解析( Anlyticity )
惯用记法
任意复数 \(z\) 记为
\begin{align*} z = x + \mathrm{i}y \end{align*}复函数 \(f(z)\) 记为
\begin{align*} f(z) = u(x,y) +\mathrm{i} v(x,y) \end{align*}例如
\begin{align*} f(z) = z + z^2 = x + \mathrm{i}y + (x + \mathrm{i}y)^2 = x + x^2 - y^2 + \mathrm{i}(y + 2xy) \end{align*}其中
\begin{align*} u(x,y) =& x + x^2 - y^2 \\ v(x,y) =& y + 2xy \end{align*}导数
由于
\begin{align*} f(z) = f(x,\mathrm{i}y) \end{align*}可以看成是一个二元函数. 那么有类似于一般的二元函数梯度
\begin{align*} \nabla g(x,y) = \frac{\partial}{\partial x} g(x,y) \vec{i}+ \frac{\partial}{\partial y} g(x,y) \vec{j} \end{align*}的定义, 复函数的导数包含两个正交的方向上的导数
\begin{align*} \frac{\partial}{\partial x} f(x,\mathrm{i}y) =&\frac{\partial u}{\partial x} + \mathrm{i}\frac{\partial v}{\partial x}\\ \frac{\partial}{\partial y} f(x,\mathrm{i}y) =&\frac{\partial u}{\partial \mathrm{i}y} + \mathrm{i} \frac{\partial v}{\partial \mathrm{i}y} =- \mathrm{i}\frac{\partial u}{\partial y} +\frac{\partial v}{\partial y} \end{align*}解析 ( Anlyticity ) 函数的定义是满足以下柯西-黎曼条件 (Cauchy-Riemann conditions)
\begin{align*} \frac{\partial u}{\partial x} =&\quad \frac{\partial v}{\partial y} \\ \frac{\partial u}{\partial y} =& -\frac{\partial v}{\partial x} \end{align*}的复函数. 可以看出, 对于解析函数, 在 \(x\) 方向和在 \(\mathrm{i}y\) 方向上的导数是一样的.
解析函数一定可以写在 \(z\) 的函数
一个复函数, 即可以看成是 \((x,y)\) 的函数, 也可以看成是 \((z,z^{*})\) 的函数. 因为这是两组独立的变量, 变换关系是
\begin{align*} \left\{ \begin{array}{c} z = x + \mathrm{i} y \\ z^{*} = x - \mathrm{i}y \end{array} \right. \quad \quad \quad \quad \left\{ \begin{array}{c} x = \frac{1}{2}(z + z^{* }) \\ y = \frac{1}{2\mathrm{i}} (z - z^{*}) \end{array} \right. \end{align*}所以, 复函数更加普遍的定义应该写成 \(f(z,z^{* })\) , 但是对于解析函数有
\begin{align*} \frac{\partial}{\partial z^{* }}f(x, y) =& \frac{\partial f}{\partial x}\frac{\partial x}{\partial z^{* }} +\frac{\partial f}{\partial y}\frac{\partial y}{\partial z^{* }} \\ =& \left(\frac{\partial u}{\partial x} + \mathrm{i}\frac{\partial v}{\partial x}\right)\left(\frac{1}{2}\right) + \left(\frac{\partial u}{\partial y} +\mathrm{i}\frac{\partial v}{\partial y}\right)\left(-\frac{1}{2\mathrm{i}}\right) \\ =&0 \end{align*}所以解析函数不包含 \(z^{* }\) .
留数定理
假设一函数函数可以写成
\begin{align*} g(z) = \frac{f(z)}{z - z_0} \end{align*}的形式, 其中 \(f(z)\) 没有奇点. 那么它在 \(z = z_0\) 处有奇点.对任意包含 \(z_0\) 点的路径 \(C\) 逆时针积分
\begin{align*} \oint_{C}\frac{f(z)}{z -z_0} \mathrm{d}z \end{align*}与在奇点附近上一个无穷小的圆的积分是相等的
\begin{align*} \oint_{C_0}\frac{f(z)}{z -z_0} \mathrm{d}z \end{align*}用振幅和辐角来表示 \(z-z_{0} = r e^{\mathrm{i}\phi}\) , 因为在圆上积分, 所以 \(r\) 是常数, 那么有
\begin{align*} \oint_{C_0}\frac{f(z)}{z -z_0} \mathrm{d}z &= \int_0^{2\pi}\frac{f(z)}{r e^{\mathrm{i}\phi}} \mathrm{i}re^{\mathrm{i}\phi}\mathrm{d}\phi \\ &=\mathrm{i}\int_0^{2\pi} f(z_0) \mathrm{d}\phi \\ &= \mathrm{i} 2\pi f(z_{0}) \end{align*}所以最终就有
\begin{align*} \oint_{C}g(z)\mathrm{d}z = \oint_{C}\frac{f_z}{z -z_0}\mathrm{d}z = 2\pi \mathrm{i} f(z_0) = 2\pi \mathrm{i}\cdot g(z)(z-z_{0})|_{z=z_0} \end{align*}\(g(z)(z-z_0)|_{z=z_0}\) 就是留数, 去掉奇点留下的数.
推广到多个奇点的情况
\begin{align*} \oint_C g(z) \mathrm{d}z = 2\pi \mathrm{i} \sum_{k=1}^N \mathrm{Res}[g(z_k)] \end{align*}参考文献
Jim, Napolitano, March 9, 2013, Notes on Complex Analysis in Physics
http://www.rpi.edu/dept/phys/Courses/PHYS6520/Spring2018/NotesOnComplexAnalysis.pdf