\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
r为z的模,\varphi为z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}及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
设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}
Comments 1 条评论
博主 lzusa
Latex记录/测试,没有任何现实意义