复变函数
作者：互联网
复数
\(z=x+iy\)所对应\(\overline{Oz}\)满足\(\tan\theta=\frac{y}{x}\)
\(\theta=Argz,-\pi<argz\leq\pi\)为主辐角
对于非零复数\(z=r(\cos\theta + i\sin\theta)\)
Euler公式
\(e^{i\theta}:=\cos\theta+i\sin\theta\)
\(z=re^{i\theta},\theta\in Arg z.\)
Example
把复数\(1-\cos \varphi+i \sin \varphi(0<\varphi \leq \pi)\)化为指数形式（\(e^{i\theta}:=\cos\theta+i\sin\theta\)）！！\(i\sin\theta\)
\(\begin{aligned} 1-\cos \varphi+i \sin \varphi &=2 \sin ^{2} \frac{\varphi}{2}+2 i \sin \frac{\varphi}{2} \cos \frac{\varphi}{2}=2 \sin \frac{\varphi}{2}\left(\sin \frac{\varphi}{2}+i \cos \frac{\varphi}{2}\right) \\ &=2 \sin \frac{\varphi}{2}\left(\cos \left(\frac{\pi}{2}-\frac{\varphi}{2}\right)++i \sin \left(\frac{\pi}{2}-\frac{\varphi}{2}\right)\right)=2 \sin \frac{\varphi}{2} e^{i\left(\frac{\pi}{2}-\frac{\varphi}{2}\right)} \end{aligned}\)
\(\operatorname{Arg}\left(z_{1} z_{2}\right)=\operatorname{Arg} z_{1}+\operatorname{Arg} z_{2}, \operatorname{Arg} \frac{z_{1}}{z_{2}}=\operatorname{Arg} z_{1}-\operatorname{Arg} z_{2}\)
De Moivre公式：
\((\cos \theta+i \sin \theta)^{n}=\cos n \theta+i \sin n \theta\)
\(z^{n}=r^{n} e^{i n \theta}=r^{n}(\cos n \theta+i \sin n \theta)\)
Example
\(\cos3\theta,\sin3\theta\)利用\(\cos\theta,\sin\theta\)来表示
\(\cos 3 \theta+i \sin 3 \theta=(\cos \theta+i \sin \theta)^{3}=\cos ^{3} \theta+3 i \cos ^{2} \theta \sin \theta-3 \cos \theta \sin ^{2} \theta-i \sin ^{3} \theta\)
Def：
\(\sqrt[n]{z}:=\left\{w \in \mathbb{C}, w^{n}=z\right\}\)
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\(\quad \quad \sqrt[n]{z}=\left\{w_{k}, k=0, \ldots, n-1\right\}\)
其中\(w_{k}=\sqrt[n]{r} e^{i \frac{\theta+2 k \pi}{n}}\)
\(|z|^{2}=z \bar{z},\) Re \(z=\frac{z+\bar{z}}{2}, \operatorname{Im} z=\frac{z-\bar{z}}{2 i}\)
复平面上的点集
点\(z_0\)的\(\rho\)邻域：\(N_{\rho}\left(z_{0}\right)=\left\{z \in \mathbb{C},\left|z-z_{0}\right| \leq \rho\right\}\)
点\(z_0\)的去心\(\rho\)邻域\(N_{\rho}\left(z_{0}\right)-\left\{z_{0}\right\}\)
考虑点集\(E\),若平面上一点\(z_0\)的任何邻域
标签：cos,right,frac,函数,varphi,复变,theta,sin 来源： https://www.cnblogs.com/zonghanli/p/12361704.html