Complex Function

hhong6@mgh.harvard.edu (Haotian Hong) — Thu, 01 Oct 2020 15:09:46 +0800

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

$f’(z_0)=\lim_{\Delta z\to 0}\cfrac{\Delta w}{\Delta z}$

Complex on Bughht Blog

Complex Function

Review

Class