Res(f(z),1))$
    - Example2: $I_2=\int_0^\infty\cfrac{1}{1+x^4}dx$
      - $=\cfrac{\int_{-\infty}^\infty\cfrac{1}{1+x^4}dx}{2}$
  - $\int_{-\infty}^\infty\cfrac{P(x)}{Q(x)}e^{iax}dx$ $(a>0)$
    - $\partial Q(x)>\partial P(x)$
    - $Q(x)\ne 0$
    - $\int_{-\infty}^\infty\cfrac{P(x)}{Q(x)}e^{iax}dx=2\pi i\sum_{i=1}^n Res[\cfrac{P(x)}{Q(z)}e^{iaz},z_k]$
      - $z_K$ is the singularity on $\cfrac{P(z)}{Q(z)}e^{iaz}$ on above part of $f(z)$
    - $A=2\pi i\sum_{i=1}^nRes[\cfrac{P(z)}{Q(z)}e^{iaz},z_k]$
    - $\int_{-\infty}^\infty \cfrac{P(x)}{Q(x)}\cos axdx=\Re A$
    - $\int_{-\infty}^\infty \cfrac{P(x)}{Q(x)}\sin axdx=\Im A$
Fourier Transform
- Dirichlet Requiremnt:
  - Continuous or has limited one-order discontinuity.
  - ?
- $f_T(t)=\cfrac{a_0}{2}\sum_{n=1}\infty a_n \cos n\omega_0 t+b_n\sin n \omega_0 t$
  - $a_n=\cfrac{2}{T}\int_{-T/2}^{T/2}f_T(t)\cos n\omega_0 tdt$
  - $b_n=\cfrac{2}{T}\int_{-T/2}^{T/2}f_T(t)\sin n\omega_0 tdt$
  - $\omega_0:$ Baseband
- using Euler Equation
  - $f_T(t)=\sum_{n=-\infty}^{\infty}c_ne^{in\omega_0 t}$
  - $c_0=\cfrac{a_0}{2}$
  - $c_n=\cfrac{1}{T}\int_{-T/2}^{T/2}f_T(t)e^{-in\omega_0 t}dt$
  - $f_T(t)=\sum_{n=-\infty}^\infty(\cfrac{1}{T}\int_{-T/2}^{T/2}f_T(t)e^{-in\omega_0 t}dt)e^{in\omega_0 t}$
  - $c_n=\cfrac{1}{T}\int_{-T/2}^{T/2}f_T(t)e^{-i\omega_nt}dt$
  - $f(t)=\lim\limits_{T\to\infty}f_T(t)=\lim\limits_{T\to\infty}\cfrac{1}{T}\sum_{n=-\infty}^\infty(\int_{-T/2}^{T/2}f_T(t)e^{-in\omega_0 t}dt)e^{i\omega_n t}$
  - $\Delta\omega=\omega_n-\omega_{n-1}=2\pi/T$
  - $\Delta\omega\to 0\Leftrightarrow T\to +\infty$
  - $f(t)=\lim\limits_{\Delta \omega\to 0}\cfrac{1}{2\pi}\sum_{n=-\infty}^\infty F_T(\omega_n)e^{i\omega_n t}\Delta\omega$
  - $f(t)=\cfrac{1}{2\pi}\int_{-\infty}^\infty[\int_{-\infty}^\infty f(t)e^{-i\omega t}dt]e^{i\omega t}d\omega$
  - $F(\omega)$: Fourier Transform of $f(t)$: $F_T(\omega)=\int_{-\infty}^{\infty}f(t)e^{-i\omega t}dt$, $T\to +\infty$
  - Fourier Integration of $f(t)$: $f(t)=\cfrac{1}{2\pi}\int_{-\infty}^{+\infty}F(\omega)e^{i\omega t}d\omega$
- Fourier Integration Exisit theorem
- LIST:
  - $\int_{-\infty}^{\infty}|f(t)|dt<+\infty$
  - $f(t)$ support Dirichlet requirement within bounded region
  - Fourier Transform of $f(t)$: $F(\omega)=\mathscr{F}[f(t)]=\int_{-\infty}^{+\infty}f(t)e^{-i\omega t}dt$
  - Inverse Fourier Transform of $F(\omega)$: $\mathscr{F}^{-1}[F(\omega)]=\cfrac{1}{2\pi}\int_{-\infty}^{+\infty}F(\omega)e^{i\omega t}d\omega$
- $\delta$ Function (Dirac Function)
  - $$\rho(x)=\lim\limits_{L\to 0}\rho_L(x)=\begin{cases} 0 & x\ne x_0\\ \infty & x=x_0\\ \end{cases}$$
  - $\int_{-\infty}^{\infty}\delta(t)dt=1$
  - $\int_{-\infty}^{\infty}\delta(t-t_0)f(t)dt=f(t_0)$
- Fourier Transformation of $\delta$ function
  - $\mathscr{F}[\delta(t)]=\int_{-\infty}^\infty\delta(t)e^{-i\omega t}dt=e^{-i\omega t}\big|_{t=0}=1$
  - $\delta(t)=\cfrac{1}{2\pi}\int_{-\infty}^\infty 1e^{i\omega t}d\omega\Rightarrow \int_{-\infty}^\infty e^{i\omega t}d\omega=2\pi \delta(t)\Rightarrow \int_{-\infty}^\infty e^{i\omega t}dt=2\pi \delta(\omega)$
  - Example $\mathscr{F}[1]$
    - $\int_{-\infty}^\infty e^{-i\omega t}dt=\int_{-\infty}^\infty e^{it(-\omega)}dt=2\pi \delta(-\omega)=2\pi\delta(\omega)$
  - Example $f(t)=e^{i\omega_0 t}$
    - $\mathscr{F}[e^{i\omega_0 t}]=\int_{-\infty}^\infty e^{it(\omega_0-\omega)}dt=2\pi\delta(\omega-\omega_0)$
  - Example $f(t)=\cos\omega_0 t$
    - $\mathscr{F}[f(t)]=\int_{-\infty}^\infty \cos\omega_0 te^{-i\omega t}=\int_{-\infty}^{\infty}=\cfrac{e^{i\omega_0t}-e^{-i\omega_0t}}{2}e^{-i\omega t}=\pi[\delta(\omega-\omega_0)+\delta(\omega+\omega_0)]$
- Characteristics
  - Linear
    - $\mathscr{F}[\alpha f+\beta g]=\alpha\mathscr{F}[f]+\beta\mathscr{F}[g]$
  - Translate
    - $\mathscr{F}[f(t-t_0)]=e^{-i\omega t_0}\mathscr{F}[f]$
    - $\mathscr{F}[e^{i\omega_0t}f(t)]=F(\omega-\omega_0)$
    - Example $\mathscr{F}[\delta(t-2)]$
      - $=e^{i\omega 2}$
    - Example $\mathscr{F}[\cos\omega_0(t-1)]=e^{i\omega}\pi(\delta(\omega-\omega_0)+\delta(\omega+\omega_0))$
    - Example $\mathscr{F}[f(t)\cos\omega_0t]=\mathscr{F}[f(t)\cfrac{e^{i\omega_0t}+e^{-i\omega_0t}}{2}]=\cfrac{1}{2}F(\omega-\omega_0)+\cfrac{1}{2}F(\omega+\omega_))$
  - Similarity
    - assume that $a\not ={0}$, $\mathscr{F}[f(at)]=\cfrac{1}{|a|}F(\cfrac{\omega}{a})$
  - differential
    - If $\lim\limits_{|t|\to\infty}f(t)=0$,then
    - $\mathscr{F}[f’(t)]=i\omega\mathscr{F}[f]$
    - $\cfrac{dF(\omega)}{d\omega}=\mathscr{F}[-itf(t)]$
    - Generally
      - $\mathscr{F}[f^{(n)}(t)]=(i\omega)^n\mathscr{F}[f]$
      - $\cfrac{d^nF(\omega)}{d\omega^n}=(-i)^n\mathscr{F}[t^nf(t)]$
    - Example $\mathscr{F}[(2t-1)f(t)]$
      - $=2\mathscr{F}[tf]-F(\omega)=$
  - integral
    - $\mathscr{F}[g(t)]=\cfrac{1}{i\omega}\mathscr{F}[f(t)]$
  - Convolution
    - $\mathscr{f_1*f_2}=\mathscr{f_1(t)}+\mathscr{f_2(t)}$
    - TODO
Preparation For Laplace Transform
- $\cfrac{Q(s)}{R(s)}$
  - $==\cfrac{c_1}{s-a}+\cfrac{c_2}{(s-a)^2}+\cdots+\cfrac{c_k}{(s-a)^k}$
  - $==\cfrac{M_1s+N_1}{(s-\alpha)^2+\beta^2}+\cfrac{M_2s+N_2}{[(s-\alpha)]^2-\beta^2}$
  - $\cdots$
- Example $\cfrac{5s-1}{(s+1)(s-2)}$
  - $=\cfrac{a}{s+1}+\cfrac{b}{s-2}$
  - $a=2$
  - $b=3$
  - $=\cfrac{2}{s+1}+\cfrac{3}{s-2}$
- Example $\cfrac{s+3}{(s+1)(s+2)^2}$
  - $=\cfrac{a}{s+1}+\cfrac{b}{s+2}+\cfrac{c}{(s+2)^2}$
  - $\cdots$
- Example $\cfrac{1}{[(s+1)^2+1]\cdot(s+2)}$
  - $=\cfrac{a}{s+2}+\cfrac{bs+c}{(s+1)^2+1}+\cfrac{ds+e}{[(s+1)^2+1]^2}$
- Example $\cfrac{s^2}{(s^2+1)^2}$
  - $=\cfrac{as+b}{(s^2+1)}+\cfrac{cs+d}{(s^2+1)^2}$
Laplace Transformation
- Definition of Laplace Transformation
  - assume that $f(t)$ is a real function defined on $[0,\infty)$ and support Dirichlet Requirements, for $s=\beta+i\omega$, the Laplace Transformation of $f(t)$ is $F(s)=\int_0^{+\infty}f(t)e^{-st}dt$ which convergent.
  - $$f_1(t)=\begin{cases} f(t)e^{-st}&t>=0\\ 0&t<0\\ \end{cases}$$
  - $\mathscr{F}[f_1(t)]=\int_{-\infty}^{+\infty}f_1(t)e^{-i\omega t}dt=\int_0^{+\infty}f(t)e^{-\beta t}e^{-i\omega t}dt=\int_0^{+\infty}f(t)e^{-(\beta+i\omega)t}dt==$
  - $\mathscr{L}[f(t)]=\int_0^{+\infty}f(t)e^{-st}dt$
  - $f(t)$ is the reverse Laplace Tranformation of $F(s)$, signed $f(t)=\mathscr{L}^{-1}[F(s)]$
  - $\mathscr{L}[f(t)]=\mathscr{F}[f_1(t)]=\mathscr{F}[f(t)e^{-\beta t}u(t)]$
- Examples:
  - $\mathscr{L}[1]=\int_0^\infty e^{-st}dt=-\cfrac{1}{s}e^{-st}\bigg|_0^{+\infty}=-\cfrac{1}{s}e^{-\beta t}e^{-i\omega t}\bigg|_0^\infty=\cfrac{1}{s}$
    - $\mathscr{L}^{-1}[\cfrac{1}{s}]=1$
  - $\mathscr{L}[e^{\alpha t}]=\int_0^\infty e^{\alpha t-st}dt=\cfrac{1}{\alpha-s}e^{(\alpha-s)t}\bigg|_0^\infty=\cfrac{1}{s-a}$
    - $\mathscr{L}^{-1}[\cfrac{1}{S-\alpha}]=e^{\alpha t}$
  - $\mathscr{L}[\cos bt]=\mathscr{L}[\cfrac{e^{ibt}+e^{-ibt}}{2}]=\cfrac{1}{2}(\cfrac{1}{s-bi}+\cfrac{1}{s+bi})=\cfrac{s}{s^2+b^2}$
    - $\mathscr{L}^{-1}[\cfrac{s}{s^2+b^2}]=\cos bt$
      - solve $\int_0^\infty e^{-3t}\cos tdt$
        $\mathscr{L}[\cos t]\bigg|{s=3}=\int_0^\infty \cos te^{-st}dt=\cfrac{s}{s^2+1}\bigg|{s=3}$
  - $\mathscr{L}[\sin bt]=\mathscr{L}[\cfrac{e^{ibt}-e^{-ibt}}{2i}]=-\cfrac{i}{2}(\cfrac{1}{s-bi}-\cfrac{1}{s+bi})=\cfrac{b}{s^2+b^2}$
    - $\mathscr{L}^{-1}[\cfrac{b}{s^2+b^2}]=\sin bt$
  - $\mathscr{L}[t]=\int_0^\infty te^{-st}dt=\cfrac{1}{s^2}$
    - $\mathscr{L}^{-1}[\cfrac{1}{s^2}]=t$
  - $\mathscr{L}[t^2]=\cfrac{2!}{s^{3}}$
    - $\mathscr{L}^{-1}[\cfrac{2!}{s^3}]=t^2$
  - $\mathscr{L}[t^m]=\cfrac{m!}{s^{m+1}}$
    - $\mathscr{L}^{-1}\cfrac{m!}{s^{m+1}}=t^m$
- Solve $\mathscr{L}^{-1}[\cfrac{Q(s)}{R(s)}]$
  - Mini-
    - $\cfrac{c_1}{s-a}$
    - $\cfrac{c_2}{(s-a)^2}$
    - $\cfrac{M_1s+N_1}{(s-\alpha)^2+\beta^2}$
    - $\cfrac{M_2s+N_2}{(s-\alpha)^2+\beta^2}$
  - 2 characteristics
    - Linear
      - $\mathscr{L}[\alpha f+\beta g]=\alpha F(s)+\beta G(s),\alpha,\beta\in C$
        $\mathscr{L}^{-1}[\alpha F(s)+\beta G(s)]=\alpha\mathscr{L}^{-1}[F(s)]+\beta\mathscr{L}^{-1}[G(s)]$
      - Example 1:
        $\mathscr{L}^{-1}[\cfrac{s}{(s^2+1)(s^2+9)}]=\cfrac{1}{8}\mathscr{L}^{-1}[(\cfrac{s}{s^2+1}-\cfrac{s}{s^2+9})]=\cfrac{1}{8}(\cos t-\cos 3t)$
      - Example 2:
        $\mathscr{L}^{-1}[\cfrac{1}{(s^2+1)(s^2+9)}]=\cfrac{1}{8}\mathscr{L}^{-1}[(\cfrac{1}{s^2+1}-\cfrac{3}{s^2+9})]=\cfrac{1}{8}(\sin t-\sin 3t)$
    - Translate
      - $\mathscr{L}[e^\alpha tf(t)]=F(s-\alpha)$
      - $\mathscr{L}^{-1}[F(s-\alpha)]=e^{\alpha t}f(t)$
      - Example 1:
        $\mathscr{L}^{-1}[\cfrac{1}{s^2+4s+8}]=\mathscr{L}^{-1}[\cfrac{1}{(s+2)^2+4}]=(\alpha=-2)=e^{-2t}\mathscr{L}^{-1}[\cfrac{1}{s^2+4}]=e^{-2t}\sin 2t$
      - Example 2:
        $\mathscr{L}^{-1}[\cfrac{c_1}{(s-a)^2}]=c_1e^{at}\mathscr{L}^{-1}[\cfrac{1}{s^2}]=c_1te^{at}$
      - $\mathscr{L}^{-1}[\cfrac{1}{(s-\alpha)^2+\beta^2}]=e^{\alpha t}\mathscr{L}^{-1}[\cfrac{1}{s^2+\beta^2}]=\cfrac{1}{\beta}e^{\alpha t}\sin\beta t$
      - $\mathscr{L}^{-1}[\cfrac{s}{(s-\alpha)^2+\beta^2}]=\mathscr{L}^{-1}[\cfrac{s-\alpha}{(s-\alpha)^2+\beta^2}]+\alpha\mathscr{L}^{-1}[\cfrac{1}{(s-\alpha)^2+\beta^2}]=\cdots$
      - $\mathscr{L}^{-1}[\cfrac{s}{[(s-\alpha)^2+\beta^2]^2}]=\mathscr{L}^{-1}[\cfrac{s-\alpha}{[(s-\alpha)^2+\beta^2]^2}]+\alpha\mathscr{L}^{-1}[\cfrac{1}{[(s-\alpha)^2+\beta^2]^2}]=???$ unknown
    - Derivative
      - Assume that $\mathscr{L}[f(t)]=F(s)$, then $F’(s)=-\mathscr{L}[tf(t)]$
      - $\Rightarrow\mathscr{L}^{-1}[F’(s)]=-tf(t)$
      - Generally, $F^{(n)}(s)=(-1)^n\mathscr{L}[t^n f(t)]$
      - $\Rightarrow\mathscr{L}^{-1}[F^{(n)}(s)]=(-1)^nt^nf(t)$
      - Example $\mathscr{L}^{-1}[\cfrac{s}{(s^2+1)^2}]$
        $=\mathscr{L}^{-1}[(-\cfrac{1}{2}\cfrac{1}{s^2+1})’]=-t\mathscr{L}^{-1}[-\cfrac{1}{2}\cfrac{1}{s^2+1}]=\cfrac{t}{2}\sin t$
      - Example $\mathscr{L}^{-1}[\cfrac{s}{(s^2+\beta^2)^2}]=\mathscr{L}^{-1}[(\cfrac{1}{2}\cfrac{1}{s^2+\beta^2})’]=\cfrac{t}{2\beta}\sin\beta t$
      - Example $\mathscr{L}[t\sin\omega t]$
        $=-(\mathscr{L}[\sin\omega t])’=-(\cfrac{\omega}{s^2+\omega^2})’=\cfrac{2s\omega}{(s^2+\omega^2)^2}$
      - Example $\mathscr{L}[e^{-3t}t\sin\omega t]$
        $=F(s+3)$
        $F(s)=\cfrac{2s\omega}{(s^2+\omega^2)^2}$
        $=F(s+s)=\cfrac{2(s+3)\omega}{((s+3)^2+\omega^2)^2}$
    - Differential
      - Assume that $\mathscr{L}[f(t)]=F(s)$, then $\mathscr{L}[f’(t)]=sF(s)-f(0)$
      - $\mathscr{L}[f^{(n)}(t)]=s^nF(s)-s^{n-1}f(0)-s^{n-2}f’(0)-\cdots-f^{(n-1)}(0)$
      - Especially when $f(0)=f’(0)=f^{(n-1)}(0)=0$, $\mathscr{L}[f^{(n)}(t)]=s^nF(s)$
      - Example Important $\begin{cases} y’’+3y’+2y=2e^{-t}\cos t\\ y(0)=y’(0)=0\\\end{cases}$
        $Y(s)=\mathscr{L}[y(t)]$
        then $\mathscr{L}[y’(t)]=sY(s)$
        $\mathscr{L}[y’’(t)]=s^2Ys$
        $\mathscr{L}[2e^{-t}\cos t]=2\cfrac{s+1}{(s+1)^2+1}$
        $(s^2+3s+2)Y(s)=\cfrac{s+1}{(s+1)^2+1}$
        $Y(s)=\cfrac{1}{((s+1)^2+1)(s+2)}$
        $\Rightarrow y(t)=\mathscr{L}^{-1}[Y(s)]=\mathscr{L}^{-1}[\cfrac{1}{s+2}-\cfrac{s}{(s+1)^2+1}]=e^{-2t}+e^{-t}(\cos t-\sin t)$
      - Example $\begin{cases}y’’(t)-y(t)=4\sin t+5\cos 2t\\y(0)=-1, y’(0)=-2\\\end{cases}$
        $Y(s)=\mathscr{L}[y(t)]$, $\mathscr{L}[y’’(t)]=s^2Y(s)+2$
        $-sy(0)-y’(0)=S^2Y(s)+s+2$
        $\mathscr{L}[4\sin t+ 5\cos 2t]=\cfrac{4}{s^2+1}+\cfrac{5s}{s^2+4}$
        $(s^2-1)Y(s)+s+2=\cfrac{4}{s^2+1}+\cfrac{5s}{s^2+4}$
        $Y(s)=\cfrac{4}{(s^2+1)(s^2-1)}+\cfrac{5s}{(s^2+4)(s^2-1)}-\cfrac{s+2}{s^2-1}=-\cfrac{2}{s^2+1}-\cfrac{s}{s^2+4}$
        $y(t)=\mathscr{L}^{-1}[Y(s)]=-2\sin t-\cos 2t$
    - Integral
      - Assume that $\mathscr{L}[f(t)]=F(s)$, then $\mathscr{L}[\int_0^t f(t)dt]=\cfrac{F(s)}{s}$
      - $\Rightarrow \mathscr{L}^{-1}[\cfrac{F(s)}{s}]=\int_0^tf(t)dt$, where $f(t)=\mathscr{L}^{-1}[F(s)]$
      - $\mathscr{L}^{-1}[F’(s)]=-tf(t)$
      - Example $\mathscr{L}^{-1}[\cfrac{1}{(s^2+\beta^2)^2}]$
        $=\mathscr{L}^{-1}[\cfrac{s}{(s^2+\beta^2)^2\cfrac{1}{s}}]=\int_0^tf(t)dt$
        $f(t)=\mathscr{L}^{-1}[\cfrac{1}{(s^2+\beta^2)^2}]=\cfrac{1}{2\beta}t\sin \beta t$
        $I=\int_0^t\cfrac{1}{2\beta}t\sin\beta\tau d\tau=\cfrac{1}{2\beta}[-\cfrac{1}{\beta}](\tau\cos\beta\tau\big|_0^t-\int_0^t\cos\beta\tau d\tau)=-\cfrac{1}{2\beta^2}[t\cos\beta t-\cfrac{1}{\beta}\sin\beta t]$
      - Example $I_1=\mathscr{L}[\int_0^t e^{-3t}t\sin 2t dt]$
        $f(t)=e^{-3t} t\sin 2t$
        $g(t)=t\sin 2t$
        $I_1=\mathscr{L}[f(t)]=\mathscr{L}[e^{-3t}t\sin 2t]=G(s+3)$
        $h(t)=\sin 2t$
        $G(s)=\mathscr{L}[g(t)]=-H’(s)$
        $H(s)=\mathscr{L}[\sin 2t]=\cfrac{2}{s^2+4}$
        $G(s)=\cfrac{4s}{s^2+4}$
        $F(s)=\cfrac{4s+3}{(s+3)^2+4}$
        $I_1=\cfrac{F(s)}{s}$
      - Example $I_2=\mathscr{L}[e^{-3t}\int_0^t t\sin 2t dt]$
        $I_2=\mathscr{L}[e^{-3t}f(t)]=F(s+3)$
        $F(s)=\mathscr{L}[f(t)]=\mathscr{L}[\int_0^tg(t)dt]=\cfrac{G(s)}{s}$
        $G(s)=\mathscr{L}[\sin 2t]=\cfrac{4s}{s^2+4}$
      - Example $I_3=\mathscr{L}[t\int_0^t e^{-3t}\sin 2t dt]$
        $F(s)=\mathscr{L}[g(t)]=\cfrac{G(s)}{s}$
        $G(s)=\mathscr{L}[e^{-3t}\sin 2t]=\cfrac{2}{(s+3)^2+4}$
      - Example $\int_0^\infty\cfrac{1-\cos t}{t}e^{-t}dt=\mathscr{L}[\cfrac{1-\cos t}{t}]\big|_{s=1}$
  - Examples:
    - $\mathscr{L}^{-1}\cfrac{s}{(s^2+1)(s^2+0)}$

Complex Function

Review

Class

The relationship between Analytic Function and Harmonic Function

Integration of Complex Function