第一章 $R^3$空间的向量分析

一、证明以下公式：

$\because r=x{e}_x+y{e}_y+z{e}_z$

$\because ▽=e_x\frac{\partial }{\partial x}+e_y\frac{\partial }{\partial y}+e_z\frac{\partial }{\partial z}$

01. $▽\cdot r=3$

$\therefore ▽\cdot r=\frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}+\frac{\partial z}{\partial z}=3$

02. $▽\times r=0$

$\because ▽\times r=\left|\begin{array}{ccc}{e}_x& e_y& e_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ x& y& z\end{array}\right|=(\frac{\partial z}{\partial y}-\frac{\partial y}{\partial z})e_x+(\frac{\partial x}{\partial z}-\frac{\partial z}{\partial x})e_y+(\frac{\partial y}{\partial x}-\frac{\partial x}{\partial y})e_z=0$

03. $▽(\phi +\psi )=▽\phi +▽\psi$

$\begin{array}{rl}\because ▽(\phi +\psi )& =\frac{\partial (\phi +\psi )}{\partial x}{e}_x+\frac{\partial (\phi +\psi )}{\partial y}{e}_y+\frac{\partial (\phi +\psi )}{\partial z}{e}_z\\ & =(\frac{\partial \phi }{\partial x}{e}_x+\frac{\partial \phi }{\partial y}{e}_y+\frac{\partial \phi }{\partial z}{e}_z)+(\frac{\partial \psi }{\partial x}{e}_x+\frac{\partial \psi }{\partial y}{e}_y+\frac{\partial \psi }{\partial z}{e}_z)\\ & =(\frac{\partial }{\partial x}{e}_x+\frac{\partial }{\partial y}{e}_y+\frac{\partial }{\partial z}{e}_z)\phi +(\frac{\partial }{\partial x}{e}_x+\frac{\partial }{\partial y}{e}_y+\frac{\partial }{\partial z}{e}_z)\psi \\ & =▽\phi +▽\psi \end{array}$

04. $▽(\phi \psi )=\phi ▽\psi +\psi ▽\phi$

$\begin{array}{rl}\because ▽(\phi \psi )& =\frac{\partial (\phi \psi )}{\partial x}{e}_x+\frac{\partial (\phi \psi )}{\partial y}{e}_y+\frac{\partial (\phi \psi )}{\partial z}{e}_z\\ & =(\phi \frac{\partial \psi }{\partial x}+\psi \frac{\partial \phi }{\partial x})e_x+(\phi \frac{\partial \psi }{\partial x}+\psi \frac{\partial \phi }{\partial x})e_y+(\phi \frac{\partial \psi }{\partial x}+\psi \frac{\partial \phi }{\partial x})e_z\\ & =\phi (\frac{\partial }{\partial x}{e}_x+\frac{\partial }{\partial y}{e}_y+\frac{\partial }{\partial z}{e}_z)\psi +\psi (\frac{\partial }{\partial x}{e}_x+\frac{\partial }{\partial y}{e}_y+\frac{\partial }{\partial z}{e}_z)\phi \\ & =\phi ▽\psi +\psi ▽\phi \end{array}$

05. $▽\cdot (A+B)=▽\cdot A+▽\cdot B$

$\because ▽\cdot (A+B)=(\frac{\partial }{\partial x}{e}_x+\frac{\partial }{\partial y}{e}_y+\frac{\partial }{\partial z}{e}_z)\cdot [(A_x+B_x)e_x+(A_y+B_y)e_y+(A_z+B_z)e_z]$

$\begin{array}{rl}\therefore ▽\cdot (A+B)& =\frac{\partial (A_x+B_x)}{\partial x}+\frac{\partial (A_y+B_y)}{\partial y}+\frac{\partial (A_z+B_z)}{\partial z}\\ & =(\frac{\partial A_x}{\partial x}+\frac{\partial B_x}{\partial x})+(\frac{\partial A_y}{\partial y}+\frac{\partial B_y}{\partial y})+(\frac{\partial A_z}{\partial z}+\frac{\partial B_z}{\partial z})\\ & =(\frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z})+(\frac{\partial B_x}{\partial x}+\frac{\partial B_y}{\partial y}+\frac{\partial B_z}{\partial z})\\ & =▽\cdot A+▽\cdot B\end{array}$

06. $▽\times (A+B)=▽\times A+▽\times B$

$\begin{array}{rl}\because ▽\times (A+B)& =\left|\begin{array}{ccc}{e}_x& e_y& e_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ (A_x+B_x)& (A_y+B_y)& (A_z+B_z)\end{array}\right|\\ & =\left|\begin{array}{ccc}{e}_x& e_y& e_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ A_x& A_y& A_z\end{array}\right|+\left|\begin{array}{ccc}{e}_x& e_y& e_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ B_x& B_y& B_z\end{array}\right|\\ & =▽\times A+▽\times B\end{array}$

07. $▽\cdot (\phi A)=A\cdot ▽\phi +\phi ▽\cdot A$

$\begin{array}{rl}\because ▽\cdot (\phi A)& =(\frac{\partial }{\partial x}{e}_x+\frac{\partial }{\partial y}{e}_y+\frac{\partial }{\partial z}{e}_z)\cdot [\phi (A_x{e}_x+A_y{e}_y+A_z{e}_z)]\\ & =\frac{\partial (u{A}_x)}{\partial x}+\frac{\partial (u{A}_y)}{\partial y}+\frac{\partial (u{A}_z)}{\partial z}\\ & =(\phi \frac{\partial A_x}{\partial x}+A_x\frac{\partial \phi }{\partial x})+(\phi \frac{\partial A_y}{\partial y}+A_y\frac{\partial \phi }{\partial y})+(\phi \frac{\partial A_z}{\partial z}+A_z\frac{\partial \phi }{\partial z})\\ & =(A_x\frac{\partial \phi }{\partial x}+A_y\frac{\partial \phi }{\partial x}+A_z\frac{\partial \phi }{\partial x})+\phi (\frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z})\\ & =A\cdot ▽\phi +\phi ▽\cdot A\end{array}$

08. $▽\times (\phi A)=A\times ▽\phi +\phi ▽\times A$

$\begin{array}{rl}\because ▽\times (\phi A)& =\left|\begin{array}{ccc}{e}_x& e_y& e_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ \phi A_x& \phi A_y& \phi A_z\end{array}\right|\\ & =(\frac{\partial (\phi A_z)}{\partial y}-\frac{\partial (\phi A_y)}{\partial z})e_x\\ & +(\frac{\partial (\phi A_x)}{\partial z}-\frac{\partial (\phi A_z)}{\partial x})e_y\\ & +(\frac{\partial (\phi A_y)}{\partial x}-\frac{\partial (\phi A_x)}{\partial y})e_z\end{array}$

$\begin{array}{rl}\therefore ▽\times (\phi A)& =[(\phi \frac{\partial A_z}{\partial y}+A_z\frac{\partial \phi }{\partial y})-(\phi \frac{\partial A_y}{\partial z}+A_y\frac{\partial \phi }{\partial z})]e_x\\ & +[(\phi \frac{\partial A_x}{\partial z}+A_x\frac{\partial \phi }{\partial z})-(\phi \frac{\partial A_z}{\partial x}+A_z\frac{\partial \phi }{\partial x})]e_y\\ & +[(\phi \frac{\partial A_y}{\partial x}+A_y\frac{\partial \phi }{\partial x})-(\phi \frac{\partial A_x}{\partial y}+A_x\frac{\partial \phi }{\partial y})]e_z\\ & =[(A_z\frac{\partial \phi }{\partial y}-A_y\frac{\partial \phi }{\partial z})e_x+(A_x\frac{\partial \phi }{\partial z}-A_z\frac{\partial \phi }{\partial x})e_y+(A_y\frac{\partial \phi }{\partial x}-A_x\frac{\partial \phi }{\partial y})e_z]\\ & +\phi [(\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z})e_x+(\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x})e_y+(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y})e_z]\\ & =A\times ▽\phi +\phi ▽\times A\end{array}$

09. $▽\cdot (A\times B)=B\cdot (▽\times A)+A\cdot (▽\times B)$

解法一：

$\begin{array}{rl}\because ▽\cdot (A\times B)& =▽\cdot \left|\begin{array}{ccc}{e}_x& e_y& e_z\\ A_x& A_y& A_z\\ B_x& B_y& B_z\end{array}\right|\\ & =\frac{\partial (A_z{B}_y-A_y{B}_z)}{\partial x}+\frac{\partial (A_z{B}_x-A_x{B}_z)}{\partial y}+\frac{\partial (A_x{B}_y-A_y{B}_x)}{\partial z}\end{array}$

$\begin{array}{rl}\therefore ▽\cdot (A\times B)& =[(A_z\frac{\partial B_y}{\partial x}+B_y\frac{\partial A_z}{\partial x})-(A_y\frac{\partial B_z}{\partial x}+B_z\frac{\partial A_y}{\partial x})]\\ & +[(A_z\frac{\partial B_x}{\partial y}+B_x\frac{\partial A_z}{\partial y})-(A_x\frac{\partial B_z}{\partial y}+B_z\frac{\partial A_x}{\partial y})]\\ & +[(A_x\frac{\partial B_y}{\partial z}+B_y\frac{\partial A_x}{\partial z})-(A_y\frac{\partial B_x}{\partial z}+B_x\frac{\partial A_y}{\partial z})]\end{array}$

$\begin{array}{rl}\therefore ▽\cdot (A\times B)& =[B_x(\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z})+B_y(\frac{\partial A_z}{\partial x}-\frac{\partial A_x}{\partial z})+B_z(\frac{\partial A_x}{\partial y}-\frac{\partial A_y}{\partial x})]\\ & -[A_x(\frac{\partial B_z}{\partial y}-\frac{\partial B_y}{\partial z})+A_y(\frac{\partial B_z}{\partial x}-\frac{\partial B_x}{\partial z})+A_z(\frac{\partial B_x}{\partial y}-\frac{\partial B_y}{\partial x})]\\ & =B\cdot (▽\times A)+A\cdot (▽\times B)\end{array}$

解法二：

$\begin{array}{rl}\because ▽\cdot (A\times B)& ={\partial }_i(A\times B{)}_i\\ & ={\partial }_i({\epsilon }_{ijk}{A}_j{B}_k)\\ & ={\epsilon }_{ijk}[({\partial }_i{A}_j)B_k+A_j({\partial }_i{B}_k)]\\ & =B_k{\epsilon }_{kij}{\partial }_i{A}_j-A_j{\epsilon }_{jik}{\partial }_i{B}_k\end{array}$ $\therefore ▽\cdot (A\times B)=B\cdot (▽\times A)+A\cdot (▽\times B)$

10. $▽\times (A\times B)=(B\cdot ▽)A-B(▽\cdot A)-(A\cdot ▽)B+A(▽\cdot B)$

解法一：

根据哈密顿算子的微分性质运算(常矢和变矢)：

$\because ▽\times (A\times B)=[▽\times (A\times B)]forB+[▽\times (A\times B)]forA$

根据二重矢量积公式：

$\therefore ▽\times (A\times B)=[A(▽\cdot B)-(A\cdot ▽)B]forB+[(B\cdot ▽)A-B(▽\cdot A)]forA$

$\therefore ▽\times (A\times B)=(B\cdot ▽)A-B(▽\cdot A)-(A\cdot ▽)B+A(▽\cdot B)$

解法二：

$\begin{array}{rl}\because ▽\times (A\times B)& ={\epsilon }_{ijk}{e}_i{\partial }_j(A\times B{)}_k\\ & ={\epsilon }_{ijk}{e}_i{\partial }_j{\epsilon }_{klm}{A}_l{B}_m\\ & =e_i{\epsilon }_{kij}{\epsilon }_{klm}{\partial }_j{A}_l{B}_m\\ & =e_i({\delta }_{il}{\delta }_{jm}-{\delta }_{im}{\delta }_{jl}){\partial }_j{A}_l{B}_m\\ & =e_i{A}_l{\partial }_j{B}_m-e_i{B}_m{\partial }_j{A}_l\\ & =A(▽\cdot B)+(B\cdot ▽)A-B(▽\cdot A)-(A\cdot ▽)B\end{array}$

$\therefore ▽\times (A\times B)=(B\cdot ▽)A-B(▽\cdot A)-(A\cdot ▽)B+A(▽\cdot B)$

11. $▽(A\cdot B)=(B\cdot ▽)A+(A\cdot ▽)B+B\times (▽\times A)+A\times (▽\times B)$

$\begin{array}{rl}\because ▽(A\cdot B)& =[▽(A\cdot B)]forB+[▽(A\cdot B)]forA\end{array}$

$\therefore ▽(A\cdot B)=[(A\cdot ▽)B+A\times (▽\times B)]forB+[(B\cdot ▽)A+B\times (▽\times A)]forA$

$\therefore ▽(A\cdot B)=(B\cdot ▽)A+(A\cdot ▽)B+B\times (▽\times A)+A\times (▽\times B)$

12. $▽\times ▽\phi =0$

$\begin{array}{rl}\because ▽\times ▽\phi & =\left|\begin{array}{ccc}{e}_x& e_y& e_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ \frac{\partial u}{\partial x}& \frac{\partial u}{\partial y}& \frac{\partial u}{\partial z}\end{array}\right|\\ & =(\frac{{\partial }^{2}u}{\partial y\partial z}-\frac{{\partial }^{2}u}{\partial y\partial z})e_x+(\frac{{\partial }^{2}u}{\partial x\partial z}-\frac{{\partial }^{2}u}{\partial x\partial z})e_y+(\frac{{\partial }^{2}u}{\partial x\partial y}-\frac{{\partial }^{2}u}{\partial x\partial y})e_z\\ & =0\end{array}$

13. $▽\cdot ▽\times A=0$

$\begin{array}{rl}\because ▽\cdot ▽\times A& =▽\cdot \left|\begin{array}{ccc}{e}_x& e_y& e_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ A_x& A_y& A_z\end{array}\right|\\ & =▽\cdot [(\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z})e_x-(\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x})e_y+(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y})e_z]\end{array}$

$\begin{array}{r}\therefore ▽\cdot ▽\times A=\frac{\partial A_z}{\partial x\partial y}-\frac{\partial A_y}{\partial x\partial y}+\frac{\partial A_x}{\partial x\partial z}-\frac{\partial A_z}{\partial y\partial x}+\frac{\partial A_y}{\partial x\partial z}-\frac{\partial A_x}{\partial y\partial z}=0\end{array}$

同理，可用哑标计算

14. $▽\times ▽\times A=▽(▽\cdot A)-{▽}^{2}A$

根据二重矢量积公式：

$\begin{array}{rl}\therefore ▽\times ▽\times A& =▽(▽\cdot A)-(▽\cdot ▽)A\\ & =▽(▽\cdot A)-{▽}^{2}A\end{array}$

同理，可用哑标计算

二、请计算${▽}^2\frac{1}{r}$，其中$r=|r|$

$\because r=|r|=\sqrt{x^2+y^2+z^2}$

$\begin{array}{rl}\therefore {▽}^2\frac{1}{r}& =\frac{{\partial }^2}{\partial x^2}\frac{1}{\sqrt{x^2+y^2+z^2}}+\frac{{\partial }^2}{\partial y^2}\frac{1}{\sqrt{x^2+y^2+z^2}}+\frac{{\partial }^2}{\partial z^2}\frac{1}{\sqrt{x^2+y^2+z^2}}\\ & =\frac{x+y+z}{(x^2+y^2+z^2{)}^{-3/2}}\end{array}$

三、由Ampere(安培)定律 ${\int }_{S}J\cdot d\sigma ={\int }_{\partial S}H\cdot dl$，其中$J$为穿过空间曲面$S$的电流密度向量，$d\sigma$为S的有向面元，其法线正向由右手定则确定，$H$为磁场强度向量，$dl$为$S$的边界$\partial S$的线元，其方向由右手定则确定，它与$d\sigma$的定向自洽。请证明：$▽\times H=J$（提示：应用Stokes公式）

解：

根据Ampere(安培)定律：${\int }_{S}J\cdot d\sigma ={\int }_{\partial S}H\cdot dl$

两边取旋度得：

$▽\times ({\int }_{S}J\cdot d\sigma )=▽\times ({\int }_{\partial S}H\cdot dl)$

${\int }_S(▽\times J)\cdot d\sigma ={\int }_{\partial S}(▽\times H)\cdot dl$

根据Stokes公式：

${\int }_S▽\times J\cdot d\sigma ={\int }_{\partial S}J\cdot dl={\int }_{\partial S}▽\times H\cdot dl$

$\therefore ▽\times H=J$

四、有一物理体系，它由三个向量场$E，H，A$和一个实标量场$V$来描述，并且满足下面关系：

$$\begin{array}{rlrl}[1]& & ▽\cdot E& =-{\mu }^{2}V\\ [2]& & E& =\frac{\partial A}{\partial t}+▽V\\ [3]& & H& =▽\times A\\ [4]& & ▽\times H& =\frac{\partial E}{\partial t}-{\mu }^{2}A\end{array}$$

其中，$\mu$为正的常数，求证：

1、$▽\cdot A+\frac{\partial V}{\partial t}=0$

2、${▽}^{2}V-{\mu }^{2}V+\frac{{\partial }^{2}V}{\partial t^2}=0$

3、当我们考虑在区域$\Omega$中，$V$与时间无关，且在$\Omega$的边界$\partial \Omega$上有$V{|}_{\partial \Omega }=0$时，请证明：在$\Omega$中$V=0$

解：

对式[4]取散度，得：

$▽\cdot (▽\times H)=\frac{\partial (▽\cdot E)}{\partial t}-{\mu }^2(▽\cdot A)=0$

根据式[1]，得：

$▽\cdot (▽\times H)=-{\mu }^2\frac{\partial V}{\partial t}-{\mu }^2(▽\cdot A)=0$

所以得

$▽\cdot A+\frac{\partial V}{\partial t}=0$

对式[2]取散度，得：

$▽\cdot (E)=\frac{\partial (▽\cdot A)}{\partial t}-{▽}^{2}V$

根据式[1]，得：

$\frac{\partial (▽\cdot A)}{\partial t}-{▽}^{2}V=-{\mu }^{2}V$

所以得：

$-\frac{{\partial }^{2}V}{{\partial }^{2}t}-{▽}^{2}V=-{\mu }^{2}V$

${▽}^{2}V-{\mu }^{2}V+\frac{{\partial }^{2}V}{{\partial }^{2}t}=0$

当我们考虑在区域$\Omega$中，$V$与时间无关，且在$\Omega$的边界$\partial \Omega$上有$V{|}_{\partial \Omega }=0$时

$\because V与时间无关$

$\therefore \frac{\partial V}{\partial t}=0$

$\therefore ▽\cdot A=0$

即区域$\Omega$是无源场

$又\because V{|}_{\partial \Omega }=0$

$\therefore 在\Omega 中V=0$

第二章 $R^3$空间曲线坐标系中的向量分析

一、在柱坐标$r(\rho ,\phi ,z)$下，试计算其坐标基$e_{\rho },e_{\phi },e_z$(可称为空间中的标架场)的散度和旋度，即$▽\cdot e_{\rho },▽\cdot e_{\phi },▽\cdot e_z,▽\times e_{\rho },▽\times e_{\phi },▽\times e_z$

在柱坐标系中：

$\begin{array}{r}▽=e_{\rho }\frac{\partial }{\partial \rho }+e_{\phi }\frac{1}{\rho }\frac{\partial }{\partial \phi }+e_z\frac{\partial }{\partial z}\end{array}$

则有：

$\begin{array}{rlrlrl}▽\cdot e_{\rho }& =\frac{\partial }{\partial \rho }& ▽\cdot e_{\phi }& =\frac{1}{\rho }\frac{\partial }{\partial \phi }& ▽\cdot e_z& =\frac{\partial }{\partial z}\\ ▽\times e_{\rho }& =e_{\phi }\frac{\partial }{\partial z}-e_z\frac{1}{\rho }\frac{\partial }{\partial \phi }& ▽\times e_{\phi }& =e_{\rho }\frac{\partial }{\partial z}-e_z\frac{\partial }{\partial \rho }& ▽\times e_z& =e_{\rho }\frac{1}{\rho }\frac{\partial }{\partial \phi }-e_{\phi }\frac{\partial }{\partial \rho }\end{array}$

二、若有一个力场在直角坐标系下的表达式为

$$F=-\frac{y}{x^2+y^2}{e}_x+\frac{x}{x^2+y^2}{e}_x$$

1、请用柱坐标系把此力场表示出来； 2、请在柱坐标系下计算次力场的散度和旋度； 3、若有一物体在此力场中沿$e_{\rho }$方向运动一单位长度时，求此力场$F$所做的功； 4、若有一物体在此力场的作用下，沿$x-y$平面的以单位圆顺时针运动一圈，求此力场$F$所做的功。并分析此功与力场$F$的旋度的关系

解：

$\because \{\begin{array}{ccc}x& =& \rho cos\phi \\ y& =& \rho sin\phi \\ z& =& z\end{array}$

$\therefore \{\begin{array}{ccc}{e}_x& =& e_{\rho }cos\phi -e_{\phi }sin\phi \\ e_y& =& e_{\rho }sin\phi +e_{\phi }cos\phi \\ e_z& =& e_z\end{array}$

$\begin{array}{rl}\therefore F& =-\frac{\rho sin\phi }{{\rho }^2}(e_{\rho }cos\phi -e_{\phi }sin\phi )+\frac{\rho cos\phi }{{\rho }^2}(e_{\rho }sin\phi +e_{\phi }cos\phi )\\ & =\frac{1}{\rho }{e}_{\phi }\end{array}$

$\begin{array}{rl}\therefore ▽\cdot F& =0\end{array}$

$\begin{array}{rl}\therefore ▽\times F& =0\end{array}$

若有一物体在此力场中沿$e_{\rho }$方向运动一单位长度时,此力场$F$所做的功为0

若有一物体在此力场的作用下，沿$x-y$平面的以单位圆顺时针运动一圈，此力场$F$所做的功为$\frac{1}{\rho }$

此功与力场$F$的旋度相等

第四章 复变函数的概念

一、写出下列复数的实部、虚部、模和幅角

二、写出下列复变函数$W=f(z)=u(x,y)+iv(x,y)$的实部$u(x,y)$、虚部$v(x,y)$、模$|f(z)|$和幅角$argf(z)$

三、求下列方程的解，并把解在复平面上用图形表示出来，说明他们的几何意义

四、对于在实变量$x$中成立的关系式$sin(x_1+x_2)=sin{x}_{1}cos{x}_2+sins{x}_{2}cos{x}_1$请证明这一关系式在复变量$z$的情况下也成立$sin(z_1+z_2)=sin{z}_{1}cos{z}_2+sins{z}_{2}cos{z}_1$

解：

$\begin{array}{rl}sin(z_1+z_2)& =sin(x_1+i{y}_1+x_2+i{y}_2)\\ & =sin[(x_1+x_2)+i(y_1+y_2)]\\ & =sin(x_1+x_2)cos(i(y_1+y_2))+sin(i(y_1+y_2))cos(x_1+x_2)\\ & =(sin{x}_{1}cos{x}_2+sin{x}_{2}cos{x}_1)(cosi{y}_{1}cosi{y}_2-sini{y}_{1}sini{y}_2)\\ & +(sini{y}_{1}cosi{y}_2+sini{y}_{2}cosi{y}_1)(cos{x}_{1}cos{x}_2-sin{x}_{1}sin{x}_2)\\ & =sin{x}_{1}cos{x}_{2}cosi{y}_{1}cosi{y}_2+sin{x}_{2}cos{x}_{1}cosi{y}_{1}cosi{y}_2\\ & -sin{x}_{1}cos{x}_{2}sini{y}_{1}sini{y}_2-sin{x}_{2}cos{x}_{1}sini{y}_{1}sini{y}_2\\ & +cos{x}_{1}cos{x}_{2}sini{y}_{1}cosi{y}_2+cos{x}_{1}cos{x}_{2}sini{y}_{2}cosi{y}_1\\ & -sin{x}_{1}sin{x}_{2}sini{y}_{1}cosi{y}_2-sin{x}_{1}sin{x}_{2}sini{y}_{2}cosi{y}_1\\ & =sin{x}_{1}cos{x}_{2}cosi{y}_{1}cosi{y}_2+cos{x}_{1}cos{x}_{2}sini{y}_{1}cosi{y}_2\\ & -sin{x}_{1}sin{x}_{2}sini{y}_{2}cosi{y}_1-sin{x}_{2}cos{x}_{1}sini{y}_{1}sini{y}_2\\ & +sin{x}_{2}cos{x}_{1}cosi{y}_{1}cosi{y}_2+cos{x}_{1}cos{x}_{2}sini{y}_{2}cosi{y}_1\\ & -sin{x}_{1}sin{x}_{2}sini{y}_{1}cosi{y}_2-sin{x}_{2}cos{x}_{1}sini{y}_{1}sini{y}_2\\ & =sin(x_1+i{y}_1)cos(x_2+i{y}_2)+sin(x_2+i{y}_2)cos(x_1+i{y}_1)\\ & =sin{z}_{1}cos{z}_2+sins{z}_{2}cos{z}_1\end{array}$