首页 > 技术文章 > 复变函数

zonghanli 2020-03-13 19:33 原文

\(x+i y, x, y \in R\longrightarrow (x, y)​\)
\(x=\operatorname{Rez} \quad y=\operatorname{Imz}​\)
\(\operatorname{Re} z=\frac{1}{2}(z+\bar{z})​\)
\(\ln z=\frac{1}{2i}(z-\bar{z})​\)

\(\operatorname{Re}\left(z_{1}+z_{2}\right)=\operatorname{Re} z_{1}+\operatorname{Re} z_{2}\)
\(I_{m} \left(z_{1}+z_{2}\right)=I_{m} z_{1}+I_{m} z_{2}\)

\(\left.\left(x_{1}, y_{2}\right) \cdot\left(x_{2}, y_{2}\right)=(x_{1} x_{2}-y_{1} y_{2}, y_{1} x_{2}+x_{1} y_{2}\right)\)

Note:\(\leq\) cannot be applied in \(C\)

Polan form

\(\theta=\tan ^{-1} \frac{y}{x}​\):argument

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