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复变函数：导数可微解析的例题
2025-09-08
复变函数
复变函数

柯西黎曼条件的应用#

eg:求参数使得函数解析#

若 f(z)=x2+axy+by2+i(cx2+dxy+y2)f(z)=x^2+axy+by^2+i(cx^2+dxy+y^2) 在整个复平面内都可导，求参数abcd

由复变函数可导的充要条件：实部虚部函数可微且满足柯西黎曼条件： ∂u∂x=∂v∂y\dfrac{\partial u}{\partial x}=\dfrac{\partial v}{\partial y}, ∂u∂y=−∂v∂x\dfrac{\partial u}{\partial y}=-\dfrac{\partial v}{\partial x}

其中u=x2+axy+by2,v=cx2+dxy+y2u=x^2+axy+by^2,v=cx^2+dxy+y^2，计算偏导数：

∂u∂x=2x+ay;∂u∂y=ax+2by∂v∂x=2cx+dy;∂v∂y=dx+2y\begin{align*} \frac{\partial u}{\partial x}&=2x+ay;& \frac{\partial u}{\partial y}&=ax+2by\\[10bp] \frac{\partial v}{\partial x}&=2cx+dy;& \frac{\partial v}{\partial y}&=dx+2y \end{align*}

代入柯西黎曼条件：

{2x+ay=dx+2yax+2by=−(2cx+dy)\left\{\begin{matrix} \begin{align*} 2x+ay&=dx+2y\\[10bp] ax+2by&=-(2cx+dy) \end{align*} \end{matrix}\right.

化简得：

a=2c=2;d=2;b=−1;a=2c=2; d=2; b=-1;

eg:求共轭调和函数#

若 u=x2+xy−y2,f(i)=−1+i,f(z)u=x^2+xy-y^2,f(i)=-1+i,f(z) 解析，求 f(z)=u+ivf(z)=u+iv

TIP

考虑用多种方法做，法一是直接用算u的共轭调和函数，法二是通过表示f的导数（用实部虚部函数偏导的四种形式），然后再积分，法三（并不算一个单独的方法，凑微分这一块）。

其中u=x2+xy−y2u=x^2+xy-y^2，直接套公式偏积分：

v(x,y)=∫0，0x,y−∂u∂ydx+∂u∂xdy=∫0x−∂u∂ydx+∫0y∂u∂xdy=∫0x−(x−2y)dx+∫0y(2x+y)dy=−12x2+xy+y2+C\begin{align*} v(x,y)&=\int_{0，0}^{x,y}{-\frac{\partial u}{\partial y}dx+\frac{\partial u}{\partial x}dy}\\ &=\int_{0}^{x}{-\frac{\partial u}{\partial y}dx}+\int_{0}^{y}{\frac{\partial u}{\partial x}dy}\\ &=\int_{0}^{x}{-(x-2y)dx}+\int_{0}^{y}{(2x+y)dy}\\ &=-\frac{1}{2}x^2+xy+y^2+C \end{align*}

由f(i)=−1+if(i)=-1+i，得初值条件 v(0,1)=1v(0,1)=1，C=0C=0，所以

f(z)=x2+xy−y2+i(−12x2+xy+y2)f(z)=x^2+xy-y^2+i(-\frac{1}{2}x^2+xy+y^2)

eg：典型的根据定义判定解析#

e.g.分析 f(z)=z‾⋅z2f(z)=\overline{z}\cdot z^2 这个函数在z=0处的解析性

f(z)=(x2+y2)(x+iy)在z=0处的可导性lim⁡z→0(x2+y2)(x+iy)−0(x+iy)−0=0在z≠0处，z的邻域里的可导性lim⁡Δz→0[(x+Δx)2+(y+Δy)2](x+Δx+i(y+Δy))−(x2+y2)(x+iy)Δz=lim⁡Δx→0,Δy→0(Δx2+Δy2+2xΔx+2yΔy)(x+iy)Δx+iΔy+x2+Δx2+2xΔx+y2+Δy2+2yΔy=lim⁡Δx→0,Δy→0(Δx2+Δy2)(x+iy)Δx+iΔy+2(xΔx+yΔy)(x+iy)Δx+iΔy+x2+y2第一坨高阶无穷小在上为0=lim⁡Δx→0,Δy→02(xΔx+yΔy)(x+iy)Δx+iΔy+x2+y2=lim⁡Δx→0,Δy→02(2xyΔxΔy+x2Δx2+y2Δy2−i(ΔxΔy(x2−y2)+xy(Δy2−Δx2)))Δx2−Δy2+x2+y2肉眼可见和xy有关\begin{align*} &f(z)=(x^2+y^2)(x+iy)\\ &\text{在$z=0$处的可导性}\\[5bp] &\lim_{z\to 0}{\frac{(x^2+y^2)(x+iy)-0}{(x+iy)-0}}=0\\[5bp] &\text{在$z\neq 0$处，z的邻域里的可导性}\\[5bp] &\lim_{\Delta z\to 0}{\frac{[(x+\Delta x)^2+(y+\Delta y)^2](x+\Delta x+i(y+\Delta y))-(x^2+y^2)(x+iy)}{\Delta z}}\\[5bp] &=\lim_{\Delta x\to 0,\Delta y\to 0}{\frac{(\Delta x^2+\Delta y^2+2x\Delta x+2y\Delta y)(x+iy)}{\Delta x+i\Delta y}+x^2+\Delta x^2 +2x\Delta x+y^2+\Delta y^2 +2y\Delta y}\\[5bp] &=\lim_{\Delta x\to 0,\Delta y\to 0}{\frac{(\Delta x^2+\Delta y^2)(x+iy)}{\Delta x+i\Delta y}+\frac{2(x\Delta x+y\Delta y)(x+iy)}{\Delta x+i\Delta y}+x^2+ y^2 }\text{第一坨高阶无穷小在上为0}\\[5bp] &=\lim_{\Delta x\to 0,\Delta y\to 0}{\frac{2(x\Delta x+y\Delta y)(x+iy)}{\Delta x+i\Delta y}+x^2+ y^2 }\\[5bp] &=\lim_{\Delta x\to 0,\Delta y\to 0}{\frac{2(2xy\Delta x\Delta y+x^2\Delta x^2+y^2\Delta y^2-i(\Delta x\Delta y(x^2-y^2)+xy(\Delta y^2-\Delta x^2)))}{\Delta x^2-\Delta y^2}+x^2+ y^2 }\text{肉眼可见和xy有关} \end{align*}

在z=0邻域内不可导，也就不解析

下列函数在何处可导，何处解析？#

(1) f(z)=x2−iyf(z)=x^2-iy

TIP

考虑用偏导+柯西黎曼条件判定

∂u∂x=2x,∂u∂y=0∂v∂x=0,∂v∂y=−1,显然一阶偏导连续当满足柯西黎曼条件时：x=−12,∀y\begin{align*} &\frac{\partial u}{\partial x}=2x,\frac{\partial u}{\partial y}=0\\[10bp] &\frac{\partial v}{\partial x}=0,\frac{\partial v}{\partial y}=-1,\text{显然一阶偏导连续}\\ &\text{当满足柯西黎曼条件时：}x=-\frac{1}{2},\forall y\\ \end{align*}

(2) f(z)=xy2+ix2yf(z)=xy^2+i x^2y

∂u∂x=y2,∂u∂y=2xy∂v∂x=2xy,∂v∂y=x2显然一阶偏导连续当满足柯西黎曼条件时：y=x=0\begin{align*} &\frac{\partial u}{\partial x}=y^2,\frac{\partial u}{\partial y}=2xy\\[10bp] &\frac{\partial v}{\partial x}=2xy,\frac{\partial v}{\partial y}=x^2\text{显然一阶偏导连续}\\ &\text{当满足柯西黎曼条件时：}y=x=0\\ \end{align*}

(3) f(z)=x+yx2+y2+ix−yx2+y2f(z)=\dfrac{x+y}{x^2+y^2}+i\dfrac{x-y}{x^2+y^2}

∂u∂x=y2−x2−2xy(x2+y2)2,∂u∂y=x2−y2−2xy(x2+y2)2∂v∂x=y2−x2+2xy(x2+y2)2,∂v∂y=y2−x2−2xy(x2+y2)2显然一阶偏导连续当满足柯西黎曼条件时：x≠0,y≠0\begin{align*} &\frac{\partial u}{\partial x}=\frac{y^2-x^2-2xy}{(x^2+y^2)^2},\frac{\partial u}{\partial y}=\frac{x^2-y^2-2xy}{(x^2+y^2)^2}\\[10bp] &\frac{\partial v}{\partial x}=\frac{y^2-x^2+2xy}{(x^2+y^2)^2},\frac{\partial v}{\partial y}=\frac{y^2-x^2-2xy}{(x^2+y^2)^2}\text{显然一阶偏导连续}\\ &\text{当满足柯西黎曼条件时：}x\neq 0,y\neq 0\\ \end{align*}

所以在复平面上除原点外处处解析

(4) f(z)=Im⁡z=yf(z)=\operatorname{Im} z=y

∂u∂x=0,∂u∂y=1∂v∂x=0,∂v∂y=0显然一阶偏导连续当满足柯西黎曼条件时：无解\begin{align*} &\frac{\partial u}{\partial x}=0,\frac{\partial u}{\partial y}=1\\[10bp] &\frac{\partial v}{\partial x}=0,\frac{\partial v}{\partial y}=0\text{显然一阶偏导连续}\\ &\text{当满足柯西黎曼条件时：无解}\\ \end{align*}

所以在复平面上处处不可导

复变函数：导数可微解析的例题
https://biscuit0613.github.io/posts/complexfunction/cmplxfunc_diffexercises/
作者
Biscuit
发布于
2025-09-08
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