6.14 使用DeMoivre理论原理的等同

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• 选择节使用 DeMoivre 定理的公式

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  使用 DeMoivre 定理的公式

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  You are given an equation in math class:
  ::您在数学课中被给定一个方程 :

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  $\begin{array}{r}{x}^4=16\end{array}$
  ::x4=16

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  and asked to solve for "x". "Excellent!" you say. "This should be easy. The answer is 2."
  ::求 解 答 "x" . 你说 、 "很好" . 你说: "这应该容易 . 答案是 2 。"

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  "Not quite so fast," says your instructor.
  ::老师说"不太快"

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  "I want you to find the complex roots as well!"
  ::"我想让你也找到复杂的根源!"

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  Equations Using De Moivre's Theorem 
  ::使用德莫伊夫尔理论原理的公式

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  We've already seen equations that we would like to solve. However, up until now, these equations have involved solutions that were real numbers. However, there is no reason that solutions need to be limited to the real number line. In fact, some equations cannot be solved completely without the use of . Here we'll explore a little more about complex numbers as solutions to equations.
  ::我们已经看到过我们想要解决的方程式。 但是,直到现在, 这些方程式都包含的是真实数字的解决方案。 但是, 我们没有理由将解决方案限制在真实数字行。 事实上, 有些方程式不能完全解决, 没有使用 。 我们将在这里探索更多关于复杂数字的解决方案, 作为方程式的解决方案 。

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  The roots of a complex number are cyclic in nature. This means that when the roots are plotted on the complex plane , the $\begin{array}{r}{n}^{th}\end{array}$ roots are equally spaced on the circumference of a circle.
  ::复杂数字的根部是周期性的,这意味着当根部在复杂的平面上绘制时,nth根部在圆圈的环绕上同样地划线。

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  Since you began Algebra, solving equations has been an extensive topic. Now we will extend the rules to include complex numbers. The easiest way to explore the process is to actually solve an equation. The solution can be obtained by using De Moivre’s Theorem.
  ::自您开始代数以来,解析方程式是一个广泛的话题。 现在我们将把规则扩展至包含复杂数字。 最简单的探索方法就是实际解析方程式。 解决方案可以通过使用德莫伊夫雷的理论来获得。

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  Using De Moivre's Theorem 
  ::使用De Moivre的定理

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  1. Consider the equation $\begin{array}{r}{x}^5-32=0\end{array}$ . The solution is the same as the solution of $\begin{array}{r}{x}^5=32\end{array}$ . In other words, we must determine the fifth roots of 32.
  ::1. 考虑方程 x5-32=0。 解决办法与X5=32的解决方法相同。 换言之,我们必须确定32的第五根。

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  $$\begin{array}{rl}{x}^5-32& =0\text{and}{x}^5=32.\\ r& =\sqrt{x^2+y^2}\\ r& =\sqrt{(32{)}^2+(0{)}^2}\\ r& =32\\ \theta & ={\tan }^{-1}\left(\frac{0}{32}\right)=0\end{array}$$

  
  ::x5-32=0和 x5=32.r=x2+y2r=(32)2+(0)2r=321(032)=0

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  Write an expression for determining the fifth roots of $\begin{array}{r}32=32+0i\end{array}$
  ::写入用于确定 32=32+0i 第五根的表达式

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  $$\begin{array}{rl}{32}^{\frac{1}{5}}& =[32(\cos (0+2\pi k)+i\sin (0+2\pi k){]}^{\frac{1}{5}}\\ & =2\left(\cos \frac{2\pi k}{5}+i\sin \frac{2\pi k}{5}\right)k=0,1,2,3,4\\ x_1& =2\left(\cos \frac{0}{5}+i\sin \frac{0}{5}\right)\to 2(\cos 0+i\sin 0)=2& & fork=0\\ x_2& =2\left(\cos \frac{2\pi }{5}+i\sin \frac{2\pi }{5}\right)\approx 0.62+1.9i& & fork=1\\ x_3& =2\left(\cos \frac{4\pi }{5}+i\sin \frac{4\pi }{5}\right)\approx -1.62+1.18i& & fork=2\\ x_4& =2\left(\cos \frac{6\pi }{5}+i\sin \frac{6\pi }{5}\right)\approx -1.62-1.18i& & fork=3\\ x_5& =2\left(\cos \frac{8\pi }{5}+i\sin \frac{8\pi }{5}\right)\approx 0.62-1.9i& & fork=4\end{array}$$

  2. Solve the equation $\begin{array}{r}{x}^3-27=0\end{array}$ .
  ::3215=[32(cos=0+2QK)+isin(0+2QK)]15=2(2(cos2QK5+isin2QK5+2QK5K=0,1,2,3,4x1=2(cos05+isin05)=2(cos0+isin0)=2k=0x2=2(cos22+5+isin}}}0.62+1.9iffor k=1x3=2(cos4}5+isin}5+isin}}}}=2(cos+isin}0=2)=0.62+1.9iffor k=42. 解说方程式 x3-27=0。

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  This is the same as the equation $\begin{array}{r}{x}^3=27\end{array}$ .
  ::这与方程式 x3=27 相同 。

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  $$\begin{array}{r}{x}^3=27\\ r& =\sqrt{x^2+y^2}\\ r& =\sqrt{(27{)}^2+(0{)}^2}\\ r& =27\\ \theta & ={\tan }^{-1}\left(\frac{0}{27}\right)=0\end{array}$$

  
  ::x3=27r=x2+y2r=(27)2+(0)2r=27}tan-1(027)=0

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  Write an expression for determining the cube roots of $\begin{array}{r}27=27+0i\end{array}$
  ::写入用于确定 27=27+0i 的立方根的表达式

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  $$\begin{array}{rl}{27}^{\frac{1}{3}}& =[27(\cos (0+2\pi k)+i\sin (0+2\pi k){]}^{\frac{1}{3}}\\ & =3\left(\cos \frac{2\pi k}{3}+i\sin \frac{2\pi k}{3}\right)k=0,1,2\\ x_1& =3\left(\cos \frac{0}{3}+i\sin \frac{0}{3}\right)\to 3(\cos 0+i\sin 0)=3& & fork=0\\ x_2& =3\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)\approx -1.5+2.6i& & fork=1\\ x_3& =3\left(\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}\right)\approx -1.5-2.6i& & fork=2\end{array}$$

  
  ::2713=[27(cos(0+2k)+isin(0+2k)]13=3(cos2K3+isin2K3)k=0,1,2x1=3(cos03+isin03))}3(cos0+isin0)=3(k=0x2=3(cos23+isin23) 1.5+2.6iffor k=1x3=3(cos43+isin43) 1.5-2.6ifor k=2

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  3. Solve the equation $\begin{array}{r}{x}^4=1\end{array}$
  ::3. 解决方程 x4=1

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  $$\begin{array}{r}{x}^4=1\\ r& =\sqrt{x^2+y^2}\\ r& =\sqrt{(1{)}^2+(0{)}^2}\\ r& =1\\ \theta & ={\tan }^{-1}\left(\frac{0}{1}\right)=0\end{array}$$

  
  ::x4=1r=x2+y2r=(1)2+(0)2r=1}111(01)=0

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  Write an expression for determining the cube roots of $\begin{array}{r}1=1+0i\end{array}$
  ::写入用于确定 1=1+0i 的立方根的表达式

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  $$\begin{array}{rl}{1}^{\frac{1}{4}}& =[1(\cos (0+2\pi k)+i\sin (0+2\pi k){]}^{\frac{1}{4}}\\ & =1\left(\cos \frac{2\pi k}{4}+i\sin \frac{2\pi k}{4}\right)k=0,1,2,3\\ x_1& =1\left(\cos \frac{0}{4}+i\sin \frac{0}{4}\right)\to 3(\cos 0+i\sin 0)=1& & fork=0\\ x_2& =1\left(\cos \frac{2\pi }{4}+i\sin \frac{2\pi }{4}\right)=0+i=i& & fork=1\\ x_3& =1\left(\cos \frac{4\pi }{4}+i\sin \frac{4\pi }{4}\right)=-1-0i=-1& & fork=2\\ x_4& =1\left(\cos \frac{6\pi }{4}+i\sin \frac{6\pi }{4}\right)=0-i=-i& & fork=3\end{array}$$

  
  ::114=[1(cos0+2k)+isin(0+2k)]14=1(cos2K4+isin2K4)k=0,1,2,3x1=1(cos04+isin04)}}3(cos0+isin0)=1k=0x2=1(cos24+isin24)=0+i=if k=1x3=1(cos444+isin4}}1-0i1fk=2x4=1(cos64+isin64)=0-iifor k=3

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked to find the complex root of  $\begin{array}{r}{x}^4=16\end{array}$ .
  ::早些时候,你被要求找到x4=16的复杂根。

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  Since you want to find the fourth root of 16, there will be four solutions in all.
  ::既然你想找到16的第四根, 就会有四种解决办法。

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  $$\begin{array}{r}{x}^4=16.\\ r& =\sqrt{x^2+y^2}\\ r& =\sqrt{(16{)}^2+(0{)}^2}\\ r& =16\\ \theta & ={\tan }^{-1}\left(\frac{0}{16}\right)=0\end{array}$$

  
  ::x4=16.r=x2+y2r=(16)2+(0)2r=16}tan -1(016)=0

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  Write an expression for determining the fourth roots of $\begin{array}{r}16=16+0i\end{array}$
  ::写入用于确定 16=16+0i 的第四根的表达式

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  $$\begin{array}{rl}{16}^{\frac{1}{4}}& =[16(\cos (0+2\pi k)+i\sin (0+2\pi k){]}^{\frac{1}{4}}\\ & =2\left(\cos \frac{2\pi k}{4}+i\sin \frac{2\pi k}{4}\right)k=0,1,2,3,4\\ x_1& =2\left(\cos \frac{0}{4}+i\sin \frac{0}{4}\right)\to 2(\cos 0+i\sin 0)=2& & fork=0\\ x_2& =2\left(\cos \frac{2\pi }{4}+i\sin \frac{2\pi }{4}\right)=2i& & fork=1\\ x_3& =2\left(\cos \frac{4\pi }{4}+i\sin \frac{4\pi }{4}\right)=-2& & fork=2\\ x_4& =2\left(\cos \frac{6\pi }{4}+i\sin \frac{6\pi }{4}\right)=-2i& & fork=3\end{array}$$

  
  ::1614 = k= 0x2= 2( cos= 1x3= 2( cos= @ @ 44+isin @ @ 24) k= 0, 1, 2, 34x1= 2( cos @ 04+isin @ 04)\2( cos @ 0+isin=0) = k= 0x2= 2( cos= @ 224+isin4) = 2if k= 1x3=2( cos= @ 44+isin4) k=2=2x4=2( 2( cos= 64+isin_ 64) 2iffor k= 3)

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  Therefore, the four roots of 16 are $\begin{array}{r}2,-2,2i,-2i\end{array}$ . Notice how you could find the two real roots if you seen complex numbers. The addition of the complex roots completes our search for the roots of equations.
  ::因此, 16 的 4 根是 2, 2, 2i, 2- i 。 请注意, 如果您看到复杂的数字, 您如何找到两个真正的根。 添加复杂的根将完成我们对方程根的搜索 。

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  Example 2
  ::例2

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  Rewrite the following in rectangular form : $\begin{array}{r}[2(\cos {315}^{\circ }+i\sin {315}^{\circ }){]}^3\end{array}$
  ::以矩形形式重写以下文字:[2(cos315isin3153]3]

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  $$\begin{array}{rl}r& =2\text{and}\theta ={315}^{\circ }\text{or}\frac{7\pi }{4}.\\ z^n& =[r(\cos \theta +i\sin \theta ){]}^n=r^n(\cos n\theta +i\sin n\theta )\\ z^3& =2^3\left[(\cos 3\left(\frac{7\pi }{4}\right)+i\sin 3\left(\frac{7\pi }{4}\right)\right]\\ z^3& =8\left(\cos \frac{21\pi }{4}+i\sin \frac{21\pi }{4}\right)\\ z^3& =8\left(-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\right)\\ z^3& =-4\sqrt{2}-4i\sqrt{2}\end{array}$$

  
  ::r=2 和 {315} 或 74.zn = [rrr(cos) {cos}}z3=23[(cos}}+isin}3}3}(74)]z3=8(cos=21}4+isin}21}4}z3=8(cos})\4}\4}z3=8(-22～i})z3\\42 -4i2]

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  $\begin{array}{r}\frac{21\pi }{4}\end{array}$ is in the third quadrant so both are negative.
  ::214是第三象限 所以两者都是负的

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  Example 3
  ::例3

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  Solve the equation $\begin{array}{r}{x}^4+1=0\end{array}$ . What shape do the roots make?
  ::解析方程式 x4+1=0。 根会产生什么形状 ?

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  $$\begin{array}{rlrl}& x^4+1=0& & r=\sqrt{x^2+y^2}\\ & x^4=-1& & r=\sqrt{(-1{)}^2+(0{)}^2}\\ & x^4=-1+0i& & r=1\\ & \theta ={\tan }^{-1}\left(\frac{0}{-1}\right)+\pi =\pi \end{array}$$

  
  ::x4+1=0r=x2+y2x41r=( - 1)2+( 0)2x41+0ir=111( 0- 1) *}

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  Write an expression for determining the fourth roots of $\begin{array}{r}{x}^4=-1+0i\end{array}$
  ::写入用于确定 x4\\\\% 1+0i 第四个根的表达式

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  $$\begin{array}{rl}(-1+0i{)}^{\frac{1}{4}}& =[1(\cos (\pi +2\pi k)+i\sin (\pi +2\pi k)){]}^{\frac{1}{4}}\\ (-1+0i{)}^{\frac{1}{4}}& =1^{\frac{1}{4}}\left(\cos \frac{\pi +2\pi k}{4}+i\sin \frac{\pi +2\pi k}{4}\right)\\ x_1& =1\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}& & \text{for}k=0\\ x_2& =1\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)=-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}& & \text{for}k=1\\ x_3& =1\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)=-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}& & \text{for}k=2\\ x_4& =1\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)=\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}& & \text{for}k=3\end{array}$$

  
  :- 1+0i) 14= [1(cos) (2k) +i22for k= 0x2=1(cos34) 22+i22for k=1(cos54) =1(cos24+i22for k=1)(cos334) 22+i22for k=1x3=1(cos54+isin54) 22-i22for k=1(cos74+isin74) =22-i22for k=3)

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  If a line segment is drawn from each root on the polar plane to its adjacent roots, the four roots will form the corners of a square.
  ::如果从极平面的每根根根抽取一条线段到其相邻根,则四根根将形成方形的角。

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  Example 4
  ::例4

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  Solve the equation $\begin{array}{r}{x}^3-64=0\end{array}$ . What shape do the roots make?
  ::解析方程式 x3- 64=0。 根会产生什么样的形状 ?

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  $\begin{array}{r}{x}^3-64=0\to x^3=64+0i\end{array}$
  ::x3-64=0x3=64+0i

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  $\begin{array}{r}64+0i=64(\cos (0+2\pi k)+i\sin (0+2\pi k))\end{array}$
  ::64+0i=64(cos(0+2k)+isin(0+2k))

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  $$\begin{array}{rl}x=(x^3{)}^{\frac{1}{3}}& =(64+0i{)}^{\frac{1}{3}}\\ & =\sqrt[3]{64}\left(\cos \left(\frac{0+2\pi k}{3}\right)+i\sin \left(\frac{0+2\pi k}{3}\right)\right)\end{array}$$

  
  ::x=( x3)13=( 64+0i)13=643( cos( 0+2k3)+isin( 0+2k3))

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  $$\begin{array}{rl}{z}_1& =4\left(\cos \left(\frac{0+2\pi 0}{3}\right)+i\sin \left(\frac{0+2\pi 0}{3}\right)\right)\\ & =4\cos 0+4i\sin 0\\ & =4\text{ for }k=0\end{array}$$

  
  ::z1=4( cos( 0+203)+isin( 0+203))=4cos=0+4isin@0=4 k=0

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  $$\begin{array}{rl}{z}_2& =4\left(\cos \left(\frac{0+2\pi }{3}\right)+i\sin \left(\frac{0+2\pi }{3}\right)\right)\\ & =4\cos \frac{2\pi }{3}+4i\sin \frac{2\pi }{3}=-2+2i\sqrt{3}\text{ for }k=1\end{array}$$

  
  ::z2=4( cos( 0+23) +isin( 0+23)) =4cos23+4isin23+23+2i3 k=1

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  $$\begin{array}{rl}{z}_3& =4\left(\cos \left(\frac{0+4\pi }{3}\right)+i\sin \left(\frac{0+4\pi }{3}\right)\right)\\ & =4\cos \frac{4\pi }{3}+4i\sin \frac{4\pi }{3}\\ & =-2-2i\sqrt{3}\text{ for }k=2\end{array}$$

  
  ::z3=4( cos( 0+43)+isin( 0+43))=4cos43+4isin4}3+4isin4322-2i3 k=2

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  If a line segment is drawn from each root on the polar plane to its adjacent roots, the three roots will form the vertices of an equilateral triangle.
  ::如果从极平面的每个根根抽取一条线段到其相邻根,则三根将形成等边三角形的顶端。

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  Review
  ::回顾

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  Solve each equation.
  ::解决每个方程式

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  1. $\begin{array}{r}{x}^3=1\end{array}$
     ::x3=1
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  2. $\begin{array}{r}{x}^5=1\end{array}$
     ::x5=1
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  3. $\begin{array}{r}{x}^8=1\end{array}$
     ::x8=1
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  4. $\begin{array}{r}{x}^5=-32\end{array}$
     ::x532
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  5. $\begin{array}{r}{x}^4+5=86\end{array}$
     ::x4+5=86
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  6. $\begin{array}{r}{x}^5=-1\end{array}$
     ::x51
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  7. $\begin{array}{r}{x}^4=-1\end{array}$
     ::x41
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  8. $\begin{array}{r}{x}^3=8\end{array}$
     ::x3=8
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  9. $\begin{array}{r}{x}^6=-64\end{array}$
     ::x664
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  10. $\begin{array}{r}{x}^3=-64\end{array}$
      ::x364
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  11. $\begin{array}{r}{x}^5=243\end{array}$
      ::x5=243x5=243
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  12. $\begin{array}{r}{x}^3=343\end{array}$
      ::x3=343xx3=343
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  13. $\begin{array}{r}{x}^7=-128\end{array}$
      ::x7128
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  14. $\begin{array}{r}{x}^{12}=1\end{array}$
      ::x12=1
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  15. $\begin{array}{r}{x}^6=1\end{array}$
      ::x6=1

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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 。