复变函数(一)

发布于 2021-10-25  58 次阅读



\overline{z_1\pm z_2} = \overline{z_1}\pm \overline{z_2}
z\overline{z}=(Rez)^2+(Imz)^2=|z|^2


欲证一个诸如\frac{z}{1+z^2}为实数,等价于证明

\frac{z}{1+z^2} = \overline{\frac{z}{1+z^2}}

有欧拉公式

e^{i\varphi} = cos\varphi + isin\varphi


rz的模,\varphiz的辐角,记作

r=|z|,\varphi=Argz.

对于一个复数,其辐角主值记作argz,argz\in(-\pi,\pi],当z=0时,辐角是无意义的。
两个三角形式的复数z_1=r_1e^{i\varphi_1}z_2=r_2e^{i\varphi_2}相等的充要条件为

r_1=r_2,\varphi_1=\varphi_2+2k\pi

两个复数共轭的条件可以表示为

|\overline{z}| = |z|,arg\overline{z} = -argz,argz\neq \pi


一个复数的乘方表示为

z^n = r^ne^{in\varphi} = r^n(cosn\varphi+isinn\varphi)

一个复数的开方表示为

\sqrt[n]{z} = \sqrt[n]{r}(cos\frac{\varphi+2k\pi}{n}+isin\frac{\varphi+2k\pi}{n}). k = 0,1,...,n-1


z_0 = x_0 + iy_0,z_n=x_n+iy_n,n=1,2,...,则

{\lim_{n \to +\infty}}z_n=z_0

\lim_{n \to+\infty} x_n = x_0,\lim_{n \to +\infty} y_n = y_0

如果z_0 \neq 0,则

\lim_{n \to +\infty} z_n = z_0

\lim_{n \to +\infty} |z_n| = |z_0|, \lim_{n \to +\infty} Argz_n = Argz_0


如果等式

\lim_{z \to z_0}f(z) = f(z_0)

\lim_{(x,y) \to (x_0,y_0)}u(x,y) = u(x_0,y_0),\lim_{(x,y) \to (x_0,y_0)}v(x,y) = v(x_0,y_0)

成立,则f(z)在点z_0连续


函数f(z) = u(x,y)+iv(x,y)在点z=x+iy可微的充要条件是:u(x,y),v(x,y)在点(x,y)可微,且满足C-R方程

\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y},\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}

足C-R方程时,有类似以下方程的四个式子

f^\prime(z)=\frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}


「雪霁融雾月,冰消凝夜雨」