Section: 引引理论论 | 6.11 引引定理

Section outline

Suppose you are given two in polar form , such as $\begin{aligned} 2 (\cos \frac{π}{3} + i \sin \frac{π}{3}) \end{aligned}$ and $\begin{aligned} 3 (\cos π + i \sin π) \end{aligned}$ and asked to divide them. Can you do this? How long will it take you?
::假设你们有两种极分形式,例如2 (cos3+isin}3 和3 (cosisin }3) 和 3 (cosisin ) , 并且要求你们分开。你能这样做吗?你们需要多长时间?

Quotient Theorem
::引引理论论

Division of complex numbers in polar form is similar to the division of complex numbers in standard form. However, to determine a general rule for division, the denominator must be rationalized by multiplying the fraction by the complex conjugate of the denominator. In addition, the trigonometric functions must be simplified by applying the sum/difference identities for cosine and sine as well as one of the .
::以极化形式划分的复杂数字与标准格式的复杂数字的划分相似,然而,为确定分法的一般规则,分母必须通过将分母乘以分母的复杂组合而合理化,此外,三角函数必须简化,对共弦和正弦和正弦应用总和/异差特性。

To obtain a general rule for the division 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 five facts: the definition $\begin{aligned} i^{2} = - 1 \end{aligned}$ , the difference identity $\begin{aligned} \cos α \cos β + \sin α \sin β = \cos (α - β) \end{aligned}$ , the difference identity $\begin{aligned} \sin α \cos β - \cos α \sin β = \sin (α - β) \end{aligned}$ , the Pythagorean identity, and the fact that the conjugate of $\begin{aligned} \cos θ_{2} + i \sin θ_{2} \end{aligned}$ is $\begin{aligned} \cos θ_{2} - i \sin θ_{2} \end{aligned}$ .
::为了获得将复数分解为极数的一般规则,第一个数字为 r1(cos1+isin1),第二个数字为 r2(cos2+isin2)。这样,产品就可以通过五个事实简化: i21 定义, 差异身份 cossísinsincos(), 区别身份 sinçosççósísin(), Pythagorean 身份, 以及 Cos2+isin2-iscos2-isin2 。

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

:cos2-isin2),r1(cos2-i2-isin2),r1(cos1+isin1),r2(cos2-isin2)(cos2-isin2),r2(cos2),r2(cos2),r2(cos2),(cos2+isin1),r2(cos2),2(cos2-i2-i2),r2(r2),(cos1),2(cos2+sin1sin}2),+i(sin2),(sin2),(cos%2),(sin2)

In general:
::总而言之:

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

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

Use this rule for the computation of two complex numbers divided by one another in the following problems.
::使用此规则计算两个复杂数字,在下列各问题中,两个复杂数字相除。

1. Find the quotient of $\begin{aligned} (\sqrt{3} - i) \div (2 - i 2 \sqrt{3}) \end{aligned}$
::1. 查找(3-i)(2-i23)的商数

Express each number in polar form.
::以极形表示的每个表达号。

$\begin{aligned} \sqrt{3} - i & 2 - i 2 \sqrt{3} \\ r_{1} = \sqrt{x^{2} + y^{2}} & r_{2} = \sqrt{x^{2} + y^{2}} \\ r_{1} = \sqrt{(\sqrt{3})^{2} + (- 1)^{2}} & r_{2} = \sqrt{(2)^{2} + (- 2 \sqrt{3})^{2}} \\ r_{1} = \sqrt{4} = 2 & r_{2} = \sqrt{16} = 4 \end{aligned}$

::3-i2-i23r1=x2+y2r2=x2+y2r2=x2+y2r1=(3)2+(-1)2r2=(2)2+(--232r1=4=2r2=16=4)

$\begin{aligned} \frac{r_{1}}{r_{2}} = .5 \\ θ_{1} = \tan^{- 1} (\frac{- 1}{\sqrt{3}}) & θ_{2} = \tan^{- 1} (\frac{- 2 \sqrt{3}}{2}) & θ = θ_{1} - θ_{2} \\ θ_{1} = 5.75959 r a d . & θ_{2} = 5.23599 r a d . & θ = 5.75959 - 5.23599 \\ θ = 0.5236 \end{aligned}$

::r2=.51=tan-1(-132=tan-1(-232125.75959 rad.2=5.23599 rad.5.75959-5.235990.5236)

Now, plug in what we found to the Quotient Theorem.
::现在,插插我们找到的引文理论。

$\begin{aligned} \frac{r_{1}}{r_{2}} [\cos (θ_{1} - θ_{2}) + i \sin (θ_{1} - θ_{2})] = .5 (\cos 0.5236 + i \sin 0.5236) \end{aligned}$

::r1r2[cos(12)+isin(12)=.5(cos0.5236+isin0.5236)

2. Find the quotient of the two complex numbers $\begin{aligned} 28 ∠ 35^{\circ} \end{aligned}$ and $\begin{aligned} 14 ∠ 24^{\circ} \end{aligned}$
::2. 找出两个复合体的商数,编号2835和1424。

$\begin{aligned} For 28 ∠ 35^{\circ} & For 14 ∠ 24^{\circ} & \frac{r_{1}}{r_{2}} = \frac{28}{14} = 2 \\ r_{1} = 28 & r_{2} = 14 & θ = θ_{1} - θ_{2} \\ θ_{1} = 35^{\circ} & θ_{2} = 24^{\circ} & θ = 35^{\circ} - 24^{\circ} = 11^{\circ} \end{aligned}$

::28=35=35=24=35=24=24=24=24=11=14=2r1=28r2=14=14=2r1=28r2=14

$\begin{aligned} \frac{r_{1} ∠ θ_{1}}{r_{2} ∠ θ_{2}} & = \frac{r_{1}}{r_{2}} ∠ (θ_{1} - θ_{2}) \\ = 2 ∠ 11^{\circ} \end{aligned}$

::112212122221111111111121121112111211211212121112121111121211122111221212212211122111212121222222222222222222222222222222222211111111111111111111111111111111111111111111111111111111111111111111112

3. Using the Quotient Theorem determine $\begin{aligned} \frac{1}{4 c i s \frac{π}{6}} \end{aligned}$
::3. 使用引引论确定 14Cis_6

Even though 1 is not a complex number , we can still change it to polar form.
::即使1不是一个复杂的数字,我们仍然可以把它变成极形。

$\begin{aligned} 1 \to x = 1, y = 0 \end{aligned}$

::1x=1,y=0

$\begin{aligned} r & = \sqrt{1^{2} + 0^{2}} = 1 \\ \tan θ & = \frac{0}{1} = 0 \to θ = 0^{\circ} \end{aligned}$

::r=12+02=1tan@01=0 @%0 @%

$\begin{aligned} So, \frac{1}{4 c i s \frac{π}{6}} = \frac{1 c i s 0}{4 c i s \frac{π}{6}} = \frac{1}{4} c i s (0 - \frac{π}{6}) = \frac{1}{4} c i s (- \frac{π}{6}) . \end{aligned}$

::所以,14cis=14cis=14cis=14cis=14cis=14cis=14cis=16。

Examples
::实例

Example 1
::例1

Earlier, you were given two complex numbers in polar form and was asked to divide them.
::早些时候,你得到两个复杂的数字以极形形式, 并被要求分割它们。

You know that the 2 numbers to divide are $\begin{aligned} 2 (\cos \frac{π}{3} + i \sin \frac{π}{3}) \end{aligned}$ and $\begin{aligned} 3 (\cos π + i \sin π) \end{aligned}$ .
::你知道要分开的2个数字是2(cos3+isin}3)和3(cosisin})。

If you consider $\begin{aligned} r_{1} = 2, r_{2} = 3, θ_{1} = \frac{π}{3}, θ_{2} = π \end{aligned}$ , you can use the formula:
::如果您考虑 r1 = 2,r2 = 3, 1, 3, 2 , 您可以使用公式 :

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

Substituting values into this equation gives:
::将数值替换为此方程给出的:

$\begin{aligned} \frac{2}{3} [\cos (\frac{π}{3} - π) + i \sin (\frac{π}{3} - π)] \\ = \frac{2}{3} [\cos (- \frac{2 π}{3}) + i \sin (- \frac{2 π}{3}) \\ = \frac{2}{3} ((- \frac{1}{2}) + i (- \frac{\sqrt{3}}{2})) \\ = - \frac{1}{3} - i \frac{\sqrt{3}}{3} \end{aligned}$

::23[cos(3)+isin(3)]=23[cos(−23)+isin(−23)=23(−12)+i(−32) 13-i33

Example 2
::例2

Divide the following complex numbers. If they are not in polar form, change them before dividing.
::将以下的复数除以。如果它们不是极形的,在除以之前先修改它们。

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

$\begin{aligned} \frac{2 ∠ 56^{\circ}}{7 ∠ 113^{\circ}} = \frac{2}{7} ∠ (56^{\circ} - 113^{\circ}) = \frac{2}{7} ∠ - 57^{\circ} \end{aligned}$

Example 3
::例3

Divide the following complex numbers. If they are not in polar form, change them before dividing.
::将以下的复数除以。如果它们不是极形的,在除以之前先修改它们。

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

$\begin{aligned} \frac{10 (\cos \frac{5 π}{3} + i \sin \frac{5 π}{3})}{5 (\cos π + i \sin π)} & = 2 (\cos (\frac{5 π}{3} - π) + i \sin (\frac{5 π}{3} - π)) \\ = 2 (\cos \frac{2 π}{3} + i \sin \frac{2 π}{3}) \end{aligned}$

::10(cos53+isin53)5(cos53})=2(cos(53)+isin(533))=2(cos23+isin23)。

Example 4
::例4

Divide the following complex numbers. If they are not in polar form, change them before dividing.
::将以下的复数除以。如果它们不是极形的,在除以之前先修改它们。

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

$\begin{aligned} \frac{2 + 3 i}{- 5 + 11 i} \to change each to polar . \end{aligned}$
$\begin{aligned} x = 2, y = 3 & x = - 5, y = 11 \\ r = \sqrt{2^{2} + 3^{2}} = \sqrt{13} \approx 3.61 & r = \sqrt{(- 5)^{2} + 11^{2}} = \sqrt{146} \approx 12.08 \\ \tan θ = \frac{3}{2} \to θ = {56.31}^{\circ} & \tan θ = - \frac{11}{5} \to θ = {114.44}^{\circ} \end{aligned}$
$\begin{aligned} \frac{3.61}{12.08} ∠ ({56.31}^{\circ} - {114.44}^{\circ}) = 0.30 ∠ - {58.13}^{\circ} \end{aligned}$
::2+3i-5+11i 改变为极点.x=2,y=3x5,y=11r=22+32=133.61r=(-5)2+112=146}12.08tan3256.3111511414443.6116.08(56.31114.44)=0.3058.13_____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Review
::回顾

Divide each pair of complex numbers. If they are not in trigonometric form, change them before dividing.
::将每对复杂数字分开。如果它们不是三角形的, 请在分解前先更改它们。

$\begin{aligned} \frac{3 (\cos 32^{\circ} + i \sin 32^{\circ})}{2 (\cos 15^{\circ} + i \sin 15^{\circ})} \end{aligned}$
::3 (cos) (cos) (cos) (cos) (15) (15) () (a) (c) (b) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (a) (c) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a)

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

$\begin{aligned} \frac{4 (\cos 45^{\circ} + i \sin 45^{\circ})}{8 (\cos 62^{\circ} + i \sin 62^{\circ})} \end{aligned}$
::4 (cos4545458) 8 (cos626262)

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

$\begin{aligned} \frac{5 (\cos 25^{\circ} + i \sin 25^{\circ})}{2 (\cos 115^{\circ} + i \sin 115^{\circ})} \end{aligned}$
::5 (cos2525252(cos115115115))

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

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

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

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

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

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

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

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

Can you divide a pair of complex numbers in standard form without converting to trigonometric form? How?
::您能否在不转换为三角形的情况下, 以标准格式分割一对复杂数字? 如何 ?

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。

Section outline

Quotient Theorem ::引引理论论

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)

Quotient Theorem
::引引理论论

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)