节: 产品定理 | 6.10 产品定理

章节大纲

What if you were given two in polar form , such as $\begin{aligned} 2 (\cos \frac{π}{2} + i \sin \frac{π}{2}), 7 (\cos \frac{3 π}{2} + i \sin \frac{3 π}{2}) \end{aligned}$ 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 ?

Product Theorem
::产品定理

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 $\begin{aligned} r_{1} (\cos θ_{1} + i \sin θ_{1}) \end{aligned}$ and the second number be $\begin{aligned} r_{2} (\cos θ_{2} + i \sin θ_{2}) \end{aligned}$ . The product can then be simplified by use of three facts: the definition $\begin{aligned} i^{2} = - 1 \end{aligned}$ , the sum identity $\begin{aligned} \cos α \cos β - \sin α \sin β = \cos (α + β) \end{aligned}$ , and the sum identity $\begin{aligned} \sin α \cos β + \cos α \sin β = \sin (α + β) \end{aligned}$ .
::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()。

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。

$\begin{aligned} r_{1} (\cos θ_{1} + i \sin θ_{1}) \cdot r_{2} (\cos θ_{2} + i \sin θ_{2}) \\ r_{1} r_{2} (\cos θ_{1} \cos θ_{2} + i \cos θ_{1} \sin θ_{2} + i \sin θ_{1} \cos θ_{2} + i^{2} \sin θ_{1} \sin θ_{2}) \\ r_{1} r_{2} [(\cos θ_{1} \cos θ_{2} - \sin θ_{1} \sin θ_{2}) + i (\sin θ_{1} \cos θ_{2} + \cos θ_{1} \sin θ_{2})] \\ r_{1} r_{2} [\cos (θ_{1} + θ_{2}) + i \sin (θ_{1} + θ_{2})] \end{aligned}$

::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

Therefore:
::因此:

$\begin{aligned} r_{1} (\cos θ_{1} + i \sin θ_{1}) \cdot r_{2} (\cos θ_{2} + i \sin θ_{2}) = r_{1} r_{2} [\cos (θ_{1} + θ_{2}) + i \sin (θ_{1} + θ_{2})] \end{aligned}$

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

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。

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

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

Use the Product Theorem, $\begin{aligned} r_{1} (\cos θ_{1} + i \sin θ_{1}) \cdot r_{2} (\cos θ_{2} + i \sin θ_{2}) = r_{1} r_{2} [\cos (θ_{1} + θ_{2}) + i \sin (θ_{1} + θ_{2})] \end{aligned}$ .
::使用产品理论,r1(cos1+isin1) r2(cos2+isin2) =r1r2[cos12] +isin1]。

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

::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 )

$\begin{aligned} ^{*} \end{aligned}$ Note: Angles are expressed $\begin{aligned} 0^{\circ} \leq θ \leq 360^{\circ} \end{aligned}$ unless otherwise stated.
::* 注:除另有说明外,括号为 0360。

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

First, calculate $\begin{aligned} r_{1} r_{2} = 5 \cdot \sqrt{3} = 5 \sqrt{3} \end{aligned}$ and $\begin{aligned} θ = θ_{1} + θ_{2} = \frac{3 π}{4} + \frac{π}{2} = \frac{5 π}{4} \end{aligned}$
::首先,计算 r1r2= 53= 53 和12= 342= 54

$\begin{aligned} 5 \sqrt{3} (\cos \frac{5 π}{4} + i \sin \frac{5 π}{4}) \end{aligned}$

::53(cos54+isin54)

3. Find the product of the numbers $\begin{aligned} r_{1} = 1 + i \end{aligned}$ and $\begin{aligned} r_{2} = \sqrt{3} - i \end{aligned}$ 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。

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

$\begin{aligned} r_{1} = \sqrt{1^{2} + 1^{2}} = \sqrt{2} \\ θ_{1} = \arctan \frac{1}{1} = \frac{π}{4} \end{aligned}$

::r1=12+12=21=arctan114

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

$\begin{aligned} r_{2} = \sqrt{{\sqrt{3}}^{2} + (- 1)^{2}} = 2 \\ θ_{2} = \arctan - \frac{1}{\sqrt{3}} = - \frac{π}{6} \end{aligned}$

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

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

$\begin{aligned} r_{1} \times r_{2} = \\ (2) (\sqrt{2}) (\cos (- \frac{π}{6} + \frac{π}{4}) + i \sin (- \frac{π}{6} + \frac{π}{4})) \\ (2) (\sqrt{2}) (\cos (\frac{π}{12}) + i \sin (\frac{π}{12})) \end{aligned}$

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

Examples
::实例

Example 1
::例1

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。

Since you want to multiply
::既然您想要乘数

$\begin{aligned} 2 (\cos \frac{π}{2} + i \sin \frac{π}{2}), 7 (\cos \frac{3 π}{2} + i \sin \frac{3 π}{2}) \end{aligned}$
::2(cos2+isin2),7(cos32+isin32)

where $\begin{aligned} r_{1} = 2, r_{2} = 7, θ_{1} = \frac{π}{2}, θ_{2} = \frac{3 π}{2} \end{aligned}$ ,
::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

you can use the equation
::您可以使用方程

$\begin{aligned} r_{1} r_{2} [\cos (θ_{1} + θ_{2}) + i \sin (θ_{1} + θ_{2})] \end{aligned}$
::r1r2[cos(12)+isin(12)

and calculate:
::计算 :

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

This simplifies to:
::这简化为:

$\begin{aligned} 14 [\cos (2 π) + i \sin (2 π)] \\ = 14 [1 + i 0] \\ = 14 \end{aligned}$

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

Example 2
::例2

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。

$\begin{aligned} 2 ∠ 56^{\circ}, 7 ∠ 113^{\circ} \end{aligned}$

$\begin{aligned} 2 ∠ 56^{\circ}, 7 ∠ 113^{\circ} = (2) (7) ∠ (56^{\circ} + 113^{\circ}) = 14 ∠ 169^{\circ} \end{aligned}$

Example 3
::例3

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。

$\begin{aligned} 3 (\cos π + i \sin π), 10 (\cos \frac{5 π}{3} + i \sin \frac{5 π}{3}) \end{aligned}$
:cosisin,10(cos53+isin53)

$\begin{aligned} 3 (\cos π + i \sin π), 10 (\cos \frac{5 π}{3} + i \sin \frac{5 π}{3}) = (3) (10) c i s (π + \frac{5 π}{3}) = 30 c i s \frac{8 π}{3} = 30 c i s \frac{2 π}{3} \end{aligned}$
::3(cosisin),10(cos53+isin53) =(3)(10)cis(53) =30cis83=30cis23

Example 4
::例4

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。

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

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

$\begin{aligned} (3.61) (12.08) ∠ ({56.31}^{\circ} + {114.44}^{\circ}) = 43.61 ∠ {170.75}^{\circ} \end{aligned}$

::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

Review
::回顾

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。

$\begin{aligned} 3 (\cos 32^{\circ} + i \sin 32^{\circ}) \cdot 2 (\cos 15^{\circ} + i \sin 15^{\circ}) \end{aligned}$
::3⁄4 ̄ ̧漯B

$\begin{aligned} 2 (\cos 10^{\circ} + i \sin 10^{\circ}) \cdot 10 (\cos 12^{\circ} + i \sin 12^{\circ}) \end{aligned}$
:cos101010101010121212)

$\begin{aligned} 4 (\cos 45^{\circ} + i \sin 45^{\circ}) \cdot 8 (\cos 62^{\circ} + i \sin 62^{\circ}) \end{aligned}$
::4 (cos454545) 8 (cos626262)

$\begin{aligned} 2 (\cos 60^{\circ} + i \sin 60^{\circ}) \cdot \frac{1}{2} (\cos 34^{\circ} + i \sin 34^{\circ}) \end{aligned}$
::12(cos343434)

$\begin{aligned} 5 (\cos 25^{\circ} + i \sin 25^{\circ}) \cdot 2 (\cos 115^{\circ} + i \sin 115^{\circ}) \end{aligned}$
::5 (cos252525) (2(cos115115115))

$\begin{aligned} - 3 (\cos 70^{\circ} + i \sin 70^{\circ}) \cdot 3 (\cos 85^{\circ} + i \sin 85^{\circ}) \end{aligned}$
::-3 -3(cos707070) -3(cos858585)

$\begin{aligned} 7 (\cos 85^{\circ} + i \sin 85^{\circ}) \cdot \sqrt{2} (\cos 40^{\circ} + i \sin 40^{\circ}) \end{aligned}$
::7(cos8585) 2(cos404040)

$\begin{aligned} (3 - 2 i) \cdot (1 + i) \end{aligned}$
:3-2(i)_________(1+1)

$\begin{aligned} (1 - i) \cdot (1 + i) \end{aligned}$
:1-一) (1+1)

$\begin{aligned} (4 - i) \cdot (3 + 2 i) \end{aligned}$
:4-一) (3+2i)

$\begin{aligned} (1 + i) \cdot (1 + 4 i) \end{aligned}$
:1+一)__________________________________(1+1+4i)

$\begin{aligned} (2 + 2 i) \cdot (3 + i) \end{aligned}$
:2+2i)(3+i)

$\begin{aligned} (1 - 3 i) \cdot (2 + i) \end{aligned}$
:1-3i)(2+i)

$\begin{aligned} (1 - i) \cdot (1 - i) \end{aligned}$
:1-一)_____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(1_______________________________________(1______________________________(1________________________________________________________________________________________________________________________________________________________

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?

Review (Answers)
::回顾(答复)

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。

章节大纲

Product Theorem ::产品定理

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)

Product Theorem
::产品定理

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Review
::回顾

Review (Answers)
::回顾(答复)