6.10 产品定理

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  产品定理

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  What if you were given two in polar form , such as $\begin{array}{r}2\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right),7\left(\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2}\right)\end{array}$ and asked to multiply them? Would you be able to do this? How long would it take you?
  ::如果给您两个极形的, 例如 2 (cos2+isin2) 、 7 (cos32+isin32) , 并请求您增加它们数量呢 ? 您能够这样做吗 ? 您需要多长时间 ?

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  Product Theorem 
  ::产品定理

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  Multiplication of complex numbers in polar form is similar to the multiplication of complex numbers in standard form. However, to determine a general rule for multiplication, the trigonometric functions will be simplified by applying the sum/difference identities for cosine and sine. To obtain a general rule for the multiplication of complex numbers in polar from, let the first number be $\begin{array}{r}{r}_1(\cos {\theta }_1+i\sin {\theta }_1)\end{array}$ and the second number be $\begin{array}{r}{r}_2(\cos {\theta }_2+i\sin {\theta }_2)\end{array}$ . The product can then be simplified by use of three facts: the definition $\begin{array}{r}{i}^2=-1\end{array}$ , the sum identity $\begin{array}{r}\cos \alpha \cos \beta -\sin \alpha \sin \beta =\cos (\alpha +\beta )\end{array}$ , and the sum identity $\begin{array}{r}\sin \alpha \cos \beta +\cos \alpha \sin \beta =\sin (\alpha +\beta )\end{array}$ .
  ::以极态表示的复杂数字的乘法与标准格式的复杂数字的乘法相似。 但是,为了确定乘法的一般规则,三角函数将通过对正弦和正弦应用总和/差异特性来简化。 为了获得关于极地复杂数字乘法的乘法的一般规则,第一个数字为r1(cos1+isin1),第二个数字为r2(cos2+isin2)。 然后,产品可以通过使用三个事实来简化: i21 定义、 总和身份 oscossinsincos(), 和总和身份 sincosçsinsin()。

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  Now that the numbers have been designated, proceed with the multiplication of these binomials.
  ::现在数字已经指定, 开始这些二进制的乘法。

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  $$\begin{array}{rl}& r_1(\cos {\theta }_1+i\sin {\theta }_1)\cdot r_2(\cos {\theta }_2+i\sin {\theta }_2)\\ & r_1{r}_2(\cos {\theta }_1\cos {\theta }_2+i\cos {\theta }_1\sin {\theta }_2+i\sin {\theta }_1\cos {\theta }_2+i^2\sin {\theta }_1\sin {\theta }_2)\\ & r_1{r}_2[(\cos {\theta }_1\cos {\theta }_2-\sin {\theta }_1\sin {\theta }_2)+i(\sin {\theta }_1\cos {\theta }_2+\cos {\theta }_1\sin {\theta }_2)]\\ & r_1{r}_2[\cos ({\theta }_1+{\theta }_2)+i\sin ({\theta }_1+{\theta }_2)]\end{array}$$

  
  ::r2(cos2+cos2+cos2+cos2+ocos1sin}2+isin1cos2+i2+i2sin2+i2+sin1sin}2r1r2[(cos1cos2+sin1sin}2+sin1sin}2)+i(sin1cos}2+cos%2+cos%1sin2+cos%1sin2+r1r1r2[cos2+sin}+isin1}2

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  Therefore:
  ::因此:

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  $$\begin{array}{r}{r}_1(\cos {\theta }_1+i\sin {\theta }_1)\cdot r_2(\cos {\theta }_2+i\sin {\theta }_2)=r_1{r}_2[\cos ({\theta }_1+{\theta }_2)+i\sin ({\theta }_1+{\theta }_2)]\end{array}$$

  
  ::r1 (cos1+isin1) r2(cos2+isin2) =r1r2 [cos12] +isin12]

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  Use this general formula for the product of complex numbers to perform computations.
  ::对复杂数字的产物使用此通用公式进行计算。

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  1. Find the product of complex numbers by using the Product Theorem
  ::1. 通过使用产品理论来查找复杂数字的产物

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  Find the product of $\begin{array}{r}3.61(\cos {56.3}^{\circ }+i\sin {56.3}^{\circ })\end{array}$ and $\begin{array}{r}1.41(\cos {315}^{\circ }+i\sin {315}^{\circ })\end{array}$
  ::查找3.61(cos56.3isin56.3)和1.41(cos315315315)的产物。

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  Use the Product Theorem, $\begin{array}{r}{r}_1(\cos {\theta }_1+i\sin {\theta }_1)\cdot r_2(\cos {\theta }_2+i\sin {\theta }_2)=r_1{r}_2[\cos ({\theta }_1+{\theta }_2)+i\sin ({\theta }_1+{\theta }_2)]\end{array}$ .
  ::使用产品理论,r1(cos1+isin1) r2(cos2+isin2) =r1r2[cos12] +isin1]。

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  $$\begin{array}{rl}& 3.61(\cos {56.3}^{\circ }+i\sin {56.3}^{\circ })\cdot 1.41(\cos {315}^{\circ }+i\sin {315}^{\circ })\\ & =(3.61)(1.41)[\cos ({56.3}^{\circ }+{315}^{\circ })+i\sin ({56.3}^{\circ }+{315}^{\circ })\\ & =5.09(\cos {371.3}^{\circ }+i\sin {371.3}^{\circ })\\ & =5.09(\cos {11.3}^{\circ }+i\sin {11.3}^{\circ })\end{array}$$

  
  ::3.61(cos56.3 56.3 56.3 )1.41(cos315 315 315 ) =(3.61) (1.41) [cos(56.3 315 ) +isin(56.3 315 ) =5.09(cos371.3 371.3 ) = 5.09(cos11.3 11.3 3 11.3 )

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  $\begin{array}{r}{}^{\ast }\end{array}$ Note: Angles are expressed $\begin{array}{r}{0}^{\circ }\le \theta \le {360}^{\circ }\end{array}$ unless otherwise stated.
  ::* 注:除另有说明外,括号为 0360。

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  2. Find the product of $\begin{array}{r}5\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)\cdot \sqrt{3}\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)\end{array}$
  ::2. 查找5(cos34+isin34)3(cos2+isin2)的产物

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  First, calculate $\begin{array}{r}{r}_1{r}_2=5\cdot \sqrt{3}=5\sqrt{3}\end{array}$ and $\begin{array}{r}\theta ={\theta }_1+{\theta }_2=\frac{3\pi }{4}+\frac{\pi }{2}=\frac{5\pi }{4}\end{array}$
  ::首先,计算 r1r2= 53= 53 和12= 342= 54

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  $$\begin{array}{r}5\sqrt{3}\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)\end{array}$$

  
  ::53(cos54+isin54)

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  3. Find the product of the numbers $\begin{array}{r}{r}_1=1+i\end{array}$ and $\begin{array}{r}{r}_2=\sqrt{3}-i\end{array}$ by first converting them to trigonometric form .
  ::3. 将数字r1=1+i和r2=3-i的产物首先转换成三角形,然后查找其产物。

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  First, convert $\begin{array}{r}1+i\end{array}$ to polar form:
  ::首先,将 1+i 转换为极形 :

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  $$\begin{array}{r}{r}_1=\sqrt{1^2+1^2}=\sqrt{2}\\ {\theta }_1=\arctan \frac{1}{1}=\frac{\pi }{4}\end{array}$$

  
  ::r1=12+12=21=arctan114

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  And now do the same with $\begin{array}{r}\sqrt{3}-i\end{array}$ :
  ::现在对 3 -i 也做同样的事情:

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  $$\begin{array}{r}{r}_2=\sqrt{{\sqrt{3}}^2+(-1{)}^2}=2\\ {\theta }_2=\arctan -\frac{1}{\sqrt{3}}=-\frac{\pi }{6}\end{array}$$

  
  ::r2=32+(- 1)2=22=arctan-136

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  And now substituting these values into the product theorem:
  ::现在用这些价值来取代产品定理:

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  $$\begin{array}{r}{r}_1\times r_2=\\ (2)(\sqrt{2})\left(\cos \left(-\frac{\pi }{6}+\frac{\pi }{4}\right)+i\sin \left(-\frac{\pi }{6}+\frac{\pi }{4}\right)\right)\\ (2)(\sqrt{2})\left(\cos \left(\frac{\pi }{12}\right)+i\sin \left(\frac{\pi }{12}\right)\right)\end{array}$$

  
  ::1×r2=(2)(2)(2)(cos) (64)+isin(64)(2)(2)(cos(12)+isin(12)))

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked to multiply two complex numbers in polar form. 
  ::早些时候,你被要求 乘以两个复杂的数字 以极形的形式。

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  Since you want to multiply
  ::既然您想要乘数

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  $\begin{array}{r}2(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}),7\left(\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2}\right)\end{array}$
  ::2(cos2+isin2),7(cos32+isin32)

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  where $\begin{array}{r}{r}_1=2,r_2=7,{\theta }_1=\frac{\pi }{2},{\theta }_2=\frac{3\pi }{2}\end{array}$ ,
  ::r1=2,r2=7,12,2=32,r2=7,r2=7,r2=7,r2=7,r12,r2=3,r2=2,r2=2,r2=2,r2=2,r2=2,r2=2,r2=7,r2=2,r2=2,r2=2,r2=2,r2=7,r2=2,r2=2,r2=2,r2=2,r2=3

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  you can use the equation
  ::您可以使用方程

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  $\begin{array}{r}{r}_1{r}_2[\cos ({\theta }_1+{\theta }_2)+i\sin ({\theta }_1+{\theta }_2)]\end{array}$
  ::r1r2[cos(12)+isin(12)

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  and calculate:
  ::计算 :

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  $\begin{array}{r}(2)(7)[\cos (\frac{\pi }{2}+\frac{3\pi }{2})+i\sin (\frac{\pi }{2}+\frac{3\pi }{2})]\end{array}$
  :2) [cos(2+32)+isin(2+32)]

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  This simplifies to:
  ::这简化为:

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  $$\begin{array}{r}14[\cos (2\pi )+i\sin (2\pi )]\\ =14[1+i0]\\ =14\end{array}$$

  
  ::14[14[1+i0]=14]

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  Example 2
  ::例2

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  Multiply together the following complex numbers. If they are not in polar form, change them before multiplying.
  ::乘以以下的复杂数字。 如果它们不是极形的, 则在乘前先修改它们 。

  $\begin{array}{r}2\mathrm{\angle }{56}^{\circ },7\mathrm{\angle }{113}^{\circ }\end{array}$

  $\begin{array}{r}2\mathrm{\angle }{56}^{\circ },7\mathrm{\angle }{113}^{\circ }=(2)(7)\mathrm{\angle }({56}^{\circ }+{113}^{\circ })=14\mathrm{\angle }{169}^{\circ }\end{array}$

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  Example 3
  ::例3

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  Multiply together the following complex numbers. If they are not in polar form, change them before multiplying.
  ::乘以以下的复杂数字。 如果它们不是极形的, 则在乘前先修改它们 。

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  $\begin{array}{r}3(\cos \pi +i\sin \pi ),10\left(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3}\right)\end{array}$
  :cosisin,10(cos53+isin53)

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  $\begin{array}{r}3(\cos \pi +i\sin \pi ),10\left(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3}\right)=(3)(10)cis\left(\pi +\frac{5\pi }{3}\right)=30cis\frac{8\pi }{3}=30cis\frac{2\pi }{3}\end{array}$
  ::3(cosisin),10(cos53+isin53) =(3)(10)cis(53) =30cis83=30cis23

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  Example 4
  ::例4

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  Multiply together the following complex numbers. If they are not in polar form, change them before multiplying.
  ::乘以以下的复杂数字。 如果它们不是极形的, 则在乘前先修改它们 。

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  $\begin{array}{r}2+3i,-5+11i\end{array}$
  ::2+3i,-5+11i

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    $\begin{array}{r}2+3i,-5+11i\to \text{change to polar}\end{array}$

  $$\begin{array}{rlrl}& x=2,y=3& & x=-5,y=11\\ & r=\sqrt{2^2+3^2}=\sqrt{13}\approx 3.61& & r=\sqrt{(-5{)}^2+{11}^2}=\sqrt{146}\approx 12.08\\ & \tan \theta =\frac{3}{2}\to \theta ={56.31}^{\circ }& & \tan \theta =-\frac{11}{5}\to \theta ={114.44}^{\circ }\end{array}$$

  $$\begin{array}{r}(3.61)(12.08)\mathrm{\angle }({56.31}^{\circ }+{114.44}^{\circ })=43.61\mathrm{\angle }{170.75}^{\circ }\end{array}$$

  
  ::2+3i,-5+11i 改为极地轴=2,y=3x5,y=11r=22+32=133.61r=(-55)2+112=146}12.08tan3256.311115114.44(3.61)(12.08)(56.31114.44)=43.61170.75

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  Review
  ::回顾

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  Multiply each pair of complex numbers. If they are not in trigonometric form, change them before multiplying.
  ::乘以每对复杂数字。 如果它们不是三角形的, 则在乘法前更改它们 。

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  1. $\begin{array}{r}3(\cos {32}^{\circ }+i\sin {32}^{\circ })\cdot 2(\cos {15}^{\circ }+i\sin {15}^{\circ })\end{array}$
     ::3⁄4 ̄ ̧漯B
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  2. $\begin{array}{r}2(\cos {10}^{\circ }+i\sin {10}^{\circ })\cdot 10(\cos {12}^{\circ }+i\sin {12}^{\circ })\end{array}$
     :cos101010101010121212)
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  3. $\begin{array}{r}4(\cos {45}^{\circ }+i\sin {45}^{\circ })\cdot 8(\cos {62}^{\circ }+i\sin {62}^{\circ })\end{array}$
     ::4 (cos454545) 8 (cos626262)
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  4. $\begin{array}{r}2(\cos {60}^{\circ }+i\sin {60}^{\circ })\cdot \frac{1}{2}(\cos {34}^{\circ }+i\sin {34}^{\circ })\end{array}$
     ::12(cos343434)
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  5. $\begin{array}{r}5(\cos {25}^{\circ }+i\sin {25}^{\circ })\cdot 2(\cos {115}^{\circ }+i\sin {115}^{\circ })\end{array}$
     ::5 (cos252525) (2(cos115115115))
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  6. $\begin{array}{r}-3(\cos {70}^{\circ }+i\sin {70}^{\circ })\cdot 3(\cos {85}^{\circ }+i\sin {85}^{\circ })\end{array}$
     ::-3 -3(cos707070) -3(cos858585)
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  7. $\begin{array}{r}7(\cos {85}^{\circ }+i\sin {85}^{\circ })\cdot \sqrt{2}(\cos {40}^{\circ }+i\sin {40}^{\circ })\end{array}$
     ::7(cos8585) 2(cos404040)
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  8. $\begin{array}{r}(3-2i)\cdot (1+i)\end{array}$
     :3-2(i)_________(1+1)
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  9. $\begin{array}{r}(1-i)\cdot (1+i)\end{array}$
     :1-一) (1+1)
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  10. $\begin{array}{r}(4-i)\cdot (3+2i)\end{array}$
      :4-一) (3+2i)
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  11. $\begin{array}{r}(1+i)\cdot (1+4i)\end{array}$
      :1+一)__________________________________(1+1+4i)
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  12. $\begin{array}{r}(2+2i)\cdot (3+i)\end{array}$
      :2+2i)(3+i)
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  13. $\begin{array}{r}(1-3i)\cdot (2+i)\end{array}$
      :1-3i)(2+i)
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  14. $\begin{array}{r}(1-i)\cdot (1-i)\end{array}$
      :1-一)_____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(1_______________________________________(1______________________________(1________________________________________________________________________________________________________________________________________________________
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  15. Can you multiply a pair of complex numbers in standard form without converting to trigonometric form?
      ::您能否在不转换成三角形的情况下,以标准格式乘乘一对复杂数字?

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  Review (Answers)
  ::回顾(答复)

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  Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
  ::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键 。