复变函数 第1次作业

Chasse_neige

5.若 $|z|=1$，请证明对任意复数 $a$ 和 $b$，均有

$$\left|\frac{az+b}{\bar{b}z+\bar{a}}\right|=1$$

证明：由于 $|z|=1$，所以 $z\bar{z}=1$

$$\left|\frac{az+b}{\bar{b}z+\bar{a}}\right|=\frac{|az+b|}{|\bar{b}z+\bar{a}|}=\frac{|az+b|}{\left|\overline{\stackrel{―}{b}z+\stackrel{―}{a}}\right|}=\frac{|az+b|}{|b\bar{z}+a|}=\frac{|az+b|}{|z\bar{z}(b\bar{z}+a)|}=\frac{|az+b|}{|\bar{z}(b+az)|}=\frac{|az+b|}{|\bar{z}||az+b|}=1$$

6.请利用单倍角的三角函数表示 $\cos (n\theta )$ 和 $\sin (n\theta )$，并给出 $\cos (3\theta )$ 和 $\sin (3\theta )$ 的具体表达式。

$$\begin{array}{c}\cos (n\theta )=\frac{e^{in\theta }+e^{-in\theta }}{2}=\frac{1}{2}((\cos \theta +i\sin \theta {)}^n+(\cos \theta -i\sin \theta {)}^n)\\ =\frac{1}{2}(\sum _{k=0}^n{i}^k{C}_n^k{\cos }^{n-k}\theta {\sin }^k\theta +\sum _{k=0}^n(-i{)}^k{C}_n^k{\cos }^{n-k}\theta {\sin }^k\theta )=\frac{1}{2}(2\sum _{k=0}^{[\frac{n}{2}]}{i}^{2k}{C}_n^{2k}{\cos }^{n-2k}{\sin }^{2k}\theta )\\ =\sum _{k=0}^{[\frac{n}{2}]}{i}^{2k}{C}_n^{2k}{\cos }^{n-2k}\theta {\sin }^{2k}\theta \end{array}$$$$\begin{array}{c}\sin (n\theta )=\frac{e^{in\theta }-e^{-in\theta }}{2i}=\frac{1}{2i}((\cos \theta +i\sin \theta {)}^n-(\cos \theta -i\sin \theta {)}^n)\\ =\frac{1}{2i}(\sum _{k=0}^n{i}^k{C}_n^k{\cos }^{n-k}\theta {\sin }^k\theta -\sum _{n=0}^k(-i{)}^k{\cos }^{n-k}\theta {\sin }^k\theta )=\frac{1}{2i}(2i\sum _{n=0}^{[\frac{n}{2}]}{i}^{2k}{C}_n^{2k+1}{\cos }^{n-2k-1}\theta {\sin }^{2k+1}\theta )\\ =\sum _{n=0}^{[\frac{n}{2}]}{i}^{2k}{C}_n^{2k+1}{\cos }^{n-2k-1}\theta {\sin }^{2k+1}\theta \end{array}$$

特别地，对于 $n=3$

$$\begin{array}{c}\cos (3\theta )=\sum _{k=0}^1{i}^{2k}{C}_3^{2k}{\cos }^{3-2k}\theta {\sin }^{2k}\theta \\ ={\cos }^3\theta -3\cos \theta {\sin }^2\theta \end{array}$$$$\begin{array}{c}\sin (3\theta )=\sum _{n=0}^1{i}^{2k}{C}_3^{2k+1}{\cos }^{3-2k-1}\theta {\sin }^{2k+1}\theta \\ =3{\cos }^2\theta \sin \theta -{\sin }^3\theta \end{array}$$

7.请利用多倍角三角函数的线性关系表达 ${\cos }^n\theta$ 和 ${\sin }^n\theta$，并给出 ${\cos }^3\theta$ 和 ${\sin }^3\theta$ 的具体表达式。

$${\cos }^n\theta ={\left(\frac{e^{i\theta }+e^{-i\theta }}{2}\right)}^n=\sum _{k=0}^{[\frac{n+1}{2}]}{C}_n^k\frac{e^{i(n-2k)\theta }+e^{-i(n-2k)\theta }}{2^n}=\frac{1}{2^{n-1}}\sum _{k=0}^{[\frac{n+1}{2}]}{C}_n^k\cos ((n-2k)\theta )$$$$\begin{gathered} {\sin }^n\theta ={\left(\frac{e^{i\theta }-e^{-i\theta }}{2i}\right)}^n=\sum _{k=0}^{[\frac{n+1}{2}]}{C}_n^k\frac{(-1{)}^k{e}^{i(n-2k)\theta }+(-1{)}^{n-k}{e}^{-i(n-2k)\theta }}{(2i{)}^n} \\ =\{\begin{array}{ll}\frac{1}{(2i{)}^{n-1}}\sum _{k=0}^{[\frac{n}{2}]}(-1{)}^k{C}_n^k\sin ((n-2k)\theta )& (\text{n为奇数})\\ \frac{1}{(2i{)}^n}\sum _{k=0}^{[\frac{n}{2}]}2(-1{)}^k{C}_n^k\cos ((n-2k)\theta )& (\text{n为偶数})\end{array} \end{gathered}$$

特别地，对于 $n=3$

$${\cos }^3\theta =\frac{1}{2^2}\sum _{k=0}^1{C}_3^k\cos ((3-2k)\theta )=\frac{1}{4}(\cos (3\theta )+3\cos \theta )$$$${\sin }^3\theta =\frac{1}{(2i{)}^2}\sum _{k=0}^1(-1{)}^k{C}_3^k\sin ((3-2k)\theta )=-\frac{1}{4}(\sin (3\theta )-3\sin \theta )$$

11 证明下列 Cauchy-Riemann 条件，其中 $w=u+iv=\rho e^{i\phi }$，$z=x+iy=r{e}^{i\theta }$：

(a)

$$\frac{\partial u}{\partial r}=\frac{1}{r}\frac{\partial v}{\partial \theta },\frac{\partial v}{\partial r}=-\frac{1}{r}\frac{\partial u}{\partial \theta }$$

考虑沿着径向的逼近方式：

$$\underset{\mathrm{\Delta }z\to 0}{lim}\frac{\mathrm{\Delta }w}{\mathrm{\Delta }z}=\underset{\mathrm{\Delta }r\to 0}{lim}\frac{\mathrm{\Delta }u+i\mathrm{\Delta }v}{e^{i\theta }\mathrm{\Delta }r}=\frac{\partial u}{\partial r}\cos \theta +\frac{\partial v}{\partial r}\sin \theta +i(\frac{\partial v}{\partial r}\cos \theta -\frac{\partial v}{\partial r}\sin \theta )$$

再考虑沿着角向的逼近方式：

$$\underset{\mathrm{\Delta }z\to 0}{lim}\frac{\mathrm{\Delta }w}{\mathrm{\Delta }z}=\underset{\mathrm{\Delta }\theta \to 0}{lim}\frac{\mathrm{\Delta }u+i\mathrm{\Delta }v}{r(i\cos \theta -\sin \theta )\mathrm{\Delta }\theta }=\frac{1}{r}((\frac{\partial v}{\partial \theta }\cos \theta -\frac{\partial u}{\partial \theta }\sin \theta )-i(\frac{\partial u}{\partial \theta }\cos \theta +\frac{\partial v}{\partial \theta }\sin \theta ))$$

对比上述二式，得到

$$\begin{array}{c}\frac{\partial u}{\partial r}\cos \theta +\frac{\partial v}{\partial r}\sin \theta =\frac{1}{r}(\frac{\partial v}{\partial \theta }\cos \theta -\frac{\partial u}{\partial \theta }\sin \theta )\\ \frac{\partial v}{\partial r}\cos \theta -\frac{\partial v}{\partial r}\sin \theta =-\frac{1}{r}(\frac{\partial u}{\partial \theta }\cos \theta +\frac{\partial v}{\partial \theta }\sin \theta )\end{array}$$

即

$$\frac{\partial u}{\partial r}=\frac{1}{r}\frac{\partial v}{\partial \theta },\frac{\partial v}{\partial r}=-\frac{1}{r}\frac{\partial u}{\partial \theta }$$

14.已知函数 $f(z)=u+iv$ 在区域 $G$ 中解析，则对于常实数 $a$ 和 $b$，$au+bv$ 在区域中也解析的条件为什么？并据此证明若在 $G$ 中 $\text{Im}f(z)=0$，则 $f(z)$ 为常函数。

也解析的条件：由于 $f(z)$ 解析，所以 $f(z)$ 的各个偏导数在区域 $G$ 中连续，即 $au+bv$ 的各个偏导数在区域 $G$ 中连续，所以 $au+bv$ 在区域中也解析的充要条件即为 Cauchy-Riemann 条件：

$$a\frac{\partial u}{\partial x}+b\frac{\partial v}{\partial x}=\frac{\partial u}{\partial y}+b\frac{\partial v}{\partial y}=0$$

若在 $G$ 中 $\text{Im}f(z)=0$，则 $v=0$，即有

$$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial y}=0$$

由于由于 $f(z)$ 解析，所以在区域 $G$ 中$u\in C^1$，所以在区域 $G$ 中 $\mathrm{d}u=0$

所以对于任意路径的积分有

$$u(\mathbf{x})-u({\mathbf{x}}_0)={\int }_{{\mathbf{x}}_0}^{\mathbf{x}}\mathrm{d}u=0$$

即$f(z)$ 为常函数。