课程： 6.12 德莫伊夫雷的定理

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选择节Demoivre 的定理

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Demoivre 的定理
Imagine that you are in math class one day, and you are given the following number:
::想象一下你有一天会上数学课, 你会得到以下数字:

$\begin{aligned} [5 (\cos 135^{\circ} + i \sin 135^{\circ})]^{4} \end{aligned}$
::[5(cos135isin1354]

Your instructor wants you to find this number. Can you do it? How long will it take you? Probably a long time, if you want to take a number to the fourth power, you'd have to multiply the number by itself, over and over again.
::您的教官希望您找到这个号码。您可以做到吗? 您需要多长时间? 您需要多长时间? 可能要很长时间, 如果您想将一个号码转到第四权力, 您必须自己一次一次一次地将数字乘以。

De Moivre's Theorem
::迪夫雷的定理

The basic operations of addition, subtraction, multiplication and division of can all be carried out, albeit with some changes in form from what you may have seen with numbers having only real components. The addition and subtraction of complex numbers lent themselves best to numbers expressed in standard form. However multiplication and division were easily performed when were in polar form . Another operation that is performed using the is the process of raising a complex number to a power.
::增加、减法、乘法和分法的基本操作都可以进行,尽管与你可能看到的只有实际组件的数字相比,形式上有一些变化。复杂数字的增减最符合标准格式的数字。不过,在极化时,很容易进行乘法和分法。使用这一方法进行的另一项操作是将一个复杂数字提高到一个电源的过程。

The is $\begin{aligned} r (\cos θ + i \sin θ) \end{aligned}$ . If we allow $\begin{aligned} z \end{aligned}$ to equal the polar form of a complex number, it is very easy to see the development of a pattern when raising a complex number in polar form to a power. To discover this pattern, it is necessary to perform some basic multiplication of complex numbers in polar form.
::r( cosisin) 。如果我们允许 z 等同一个复杂数字的极性形式, 很容易看到在以极性形式将一个复杂数字增殖到一个电源时, 模式的发展。要发现这个模式, 就需要以极性形式对复杂数字进行某些基本的乘法。

If $\begin{aligned} z = r (\cos θ + i \sin θ) \end{aligned}$ and $\begin{aligned} z^{2} = z \cdot z \end{aligned}$ then:
::如果z=r(cosisin)和z2=zz,那么:

$\begin{aligned} z^{2} & = r (\cos θ + i \sin θ) \cdot r (\cos θ + i \sin θ) \\ z^{2} & = r^{2} [\cos (θ + θ) + i \sin (θ + θ)] \\ z^{2} & = r^{2} (\cos 2 θ + i \sin 2 θ) \end{aligned}$

::z2=r2(cosisin),z2=r2[cos()+isin(),z2=r2(cos2),

Likewise, if $\begin{aligned} z = r (\cos θ + i \sin θ) \end{aligned}$ and $\begin{aligned} z^{3} = z^{2} \cdot z \end{aligned}$ then:
::同样,如果z=r(cosisin)和z3=z2z,那么:

$\begin{aligned} z^{3} & = r^{2} (\cos 2 θ + i \sin 2 θ) \cdot r (\cos θ + i \sin θ) \\ z^{3} & = r^{3} [\cos (2 θ + θ) + i \sin (2 θ + θ)] \\ z^{3} & = r^{3} (\cos 3 θ + i \sin 3 θ) \end{aligned}$

::z3=r3[cos22222222333]

Again, if $\begin{aligned} z = r (\cos θ + i \sin θ) \end{aligned}$ and $\begin{aligned} z^{4} = z^{3} \cdot z \end{aligned}$ then
::再次,如果z=r(cosisin)和z4=z3z

$\begin{aligned} z^{4} = r^{4} (\cos 4 θ + i \sin 4 θ) \end{aligned}$

::z4=r4(cos44isin4)

These examples suggest a general rule valid for all powers of $\begin{aligned} z \end{aligned}$ , or $\begin{aligned} n \end{aligned}$ . We offer this rule and assume its validity for all $\begin{aligned} n \end{aligned}$ without formal proof, leaving that for later studies. The general rule for raising a complex number in polar form to a power is called De Moivre’s Theorem, and has important applications in engineering, particularly circuit analysis. The rule is as follows:
::这些例子表明了一条适用于z或n等所有权力的一般规则。我们提出这一规则,并在没有正式证据的情况下假定它对所有n等权力的有效性,留待以后的研究研究。将极形的复杂数字提高到一个权力的一般规则叫做德莫伊弗雷的理论,在工程学,特别是电路分析方面有着重要的应用。规则如下:

$\begin{aligned} z^{n} = [r (\cos θ + i \sin θ)]^{n} = r^{n} (\cos n θ + i \sin n θ) \end{aligned}$

::zn=[r(cosisin)]n(cosnisinn)

Where $\begin{aligned} z = r (\cos θ + i \sin θ) \end{aligned}$ and let $\begin{aligned} n \end{aligned}$ be a positive integer.
::z=r(cosisin) 并让 n 成为正整数。

Notice what this rule looks like geometrically. A complex number taken to the $\begin{aligned} n \end{aligned}$ th power has two motions: First, its distance from the origin is taken to the $\begin{aligned} n \end{aligned}$ th power; second, its angle is multiplied by $\begin{aligned} n \end{aligned}$ . Conversely, the roots of a number have angles that are evenly spaced about the origin.
::注意这个规则的几何性质。进入 nth 权力的复杂数字有两个动作: 第一, 其与起源的距离被带到 nth 权力; 第二, 其角度乘以 n。相反, 数字的根部对起源有均匀的角间距。

Let's use De Moivre's Theorem to solve the following problems
::让我们利用德莫伊夫雷的理论来解决以下问题

1. Find $\begin{aligned} [2 (\cos 120^{\circ} + i \sin 120^{\circ})]^{5} \end{aligned}$
::1. 查找[2(cos120isin1205)]

$\begin{aligned} θ = 120^{\circ} = \frac{2 π}{3} rad \end{aligned}$ , using De Moivre’s Theorem:
::12023 rad,使用德莫伊夫雷的理论:

$\begin{aligned} z^{n} = [r (\cos θ + i \sin θ)]^{n} & = r^{n} (\cos n θ + i \sin n θ) \\ [2 (\cos 120^{\circ} + i \sin 120^{\circ})]^{5} & = 2^{5} [\cos 5 \frac{2 π}{3} + i \sin 5 \frac{2 π}{3}] \\ = 32 (\cos \frac{10 π}{3} + i \sin \frac{10 π}{3}) \\ = 32 (- \frac{1}{2} + - \frac{i \sqrt{3}}{2}) \\ = - 16 - 16 i \sqrt{3} \end{aligned}$
2. Find $\begin{aligned} {(- \frac{1}{2} + \frac{i \sqrt{3}}{2})}^{10} \end{aligned}$
::[(cos1201201201205] 5=25[cos523+isin523] =32(cos103+isin103] =32(12i32) 16-16i32。查找(-12+i32) 10。

Change into polar form.
::变成极形

$\begin{aligned} r & = \sqrt{x^{2} + y^{2}} & θ = \tan^{- 1} (\frac{\sqrt{3}}{2} \cdot - \frac{2}{1}) = - \frac{π}{3} + π = \frac{2 π}{3} \\ r & = \sqrt{{(\frac{- 1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} \\ r & = \sqrt{\frac{1}{4} + \frac{3}{4}} \\ r & = \sqrt{1} = 1 \end{aligned}$

::rx2+y2+y2} -1(3221)3323r=(- 12)2+( 322)2+( 322r=14+34r=1=1)

The polar form of $\begin{aligned} (- \frac{1}{2} + \frac{i \sqrt{3}}{2}) \end{aligned}$ is $\begin{aligned} 1 (\cos \frac{2 π}{3} + i \sin \frac{2 π}{3}) \end{aligned}$
::极( - 12+i32) 的极形式为 1( cos23+isin23) 。

Now use De Moivre’s Theorem:
::现在使用德莫伊夫雷的定理:

$\begin{aligned} z^{n} = [r (\cos θ + i \sin θ)]^{n} & = r^{n} (\cos n θ + i \sin n θ) \\ {(- \frac{1}{2} + \frac{i \sqrt{3}}{2})}^{10} & = 1^{10} [\cos 10 (\frac{2 π}{3}) + i \sin 10 (\frac{2 π}{3})] \\ {(- \frac{1}{2} + \frac{i \sqrt{3}}{2})}^{10} & = 1 (\cos \frac{20 π}{3} + i \sin \frac{20 π}{3}) \\ {(- \frac{1}{2} + \frac{i \sqrt{3}}{2})}^{10} & = - \frac{1}{2} + i \frac{\sqrt{3}}{2} \to Standard Form \end{aligned}$

::{(cosisin}}}n=rn(cosnisin}}(-12+i32}10=110[cos10(23)+isin10(23)](-12+i32)](-12+i32}10)10=1(cos203+isin203)(-12+i32)10}标准表格

3. Find $\begin{aligned} [3 (\cos 45^{\circ} + i \sin 45^{\circ})]^{3} \end{aligned}$
::3. 查找[[(cos45isin453)]3

$\begin{aligned} θ = 45^{\circ} = \frac{π}{4} rad \end{aligned}$ , using De Moivre’s Theorem:
::使用De Moivre的定理:

$\begin{aligned} z^{n} = [r (\cos θ + i \sin θ)]^{n} & = r^{n} (\cos n θ + i \sin n θ) \\ [3 (\cos 45^{\circ} + i \sin 45^{\circ})]^{3} & = 3^{3} [\cos 3 \frac{π}{4} + i \sin 3 \frac{π}{4}] \\ = 27 (\cos \frac{3 π}{4} + i \sin \frac{3 π}{4}) \\ = 27 (- \frac{\sqrt{2}}{2} + \frac{i \sqrt{2}}{2}) \\ = - 13.5 + 13.5 i \sqrt{2} \end{aligned}$

::-=============================================================================================================================================================================================================================================================================================================================================================================================================================================================================================================================

Examples
::实例

Example 1
::例1

Earlier, you were asked to solve $\begin{aligned} [5 (\cos 135^{\circ} + i \sin 135^{\circ})]^{4} \end{aligned}$ .
::早些时候,你被要求解决 [5(cos135isin1354)]4。

In this problem, $\begin{aligned} r = 5, θ = \frac{3 π}{4} \end{aligned}$ .
::在这个问题上, r=5, 34。

Using
$\begin{aligned} z^{n} = [r (\cos θ + i \sin θ)]^{n} = r^{n} (\cos n θ + i \sin n θ), \end{aligned}$
you can substitute to get:
::使用zn=[r(cosisin)]nrn(cosnisinnn),您可以替代 :

$\begin{aligned} [5 (\cos \frac{3 π}{4} + i \sin \frac{3 π}{4})]^{4} \\ = 5^{4} (\cos 4 \frac{3 π}{4} + i \sin 4 \frac{3 π}{4}) \\ = 625 (\cos (3 π) + i \sin (3 π) \\ = 625 (- 1 + i 0) \\ = - 625 \end{aligned}$

::[(cos34+isin34]4=54(cos434+isin434)=625(cos(3)+isin(3)=625(1+i0)_625]

Example 2
::例2

Evaluate: $\begin{aligned} {[\frac{\sqrt{2}}{2} (\cos \frac{π}{4} + i \sin \frac{π}{4})]}^{8} \end{aligned}$
::评价:[22(cos4+isin4)]8]

$\begin{aligned} {[\frac{\sqrt{2}}{2} (\cos \frac{π}{4} + i \sin \frac{π}{4})]}^{8} = {(\frac{\sqrt{2}}{2})}^{8} (\cos \frac{8 π}{4} + i \sin \frac{8 π}{4}) = \frac{1}{16} \cos 2 π + \frac{i}{16} \sin 2 π = \frac{1}{16} \end{aligned}$
::[22(cos_4+isin_4)]8=(22)8(cos_8_4+isin_8__4)=116cos_2i16sin_2116

Example 3
::例3

Evaluate: $\begin{aligned} {[3 (\sqrt{3} - i \sqrt{3})]}^{4} \end{aligned}$
::评价:[3(3-i3)]4

$\begin{aligned} {[3 (\sqrt{3} - i \sqrt{3})]}^{4} & = (3 \sqrt{3} - 3 i \sqrt{3})^{4} \\ r & = \sqrt{(3 \sqrt{3})^{2} + (3 \sqrt{3})^{2}} = 3 \sqrt{6}, \tan θ = \frac{3 \sqrt{3}}{3 \sqrt{3}} = 1 \to 45^{\circ} \\ = {(3 \sqrt{6} (\cos \frac{π}{4} + i \sin \frac{π}{4}))}^{4} = (3 \sqrt{6})^{4} (\cos \frac{4 π}{4} + i \sin \frac{4 π}{4}) \\ = 81 (36) [- 1 + i (0)] = - 2936 \end{aligned}$

::[(3-3-i3)]4=(33-33-34r=(332+(333)2=36,tan*33=145(36(cos4+isin4)4)4=(364(cos4+isin4)4=81(36)[-1+i(0)]____________________________________________________________________________________________________________________________________________________________________________________(36[1+i(0)]________________________________________(36(36(36)(4)(4)(4)(4)(36(c___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Example 4
::例4

Evaluate: $\begin{aligned} (\sqrt{5} - i)^{7} \end{aligned}$
::评价5-i)7

$\begin{aligned} (\sqrt{5} - i)^{7} \to r & = \sqrt{(\sqrt{5})^{2} + (- 1)^{2}} = \sqrt{6}, \tan θ = - \frac{1}{\sqrt{5}} \to θ = {335.9}^{\circ} \\ [\sqrt{6} (\cos {335.9}^{\circ} + i \sin {335.9}^{\circ})]^{7} & = (\sqrt{6})^{7} (\cos (7 \cdot {335.9}^{\circ}) + i \sin (7 \cdot {335.9}^{\circ})) \\ = 216 \sqrt{6} (\cos {2351.3}^{\circ} + i \sin {2351.3}^{\circ}) \\ = 216 \sqrt{6} (- 0.981 - 0.196 i) \\ = - 519.04 - 103.7 i \end{aligned}$

:5-i)7r=(5)2+(-1-1)2=6,tann153335.9335.99)7=(6)7(cos(7335.9)+isin(7335.9))=2166(cos2351.332351.33))=2166(0.981-0.96i)519.04-103.7i

Review
::回顾

Use DeMoivre's Theorem to evaluate each expression. Write your answer in standard form.
::使用 DeMoivre 的定理来评估每个表达式。以标准格式写入您的答复。

$\begin{aligned} (\cos \frac{π}{4} + i \sin \frac{π}{4})^{3} \end{aligned}$
:cos+4+isin+4)3 (cos+4+isin+4)3

$\begin{aligned} [2 (\cos \frac{π}{6} + i \sin \frac{π}{6})]^{2} \end{aligned}$
::[2(cos6+isin6)]2

$\begin{aligned} [3 (\cos \frac{3 π}{2} + i \sin \frac{π}{2})]^{5} \end{aligned}$
::[(cos32+isin2]5]

$\begin{aligned} (1 + i)^{5} \end{aligned}$
:1+一)5

$\begin{aligned} (1 - \sqrt{3} i)^{3} \end{aligned}$
:1-3)(一)3

$\begin{aligned} (1 + 2 i)^{6} \end{aligned}$
:1+2i)6

$\begin{aligned} (\sqrt{3} - i)^{5} \end{aligned}$
:3-1)5

$\begin{aligned} (\frac{1}{2} + \frac{i \sqrt{3}}{2})^{4} \end{aligned}$
:12+i32)4

$\begin{aligned} [3 (\cos \frac{π}{4} + i \sin \frac{π}{4})]^{5} \end{aligned}$
::[(cos4+isin4]5]

$\begin{aligned} (2 - \sqrt{5} i)^{5} \end{aligned}$
:2-5i)5

$\begin{aligned} (\sqrt{2} + \sqrt{2} i)^{4} \end{aligned}$
:2+2i)4

$\begin{aligned} [2 (\cos \frac{π}{12} + i \sin \frac{π}{12})]^{8} \end{aligned}$
::[2(cos12+isin12]8]

$\begin{aligned} (- 1 + \sqrt{2} i)^{6} \end{aligned}$
:-1+2i)6

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

$\begin{aligned} (3 - 4 i)^{6} \end{aligned}$
:3-4)(一)6

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

常规

Demoivre 的定理

De Moivre's Theorem ::迪夫雷的定理

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)

De Moivre's Theorem
::迪夫雷的定理

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)