12.8 产品和引引理论

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  产品和引引理论

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  Introduction
  ::导言

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  are found in real-world calculations involving quantum mechanics, signal analysis, fluid dynamics, control theory, and many other fields.
  ::存在于现实世界的计算中,涉及量子力学、信号分析、流体动力学、控制理论和许多其他领域。

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  In electrical engineering, complex numbers are used for calculations involving impedance   (the resistance to electric flow in a circuit). Electrical engineers are familiar with the formula
  ::在电气工程中,复杂数字用于计算阻力(对电路电流的阻力)。

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  $$\begin{align*} V = V_0 e^{j \omega t} = V_0 \left (\cos \omega t + j \sin\omega t \right ).\end{align*}$$
  ::V=V0ejt=V0(cost+jsint)。

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  By comparing it to the similar expression below, which is explored in this lesson, can you identify the variable $\begin{align*}j\end{align*}$ ?

  $$\begin{align*}r_2 (cos \theta _2 + i sin \theta _2)\end{align*}$$
  ::将它与本课中探讨的以下类似表达式进行比较,可以识别变量j?r2(cos2+isin2)

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  Product Theorem
  ::产品定理

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  Since complex numbers can be transformed to polar form , the operation of multiplication is also possible when  complex numbers are  in polar form.  
  ::由于复数可以转换成极形,当复数为极形时,乘法也可以发挥作用。

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  Given two complex numbers,  $\begin{align*}z_1 = r_1 (\mbox{cos}\ \theta_1 + i\ \mbox{sin}\ \theta_1) \end{align*}$  and  $\begin{align*}z_2 =r_2(\mbox{cos}\ \theta_2 + i\ \mbox{sin}\ \theta_2),\end{align*}$
  ::根据两个复杂的数字,z1=r1(cos +1+i sin _1)和z2=r2(cos +2+i sin _2),

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  multiply the two complex numbers and apply the commutative property of multiplication.
  ::乘以两个复数,并应用乘法的通量属性。

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  $$\begin{align*}z_1 \cdot z_2 &= r_1 (\cos \theta_1 + i \sin \theta_1) \cdot r_2 (\cos \theta_2 + i \sin \theta_2) \\ &= r_1r_2 (\cos \theta_1 + i \sin \theta_1)(\cos \theta_2 + i \sin \theta_2) \end{align*}$$
  ::z2=r1 (cos%1+isin%1)1,1,1,2(cos%2+isin%2)=r1r2(cos%1+isin%1)(cos%2+isin%2)

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  Next, apply the distributive property of multiplication over addition.
  ::其次,适用乘以增加的分配属性。

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  $$\begin{align*}z_1 \cdot z_2= r_1 r_2 (\cos \theta_1\cdot \cos \theta_2 + i \cos \theta_1\cdot \sin \theta_2 + i \sin \theta_1\cdot \cos \theta_2 + i^2 \sin \theta_1\cdot \sin \theta_2)\end{align*}$$
  ::z2=r1r2(cos%1cos%2+icos%1sin%2+isin%1cos%2+i2+i2sin%1sin%2)

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  Recall   $\begin{align*}{i}^{2}=-1\end{align*}$ .

  $$\begin{align*}z_1 \cdot z_2 &= r_1 r_2 (\cos \theta_1\ \cos \theta_2\ + i \cos \theta_1\sin \theta_2 + i\sin \theta_1\cos \theta_2\ - \sin \theta_1\sin \theta_2) \\ &= r_1 r_2 (\cos \theta_1\ \cos \theta_2\ - \sin \theta_1\sin \theta_2+ i \cos \theta_1\sin \theta_2 + i\sin \theta_1\cos \theta_2\ ) \\ &= r_1 r_2 \left[(\cos \theta_1\ \cos \theta_2\ - \sin \theta_1\sin \theta_2)+i( \cos \theta_1\sin \theta_2 + \sin \theta_1\cos \theta_2\ )\right]\end{align*}$$ Recall
  ::回顾 i21.z1z2=r1r2(cos%1 cos%2+icos%1sin2+isin%1cs%2+isin1cos%2=r1r2(cos%1csin2+sin%1sin2+icos%1sin2+isin1sin%2+isin1sin%2+isin1cos%2)=r1r2[(cos%1 COs%2-sin%2-sin%1sin2+sin1cos%2]

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  $$\begin{align*}cis \ \theta = \cos \theta + i\sin \theta.\end{align*}$$
  ::确实如此

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  Also, note that the product of two complex numbers in trigonometric form is
  

  $$\begin{align*}(r_1 \ cis \ \theta_1)(r_2 \ cis \ \theta_2)=r_1r_2 \ cis \ (\theta_1+\theta_2).\end{align*}$$  
  ::另外,请注意,以三角测量形式呈现的两个复杂数字的产物是(r1cis =1cis) (r2cis =r1r2cis (12)。

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     Product of Two Complex Numbers
  ::两个复合数字产品

  $$\begin{align*}z_1 \cdot z_2 = r_1 r_2 \ cis \ ( \theta_1 + \theta_2)\end{align*}$$

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  Quotient Theorem 
  ::引引理论论

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  Dividing complex numbers in polar form can be shown using a similar proof that was used to show multiplication of complex numbers. Given two complex numbers,  $\begin{align*}z_1 = r_1 (\mbox{cos}\ \theta_1 + i\ \mbox{sin}\ \theta_1) \end{align*}$  and  $\begin{align*}z_2 =r_2(\mbox{cos}\ \theta_2 + i\ \mbox{sin}\ \theta_2)\end{align*}$ , then $\begin{align*}\frac{z_1} {z_2} = \frac{r_1} {r_2} \cdot \ cis \ (\theta_1 - \theta_2).\end{align*}$
  ::以极形表示的相异复杂数字可以用类似的证据来显示复杂数字的乘法。在两个复杂数字中,z1=r1(cos =1+i sin =%1)和z2=r2(cos =2+i sin =%2),然后z1z2=r1r2\cis(12)。

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     Quotient Theorem
  ::引引理论论

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  For   $\begin{align*}z_1 = r_1(cos θ_1 + i sin θ_1) \end{align*}$ and  $\begin{align*}z_1 = r_1(cos θ_1 + i sin θ_1), \end{align*}$
  ::z1=r1(cos%1+isin%1)和z1=r1(cos%1+isin%1),

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  $$\begin{align*}\frac{z_1} {z_2} = \frac{r_1} {r_2} {cis} \ (\theta_1 - \theta_2).\end{align*}$$
  ::z1z2=r1r2cis (12)

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  The following video demonstrates  how to multiply and divide complex numbers in trigonometric form:
  ::以下视频展示如何以三角形形式乘以和分离复杂数字:

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  The following video demonstrates  examples of using Ohm's Law to compute voltage, currents, and power in resistors:
  ::以下影片展示了使用《奥姆法》计算电压、电流和抵抗者力量的例子:

    

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  Examples
  ::实例

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  Example 1
  ::例1

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  Multiply  $\begin{align*}z_1 = 2 + 2i\end{align*}$ and $\begin{align*}z_2 = 1 - \sqrt{3}i\end{align*}$ .
  ::乘以z1=2+2i和z2=13i。

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  Solution:
  ::解决方案 :

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  For $\begin{align*}z_1,\end{align*}$
  ::z1, 用于 z1 。

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  $$\begin{align*}r_1 = \sqrt{2^2 + 2^2} = \sqrt{8}=2\sqrt{2}\end{align*}$$
  ::122+228=22

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  and

  $$\begin{align*}\tan \theta_1 &= \left|\frac{2} {2} \right|=1 \\ \theta_1 &= \frac{\pi}{4}.\end{align*}$$
  ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}还有... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}还有... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}还有 {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}

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  Note that $\begin{align*}\theta_1\end{align*}$ is in the 1st quadrant since a  and b > 0.
  ::请注意,% 1 位于自 a 和 b > 0 以来的第一个象限内。

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  For $\begin{align*}z_2\end{align*}$ ,
  ::对于z2, z2,

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  $$\begin{align*}r_2 = \sqrt{1^2 + (-\sqrt{3})^2}=\sqrt{4}=2\end{align*}$$
  ::r212+(3)24=2

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  and

  $$\begin{align*}\tan \theta_{\text{ref}} &= \left| \frac{-\sqrt{3}} {1}\right| = \sqrt{3} \\ \theta_{\text{ref}} &= \frac{\pi} {3}.\end{align*}$$
  ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}还有... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}...

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  Note that  $\begin{align*}\theta_2\end{align*}$  is in the 4th quadrant, since  a > 0  and  b  < 0, so  $\begin{align*}\theta_2=\frac{5\pi}{3}.\end{align*}$  
  ::请注意,% 2 位于第 4 象限内, 因为 a > 0 和 b < 0, 所以 % 2= 5 3 。

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  Use the formula  $\begin{align*}z_1 \cdot z_2 = r_1 \cdot r_2 cis \left( \theta_1 +\theta_2 \right)\end{align*}$ .
  ::使用公式z1z2=r1°r2cis(12) 。

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  $$\begin{align*}z_1 \cdot z_2 = 2\sqrt{2} \cdot 2\ cis\ \left [\frac{\pi} {4} + \frac{5\pi} {3}\right ] = 4\sqrt{2} \ cis\ \left [\frac{23\pi} {12} \right ]\end{align*}$$
  ::[4+5}3]=4__2 cis [23_12]

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  Then,  $\begin{align*}z_1 \cdot z_2 = 4\sqrt{2} \left (\mbox{cos}\ \frac{23\pi} {12} + i\ \mbox{sin}\ \frac{23 \pi} {12}\right )\end{align*}$ .
  ::然后,z1z2=4/2(2312+i罪2312)。

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  If the problem is done using rectangular units,
  ::如果使用矩形单位解决问题,

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  $$\begin{align*}z_1 \cdot z_2 = (2 + 2i)(1 - \sqrt{3}i) =2-2\sqrt{3}+2i-2\sqrt{3}i=(2 + 2\sqrt{3})-(2+ 2\sqrt{3})i.\end{align*}$$
  ::z2=(2+2i)(13i)=2-23+2i-23i=(2+23)-(2+23)。

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  Both results gives $\begin{align*}5.46 - 1.46i\end{align*}$  in decimal form.
  ::这两个结果以小数点为单位显示5.46-1.46i。

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  Example 2
  ::例2

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  Using polar multiplication, determine the product:  $\begin{align*}(6 - 2\sqrt{3}i)(4 + 4\sqrt{3}i)\end{align*}$ .
  ::使用极化乘法确定产品6-23i)(4+43i)

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  Solution:

  $$\begin{align*}r_1 &= \sqrt{(6)^2 - (2\sqrt{3})^2}= \sqrt{48}=4\sqrt{3} \\ \tan \theta_{\text{ref}} &= \left| \frac{2\sqrt{3}}{6} \right| = \frac{\sqrt{3}}{3} \\ \theta_{\text{ref}} &= \frac{\pi}{6} \end{align*}$$
  ::解答:r1(6)2-(23)248=43tanèref2363396

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  Since a  > 0 and b  < 0,  θ ₁ is in the 4th quadrant:  $\begin{align*}\theta_1 = \frac{11\pi} {6}\end{align*}$ .
  ::由于 a > 0 和 b < 0, 1 位于第 4 象限: # 1 = 11 6 。

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  $$\begin{align*}r_2 &= \sqrt{(4)^2 + (4\sqrt{3})^2}= \sqrt{64}=8 \\ \tan \theta_{\text{ref}} &= \left| \frac{4\sqrt{3}}{4} \right| = \sqrt{3} \\ \theta_{\text{ref}} &= \frac{\pi}{3}\end{align*}$$
  ::2444433264=894434333333

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  Since  a,   b  > 0,  θ ₂ is in the 1st quadrant:  $\begin{align*}\theta_2 = \frac{\pi} {3}\end{align*}$ .
  ::由于 a, b > 0, 2 位于第 1 象限中 : 2 3 。

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  Using polar multiplication,
  ::利用两极乘法,

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  $$\begin{align*}z_1 \cdot z_2 = 4\sqrt{3} \cdot 8 \ cis \ \left(\frac{11\pi} {6} + \frac{\pi} {3}\right ) = 32\sqrt{3} \ cis \ \frac{13\pi} {6}.\end{align*}$$
  ::z2=438cis (1163) = 323cis 136。

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  Since $\begin{align*}\theta > 2\pi\end{align*}$ , subtract  $\begin{align*}2\pi\end{align*}$  from the argument:
  ::自\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\"\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\2\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\2\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  $$\begin{align*}z_1 \cdot z_2 = 32\sqrt{3} \ cis \ \frac{\pi} {6}.\end{align*}$$
  ::z1z2=323cis =6 z1z2=323cis cis =6。

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  Example 3
  ::例3

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  Use polar division for   $\begin{align*}z_1 = 5 - 5i\end{align*}$ and $\begin{align*}z_2 = -2\sqrt{3} - 2i\end{align*}$ .
  ::z1=5-5i和z2=23-2i使用极分。

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  Solution:
  ::解决方案 :

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  $$\begin{align*}r_1 = \sqrt{5^2 + (-5)^2}=5\sqrt{2} \quad \text{and} \quad \tan \theta_{\text{ref}} = \left| \frac{-5}{5} \right|=1 \to \theta_1 =\frac{7\pi}{4}\end{align*}$$  
  $$\begin{align*}r_2 = \sqrt{(-2\sqrt{3})^2 + (-2)^2}=4 \quad \text{and} \quad \tan \theta_{\text{ref}} = \left| \frac{-2}{-2\sqrt{3}} \right|=\frac{\sqrt{3}}{3} \to \theta_2 =\frac{7\pi}{6}\end{align*}$$ Use the formula  $\begin{align*}\frac{z_1} {z_2} = \frac{r_1} {r_2} \cdot \mbox{cis}\ [\theta_1 - \theta_2]\end{align*}$ .
  ::-52+(-5)2=52andtanèref5511=74 r2(-23)2+(-2)2=4andtanäref2-2-23332=76 使用公式z1z2=r1r2}[12]。

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  $$\begin{align*}\frac{z_1}{z_2} &= \frac{5\sqrt{2}} {4} \cdot \ cis\ \left (\frac{7\pi} {4} - \frac{7\pi} {6}\right) \\ &= \frac{5\sqrt{2}}{4} \cdot \ cis\ \frac{7\pi} {12}\end{align*}$$
  ::z2=524Cis(74-76)=524Cis 712

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  Example 4
  ::例4

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  Recall the question from the Introduction: C an you identify the variable j  when comparing the electrical engineering formula
  ::回顾导言中的问题:在比较电气工程公式时,你能辨别变量j吗?

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  $$\begin{align*} V = V_0 e^{j \omega t} = V_0 \left (\cos \omega t + j \sin\omega t \right )\end{align*}$$
  ::V=V0ejt=V0(cost+jsint)

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  with  the similar expression

  $$\begin{align*}r (\cos \theta + i \sin \theta)?\end{align*}$$
  ::使用类似表达式 r(cosisin}) 吗 ?

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  Solution:
  ::解决方案 :

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  In electrical calculations, the letter $\begin{align*}I\end{align*}$  is commonly used to denote current.  Therefore, are identified with a $\begin{align*}j\end{align*}$ .
  ::在电算中,通常使用I字母表示当前值,因此与j(j)表示。

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  Example 5
  ::例5

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  Find the product: $\begin{align*}\left( 7, \frac{\pi}{6} \right) \cdot \left( 5, \frac{-\pi}{4} \right)\end{align*}$ .
  ::寻找产品: (7, 6) (5, 4)。

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  Solution:

  $$\begin{align*}r_1 \cdot r_2 &\to 7 \cdot 5 = 35 \\ \theta_1 + \theta_2 &\to \frac{\pi}{6} + \frac{-\pi}{4} =- \frac{\pi}{12} \end{align*}$$ Therefore, the product is  $\begin{align*}35 \ cis \ \left( \frac{-\pi}{12} \right)\end{align*}$ .
  ::解答:r1r275=35126412 因此,该产品为35cis(12)。

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  Example 6
  ::例6

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  Find the quotient  $\begin{align*}\frac{1 + 2i}{2 - i}\end{align*}$ .
  ::查找商数 1+2i2-i。

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  Solution:
  ::解决方案 :

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  $$\begin{align*}r_1 = \sqrt{(1)^2 + (2)^2} = \sqrt{5} \qquad \qquad r_2 = \sqrt{(2)^2 + (-1)^2} = \sqrt{5}\end{align*}$$
  ::============================================================================================================== =================================================================================================================================================================================================================================================================================================================

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  $$\begin{align*}\tan \theta_1 = \left| \frac{2} {1} \right|=2 \to \theta_1 = 1.107 \text{ radians} \end{align*}$$
  ::~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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  Since $\begin{align*}a,b>0\end{align*}$ , $\begin{align*}\theta_1\end{align*}$  is in the 1st quadrant.
  ::a,b>0, 1位于第1象限内。

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  $$\begin{align*}\tan \theta_{\text{ref}} = \left|\frac{-1}{2} \right| = \frac{1}{2} \to \theta_{\text{ref}}=0.464 \text{ radians} \end{align*}$$
  ::tanäref121212ref=0.464弧度

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  Since $\begin{align*}a>0 \text{ and } b<0,\end{align*}$   $\begin{align*}\theta_2\end{align*}$  is in the 4th quadrant, so $\begin{align*}\theta_2 = 5.820 \text{ radians}.\end{align*}$
  ::a>0和b<0,%2位于第4象限内,因此2=5.820弧度。

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  $$\begin{align*}\frac{z_1} {z_2} &= \frac{\sqrt{5}} {\sqrt{5}} \ cis \ (1.107 - 5.820) \\ &= cis\ (-4.713)\\ &= \cos (-4.713)+i\sin (-4.713)\\ &= \cos (1.570)+i\sin (1.570)\end{align*}$$
  ::1z255cis (1.107-5.820) =cis(-4.713) =cos(-4.713) +isin(-4.713) =cos(1.570) +isin(1.570)

   

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  Summary
  ::摘要

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  • The Product theorem can be used to multiply complex numbers:  $$\begin{align*}z_1 \cdot z_2 = r_1 \cdot r_2 cis \left( \theta_1 +\theta_2 \right).\end{align*}$$
    ::产品定理可用于乘以复数: z1z2=r1r2cis(12)。
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  • The Quotient theorem can be used to divide complex  numbers:
    ::引号定理可用于分隔复杂数字:

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  $$\begin{align*}\frac{z_1} {z_2} = \frac{r_1} {r_2} {cis} \ (\theta_1 - \theta_2).\end{align*}$$
  ::z1z2=r1r2cis (12)

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  Review
  ::回顾

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  1. Find the product using polar form: $\begin{align*}(2 + 2i)(\sqrt{3} - i).\end{align*}$
     ::使用极形(2+2i)(3-i)查找产品。
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  2.   $\begin{align*} 2 cis \ 40^\circ \cdot 4cis \ 20^\circ\end{align*}$
     ::2cis 404cis 20________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
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  3. $\begin{align*}\frac{2 cis (80)}{6 cis (200)}\end{align*}$
     ::Cis( 806Cis( 200) )
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  4. By how many degrees would the radius from the origin to  $\begin{align*}(a,b)\end{align*}$   need to be rotated to coincide with the radius from the pole to the graph of the product of  $\begin{align*}a+bi\end{align*}$  and  $\begin{align*}1+i\sqrt{3}\end{align*}$  ?
     ::从来源到(a,b)的半径需要旋转多少度才能与极的半径相吻合,从极到+Bi和1+i3的产品图?

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  Let  $\begin{align*} z_1 = \left(7, \frac{\pi}{6} \right)\end{align*}$ and $\begin{align*}z_2 = \left(5, \frac{-\pi}{4} \right).\end{align*}$  Find:
  ::Letz1=( 7, 6) 和z2=( 5, 4)。 查找 :

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  5. $\begin{align*}z_1 \cdot z_2\end{align*}$
     ::z2 z1z2 z1z2
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  6. $\begin{align*}\left( \frac{z_1}{z_2} \right)\end{align*}$
     :z1z2)
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  7. $\begin{align*}\left( \frac{z_2}{z_1} \right)\end{align*}$
     :z2z1)

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  Let  $\begin{align*}z_1 = \left( 8,\frac{\pi}{3} \right)\end{align*}$ and $\begin{align*} z_2 = \left(5, \frac{\pi}{6} \right)\end{align*}$ . Find:
  ::Letz1=( 8) 3和z2=( 5) 6 查找 :

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  8. $\begin{align*}z_1 \cdot z_2\end{align*}$
     ::z2 z1z2 z1z2
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  9. $\begin{align*}\left( \frac{z_1}{z_2} \right)\end{align*}$
     :z1z2)
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  10. $\begin{align*}\left( \frac{z_2}{z_1} \right)\end{align*}$
      :z2z1)
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  11. $\begin{align*}(z_1)^2\end{align*}$
      :z1)2
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  12. $\begin{align*}(z_2)^3\end{align*}$
      :z2)3

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  Find the products and put your answer in complex number form:
  ::查找产品,将您的答复填入复数格式 :

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  13. Find the product using poorm: $\begin{align*}(2 + 2i)(\sqrt{3} - i).\end{align*}$
      ::使用polm2+2i)(3-i)查找产品。
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  14. $\begin{align*}2 (\cos 40^\circ + i \sin 40^\circ) \cdot 4(\cos 20^\circ + i \sin 20^\circ)\end{align*}$
      :cos40isin40) 44(cos20isin20)
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  15. $\begin{align*} 2 \left(\cos \frac{\pi}{8} + i \sin \frac{\pi}{8} \right) \cdot 2 \left(\cos \frac{\pi}{10} + i \sin \frac{\pi}{10} \right)\end{align*}$
      ::2(cos__8+isin__8) (8)(2)(cos__10+isin__10)

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  Find the quotients and put your answer in complex number form:
  ::查找商数并将您的回答以复数形式显示:

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  16. $\begin{align*} 2(\cos 80^\circ + i \sin 80^\circ) \div 6(\cos 200^\circ + i \sin 200^\circ)\end{align*}$
      ::6(cos200)\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\8\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
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  17. $\begin{align*} 3 \ cis \ 130^\circ \div 4 \ cis \ 270^\circ\end{align*}$
      ::3 Cis 1304 Cis 270

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  Apply your knowledge. Remember Ohm's Law:  $\begin{align*}E=I\cdot Z.\end{align*}$
  ::运用你的知识 记住 Ohm 法则: E=I。

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  18. The voltage of a circuit can be represented by  $\begin{align*}160(\cos{330^\circ}+i\sin{330^\circ})\end{align*}$ ohms. Find the polar form of the current if the polar form of the impedance is  $\begin{align*}35(\cos{300^\circ}+i\sin{300^\circ})\end{align*}$  ohms.
      ::电路的电压为 160 (cos330 isin330 ) ohms 。 如果阻力的极形式为 35 (cos300 isin300 ) ohms , 请查找电流的极化形式 。
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  19. What is the voltage of a circuit with an impedance of  $\begin{align*}14(\cos{(-10^\circ)}+i\sin{(-10^\circ)})\end{align*}$  and current of  $\begin{align*}6(\cos{(47^\circ)}+i\sin{(47^\circ)})\end{align*}$  ?
      ::阻碍14(cos(-10)+isin(-10)和6(cos(47)+isin(47))的电路电压是多少?

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  Review (Answers)
  ::回顾(答复)

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  Please see the Appendix.
  ::请参看附录。