Section: 莫伊夫雷的理论和Nth根 | 11.4 德莫伊夫雷理论和nth根

Section outline

You know how to multiply two together and you’ve seen the advantages of using trigonometric polar form , especially when multiplying more than two complex numbers at the same time. Because raising a number to a whole number power is repeated multiplication, you also know how to raise a complex number to a whole number power.
::您知道如何将两个乘在一起,您也看到了使用三角极形式的优势,特别是同时乘以两个复杂数字的优势。因为将一个数字加到一个整数的功率是重复乘法,您也知道如何将一个复杂数字加到一个整数的功率。

What is a geometric interpretation of squaring a complex number?
::如何几何解释一个复杂数字的对称?

De Moivre's Theorem and nth Roots
::莫伊夫雷的理论和Nth根

Recall that if $\begin{aligned} z_{1} = r_{1} \cdot cis θ_{1} \end{aligned}$ and $\begin{aligned} z_{2} = r_{2} \cdot cis θ_{2} \end{aligned}$ with $\begin{aligned} r_{2} \neq 0 \end{aligned}$ , then $\begin{aligned} z_{1} \cdot z_{2} = r_{1} \cdot r_{2} \cdot cis (θ_{1} + θ_{2}) \end{aligned}$ .
::回顾,如果z1=r1Icis%1和z2r2Icis%2,加上r2°0,则z1Z2=r1R2Icis(1)2。

If $\begin{aligned} z_{1} = z_{2} = z = r cis θ \end{aligned}$ then you can determine $\begin{aligned} z^{2} \end{aligned}$ and $\begin{aligned} z^{3} \end{aligned}$ :
::如果z1=z2=z=z=rcis ,那么您可以确定z2和z3:

\begin{aligned} z^{2} & = r \cdot r \cdot cis (θ + θ) = r^{2} cis (2 \cdot θ) \\ z^{3} & = r^{3} cis (3 \cdot θ) \end{aligned}

::z2=rrcis()=r2cis(2)z3=r3cis(□)

De Moivre’s Theorem simply generalizes this pattern to the power of any positive integer.
::Deivre 的理论只是将这种模式概括为任何正整数的力量。

$\begin{aligned} z^{n} = r^{n} \cdot cis (n \cdot θ) \end{aligned}$
::zn=破碎

In addition to raising a complex number to a power, you can also take square roots, cube roots and $\begin{aligned} n^{t h} \end{aligned}$ roots of complex numbers. Suppose you have complex number $\begin{aligned} z = r cis θ \end{aligned}$ and you want to take the $\begin{aligned} n^{t h} \end{aligned}$ root of $\begin{aligned} z \end{aligned}$ . In other words, you want to find a number $\begin{aligned} v = s \cdot cis β \end{aligned}$ such that $\begin{aligned} v^{n} = z \end{aligned}$ . Do some substitution and manipulation:
::除了向电源添加复数外, 您还可以从复数的平方根、立方根和 nth 根根。假设您有复数 z=r cis , 您想要从 z 的 nth 根中取出。换句话说, 您想要找到数字 v=sQis β, 这样Vn=z。做一些替换和操作 :

\begin{aligned} v^{n} & = z \\ (s \cdot cis β)^{n} & = r \cdot cis θ \\ s^{n} \cdot cis (n \cdot β) & = r \cdot cis θ \end{aligned}

::vn=z(s-cis β)n=r-cis °sn-cis

=r-cis

You can see at this point that to find $\begin{aligned} s \end{aligned}$ you need to take the $\begin{aligned} n^{t h} \end{aligned}$ root of $\begin{aligned} r \end{aligned}$ . The trickier part is to find the angles, because $\begin{aligned} n \cdot β \end{aligned}$ could be any angle coterminal with $\begin{aligned} θ \end{aligned}$ . This means that there are $\begin{aligned} n \end{aligned}$ different $\begin{aligned} n^{t h} \end{aligned}$ roots of $\begin{aligned} z \end{aligned}$ .
::您可以在这一点上看到, 要找到 s, 您需要选择 r 的 nth 根。最棘手的部分是找到角度, 因为 n 可以是的任何角度的 Coterminal 。这意味着 z 有 n 不同的 nw 根。

\begin{aligned} n \cdot β & = θ + 2 π k \\ β & = \frac{θ + 2 π k}{n} \end{aligned}

::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不

The number $\begin{aligned} k \end{aligned}$ can be all of the counting numbers including zeros up to $\begin{aligned} n - 1 \end{aligned}$ . So if you are taking the $\begin{aligned} 4^{t h} \end{aligned}$ root, then $\begin{aligned} k = 0, 1, 2, 3 \end{aligned}$ .
::数字 k 可以是所有计数数, 包括n- 1 之前的零。所以, 如果您选择了 4 根, 那么 k= 0, 1, 2, 3 。

Thus the $\begin{aligned} n^{t h} \end{aligned}$ root of a complex number requires $\begin{aligned} n \end{aligned}$ different calculations, one for each root:
::因此,复数的 nth 根需要 n 不同的计算, 每个根需要一个 :

$\begin{aligned} v = \sqrt[n]{r} \cdot cis (\frac{θ + 2 π k}{n}) f o r {k ϵ I | 0 \leq k \leq n - 1} \end{aligned}$
::v=rnQICIS (2kn) 用于 {k} {I0k} 的 v=rnQICIS (2kn)

To apply this formula, find the cube root of the number 8. Most students know that $\begin{aligned} 2^{3} = 8 \end{aligned}$ and so know that 2 is the cube root of 8. However, they don’t realize that there are two other cube roots that they are missing. Remember to write out $\begin{aligned} k = 0, 1, 2 \end{aligned}$ and use the unit circle whenever possible to help you to find all three cube roots.
::要应用此公式, 请找到数字 8 的立方根。大多数学生知道 23 = 8 , 所以知道 2 是 8 的立方根。但是, 他们不知道他们缺少了另外两个立方根。记住要写入 k= 0. 1, 2 并尽可能使用单位圆来帮助您找到所有三个立方根。

\begin{aligned} 8 & = 8 cis 0 = (s \cdot cis β)^{3} \\ z_{1} & = 2 \cdot cis (\frac{0 + 2 π \cdot 0}{3}) = 2 cis 0 = 2 (\cos 0 + i \cdot \sin 0) = 2 (1 + 0) = 2 \\ z_{2} & = 2 \cdot cis (\frac{0 + 2 π \cdot 1}{3}) = 2 cis (\frac{2 π}{3}) \\ = 2 (\cos (\frac{2 π}{3}) + i \cdot \sin (\frac{2 π}{3})) = 2 (- \frac{1}{2} + \frac{\sqrt{3}}{2} i) = - 1 + i \sqrt{3} \\ z_{3} & = 2 \cdot cis (\frac{0 + 2 π \cdot 2}{3}) = 2 cis (\frac{4 π}{3}) \\ = 2 (\cos (\frac{4 π}{3}) + i \cdot \sin (\frac{4 π}{3})) = 2 (- \frac{1}{2} - \frac{\sqrt{3}}{2} i) = - 1 - i \sqrt{3} \end{aligned}

::8=8 Cis0 = 2Cis0 = 2Cis0= 2Cis0= 2Cis0= 2Cos0+0+isin=0=0+0=0+0=2Z2=2Cis2=2Cis2=2Cis2(2+23)=2Cis(23)+isin(23))=2(12+32i)=1+i3Z3=2Cis(0+223)=2Cis3=2Cis0=2Z2=2(43)+isin(43))=2Cis(2=2(12-32i) _1-i

The cube roots of 8 are 2, $\begin{aligned} - 1 + i \sqrt{3}, - 1 - i \sqrt{3} \end{aligned}$ .
::8的立方根是 2 - 1+3 - 1 - i3 。

To check, that they are the cube roots, cube them all simplify.
::要检查,它们是立方根, 立方体所有简化。

\begin{aligned} z_{1}^{3} & = 2^{3} = 8 \\ z_{2}^{3} & = {(- 1 + i \sqrt{3})}^{3} \\ = (- 1 + i \sqrt{3}) \cdot (- 1 + i \sqrt{3}) \cdot (- 1 + i \sqrt{3}) \\ = (1 - 2 i \sqrt{3} - 3) \cdot (- 1 + i \sqrt{3}) \\ = (- 2 - 2 i \sqrt{3}) \cdot (- 1 + i \sqrt{3}) \\ = 2 - 2 i \sqrt{3} + 2 i \sqrt{3} + 6 \\ = 8 \end{aligned}

::z13=23=23=8z23=(-1+1)3=(-1+1)3=(-1+3)____(-1+3)____(-1+3)____(-1+3)____(-1+3)=(1-2-3)____(-1+3)____(-1+3)=(-2-2-2-2)____(-1+3)____(-1+3)=2-2i3+23+6=8)

Note how many steps and opportunities there are for making a mistake when multiplying multiple terms in rectangular form . When you check $\begin{aligned} z_{3} \end{aligned}$ , use trigonometric polar form.

\begin{aligned} z_{3}^{3} & = 2^{3} cis (3 \cdot \frac{4 π}{3}) \\ = 8 (\cos 4 π + i \cdot \sin 4 π) \\ = 8 (1 + 0) \\ = 8 \end{aligned}

::注意在以矩形形式乘以多个条件时, 错误的步数和机会有多少。当检查 z3 时, 请使用三角极表。 z33= 23 cis (343) = 8 (cos @ 44 isin4) = 8(1+0) =8

Examples
::实例

Example 1
::例1

Earlier, you were asked what a geometric interpretation of squaring a complex number is. Squaring a complex number produces a new complex number. The angle gets doubled and the magnitude gets squared, so geometrically you see a rotation.
::早些时候,有人问您如何对组合数字进行几何解释。对复杂数字进行几何解释会产生新的复杂数字。角度翻倍,大小平方,因此您可以看到旋转的几何解释。

Example 2
::例2

Plot the roots of 8 graphically and discuss any patterns you notice.
::用图形绘制 8 的根, 并讨论您注意到的任何模式。

The three points are equally spaced around a circle of radius 2. Only one of the points, $\begin{aligned} 2 + 0 i \end{aligned}$ , is made up of only real numbers. The other two points have both a real and an imaginary component which is why they are off of the $\begin{aligned} x \end{aligned}$ axis.
::三个点在半径 2 的圆圈上都一样。只有一个点, 2+0i, 由实际数字组成。另外两个点既有真实的, 也有虚构的成分, 这也是为什么它们不在 x 轴外的原因。

As you become more comfortable with roots, you can just determine the number of points that need to be evenly spaced around a certain radius circle and find the first point. The rest is just logic.
::随着你对根变得更加舒适,你就可以确定在某个半径圆周围需要平均间距的点数,然后找到第一点。剩下的只是逻辑。

Example 3
::例3

What are the fourth roots of $\begin{aligned} 16 cis 48^{\circ} \end{aligned}$ ?
::16CS48的第四根是什么?

There will be 4 points, each $\begin{aligned} 90^{\circ} \end{aligned}$ apart with the first point at $\begin{aligned} 2 cis (12^{\circ}) \end{aligned}$ .
::将会有4个点,每90个点,除第1点2Cis (12个)外。

$\begin{aligned} 2 cis (12^{\circ}), 2 cis (102^{\circ}), 2 cis (192^{\circ}), 2 cis (282^{\circ}) \end{aligned}$
::2cis(12)、2cis(102)、2cis(192)、2cis(282)

Example 4
::例4

Solve for $\begin{aligned} z \end{aligned}$ by finding the nth root of the complex number.
::通过找到复数的 nth 根来解决 z 。

$\begin{aligned} z^{3} = 64 - 64 \sqrt{3} i \end{aligned}$
::z3=64-643i

First write the complex number in $\begin{aligned} cis \end{aligned}$ form. Remember to identify $\begin{aligned} k = 0, 1, 2 \end{aligned}$ . This means the roots will appear every $\begin{aligned} \frac{360^{\circ}}{3} = 120^{\circ} \end{aligned}$ .
::先在 Cis 格式中写入复数。记住要识别 k= 0, 1, 2 。这意味着根将显示每360\\ 3= 120\ 。

\begin{aligned} z^{3} & = 64 - 64 \sqrt{3} i = 128 \cdot cis 300^{\circ} \\ z_{1} & = 128^{\frac{1}{3}} \cdot cis (\frac{300}{3} \circ) = 128^{\frac{1}{3}} \cdot cis (100^{\circ}) \\ z_{2} & = 128^{\frac{1}{3}} \cdot cis (220^{\circ}) \\ z_{3} & = 128^{\frac{1}{3}} \cdot cis (340^{\circ}) \end{aligned}

::z3=64-643i=128-Cis 300-Z1=12813-Cis(3003)=12813-Cis(100)z2=12813-Cis(220)z3=12813-Cis(340)

Example 5
::例5

Use De Moivre’s Theorem to evaluate the following power.
::使用德莫伊夫雷的理论来评估以下权力。

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

First write the number in trigonometric polar form, then apply De Moivre’s Theorem and simplify.
::先以三角极形式写入数字, 然后应用德莫伊夫雷的定理并简化。

\begin{aligned} {(\sqrt{2} - \sqrt{2} i)}^{6} & = (2 cis 315^{\circ})^{6} \\ = 2^{6} \cdot cis (6 \cdot 315^{\circ}) \\ = 64 \cdot cis (1890^{\circ}) \\ = 64 \cdot cis (1890^{\circ}) \\ = 64 \cdot cis (90^{\circ}) \\ = 64 (\cos 90^{\circ} + i \cdot \sin 90^{\circ}) \\ = 64 (0 + i) \\ = 64 i \end{aligned}

2-2i)6=(2cis 315)6=(26cis (6}315)=(64)cis(1890)=(64)cis(1890)=(64)cis(90)=(64)(64)(cos 90)-(i) i(sin)-(90)=(64)(0+(i)=64i)

Summary

De Moivre's Theorem generalizes the pattern of raising a complex number to the power of any positive integer: $\begin{aligned} z^{n} = r^{n} cis (n \cdot θ) \end{aligned}$
::Deivre 的理论理论概括了将一个复杂数字提高到任何正整数的功率的模式:zn=rncis

There are $\begin{aligned} n \end{aligned}$ different nth roots of a complex number, each corresponding to a different angle.
::复数有 n 不同的 nn 根, 每一根对应不同的角度。

To find the nth root of a complex number, use the formula: $\begin{aligned} v = \sqrt[n]{r} \cdot cis (\frac{θ + 2 π k}{n}) for {k \in I | 0 \leq k \leq n - 1} \end{aligned}$
::要找到复数的 nth 根, 请使用公式 : v=rnQis(\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Review
::回顾

Use De Moivre’s Theorem to evaluate each expression. Write your answers in rectangular form.
::使用 De Moivre 的定理来评估每个表达式。以矩形形式写入您的答复。

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

2. $\begin{aligned} {(1 - \sqrt{3} i)}^{3} \end{aligned}$
::2. (1-3i)3

3. $\begin{aligned} (1 + 2 i)^{6} \end{aligned}$
::3. (1+2i)6

4. $\begin{aligned} {(\sqrt{3} - i)}^{5} \end{aligned}$
::4. (3-1)5

5. $\begin{aligned} {(\frac{1}{2} + \frac{i \sqrt{3}}{2})}^{4} \end{aligned}$
::5. (12+i32)4

6. Find the cube roots of $\begin{aligned} 3 + 4 i \end{aligned}$ .
::6. 找出3+4i的立方根。

7. Find the $\begin{aligned} 5^{t h} \end{aligned}$ roots of $\begin{aligned} 32 i \end{aligned}$ .
::7. 找出32i的第5根根。

8. Find the $\begin{aligned} 5^{t h} \end{aligned}$ roots of $\begin{aligned} 1 + \sqrt{5} i \end{aligned}$ .
::8. 找出1+5i的第5根根。

9. Find the $\begin{aligned} 6^{t h} \end{aligned}$ roots of - 64 and plot them on the complex plane.
::9. 找出64的第六根根,在复杂的平面上绘制。

10. Use your answers to #9 to help you solve $\begin{aligned} x^{6} + 64 = 0 \end{aligned}$ .
::10. 使用对 #9 的回答来帮助解析 x6+64=0 。

For each equation: a) state the number of roots, b) calculate the roots, and c) represent the roots graphically.
::每个方程a) 说明根数,(b) 计算根数,(c) 以图形表示根数。

11. $\begin{aligned} x^{3} = 1 \end{aligned}$
::11. x3=1

12. $\begin{aligned} x^{8} = 1 \end{aligned}$
::12. x8=1

13. $\begin{aligned} x^{12} = 1 \end{aligned}$
::13. x12=1

14. $\begin{aligned} x^{4} = 16 \end{aligned}$
::14. x4=16

15. $\begin{aligned} x^{3} = 27 \end{aligned}$
::15. x3=27

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

De Moivre's Theorem and nth Roots ::莫伊夫雷的理论和Nth根

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Review ::回顾

Review (Answers) ::回顾(答复)