6.13 德莫伊夫雷的理论和nth根

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  Demoivre 的理论和 nth 根

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  You are in math class one day when your teacher asks you to find $\begin{array}{r}\sqrt{3i}\end{array}$ . Are you able to find roots of ? 
  ::有一天,当你的老师要求你找到3i的时候,你上数学课。你能找到根吗?

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

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  Other sections in this course have explored all of the basic operations of arithmetic as they apply to complex numbers in standard form and in polar form . The last discovery is that of taking roots of complex numbers in polar form. Using De Moivre’s Theorem we can develop another general rule – one for finding the $\begin{array}{r}{n}^{th}\end{array}$ root of a complex number written in polar form.
  ::本课程中的其他章节探讨了所有基本的算术操作,因为它们适用于标准形式和极形式的复杂数字。 最后一个发现是以极的形式产生复杂数字。 利用德莫伊弗尔的理论,我们可以制定另一个一般规则 — — 一个用于找到以极形式书写复杂数字的nth根的规则。

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  As before, let
  ::和以前一样,让我们

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  $\begin{array}{r}z=r(\cos \theta +i\sin \theta )\end{array}$
  :cosisin) z=r(cosisin)

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  and let the
  ::并让

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  $\begin{array}{r}{n}^{th}\end{array}$
  ::nnth 北

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  root of
  ::根根

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  $\begin{array}{r}z\end{array}$
  ::z 兹

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  be
  ::be be be 的, 是

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  $\begin{array}{r}v=s(\cos \alpha +i\sin \alpha )\end{array}$
  :cosisin) v=(cosisin)

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  . So, in general,
  ::。因此,一般而言,

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  $\begin{array}{r}\sqrt[n]{z}=v\end{array}$
  ::zn=v

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  and
  ::和

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  $\begin{array}{r}{v}^n=z\end{array}$
  ::vn=z (n=z)

  .

   

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  $$\begin{array}{rl}\sqrt[n]{z}& =v\\ \sqrt[n]{r(\cos \theta +i\sin \theta )}& =s(\cos \alpha +i\sin \alpha )\\ [r(\cos \theta +i\sin \theta ){]}^{\frac{1}{n}}& =s(\cos \alpha +i\sin \alpha )\\ r^{\frac{1}{n}}\left(\cos \frac{1}{n}\theta +i\sin \frac{1}{n}\theta \right)& =s(\cos \alpha +i\sin \alpha )\\ r^{\frac{1}{n}}\left(\cos \frac{\theta }{n}+i\sin \frac{\theta }{n}\right)& =s(\cos \alpha +i\sin \alpha )\end{array}$$

  
  ::zn=vr(cosisin)n=s(cosisin)1n=s(cosisin)r1n(cos1nisin1n)r1n(cosn+isinn)=s(cosisin)

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  From this derivation, we can conclude that $\begin{array}{r}{r}^{\frac{1}{n}}=s\end{array}$ or $\begin{array}{r}{s}^n=r\end{array}$ and $\begin{array}{r}\alpha =\frac{\theta }{n}\end{array}$ . Therefore, for any integer $\begin{array}{r}k(0,1,2,\dots n-1)\end{array}$ , $\begin{array}{r}v\end{array}$ is an $\begin{array}{r}{n}^{th}\end{array}$ root of $\begin{array}{r}z\end{array}$ if $\begin{array}{r}s=\sqrt[n]{r}\end{array}$ and $\begin{array}{r}\alpha =\frac{\theta +2\pi k}{n}\end{array}$ . Therefore, the general rule for finding the $\begin{array}{r}{n}^{th}\end{array}$ roots of a complex number if $\begin{array}{r}z=r(\cos \theta +i\sin \theta )\end{array}$ is: $\begin{array}{r}\sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right)\end{array}$ .
  ::从这一推论中,我们可以得出r1n=s或 sn=r和n。 因此,对于任何整数 k( 0, 1, 2,...n-1) , v 是 z 的 nth 根, 如果 s=rn 和 2kn。 因此, 如果 z=r( cosisin) , 找到复数的 n 根的一般规则是 rn( cos2kn+isin2kn) 。

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  Let's solve the following problems and  leave $\begin{array}{r}\theta \end{array}$ in degrees.
  ::让我们解决以下的问题 然后在度上留下

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  1. Find the two square roots of $\begin{array}{r}2i\end{array}$ .
  ::1. 找出2i的两根根根。

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  Express $\begin{array}{r}2i\end{array}$ in polar form.
  ::2号快车以极地形式出现

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  $$\begin{array}{rlrl}& r=\sqrt{x^2+y^2}& & \cos \theta =0\\ & r=\sqrt{(0{)}^2+(2{)}^2}& & \theta ={90}^{\circ }\\ & r=\sqrt{4}=2\end{array}$$

  
  ::r= rx2+y2cos=0r=( 0)2+(2)2=90@r=4=2

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  $$\begin{array}{r}(2i{)}^{\frac{1}{2}}=2^{\frac{1}{2}}\left(\cos \frac{{90}^{\circ }}{2}+i\sin \frac{{90}^{\circ }}{2}\right)=\sqrt{2}(\cos {45}^{\circ }+i\sin {45}^{\circ })=1+i\end{array}$$

  
  :2)(一)12=212(cos902+isin902)=2(cos45454545)=1+i

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  To find the other root, add $\begin{array}{r}{360}^{\circ }\end{array}$ to $\begin{array}{r}\theta \end{array}$ .
  ::要找到另一根根, 请在 __ 中添加 360 。

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  $$\begin{array}{r}(2i{)}^{\frac{1}{2}}=2^{\frac{1}{2}}\left(\cos \frac{{450}^{\circ }}{2}+i\sin \frac{{450}^{\circ }}{2}\right)=\sqrt{2}(\cos {225}^{\circ }+i\sin {225}^{\circ })=-1-i\end{array}$$

  
  :2)(一)12=212(cos450)2+isin450)2=2(cos225)225)1-i

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  2. Express $\begin{array}{r}-2-2i\sqrt{3}\end{array}$ in polar form:
  ::2. 极形式的快报-2-2-2i3:

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  $$\begin{array}{rl}r& =\sqrt{x^2+y^2}\\ r& =\sqrt{(-2{)}^2+(-2\sqrt{3}{)}^2}\\ r& =\sqrt{16}=4& & \theta ={\tan }^{-1}\left(\frac{-2\sqrt{3}}{-2}\right)=\frac{4\pi }{3}\end{array}$$

  
  ::r=x2+y2r=(-2)2+(-232r=16=4tan-1(-23-2)=43

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  $$\begin{array}{rl}& \sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right)\\ \sqrt[3]{-2-2i\sqrt{3}}& =\sqrt[3]{4}\left(\cos \frac{\frac{4\pi }{3}+2\pi k}{3}+i\sin \frac{\frac{4\pi }{3}+2\pi k}{3}\right)k=0,1,2\end{array}$$

  
  ::rn(cos2kn+isin22kn)-2-2i33=43(cos43+2k3+isin43+2k3k3k=0,1,2)k=0,2

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  $$\begin{array}{rlrl}{z}_1& =\sqrt[3]{4}\left[\cos \left(\frac{4\pi }{9}+\frac{0}{3}\right)+i\sin \left(\frac{4\pi }{9}+\frac{0}{3}\right)\right]& & k=0\\ & =\sqrt[3]{4}\left[\cos \frac{4\pi }{9}+i\sin \frac{4\pi }{9}\right]\\ z_2& =\sqrt[3]{4}\left[\cos \left(\frac{4\pi }{9}+\frac{2\pi }{3}\right)+i\sin \left(\frac{4\pi }{9}+\frac{2\pi }{3}\right)\right]& & k=1\\ & =\sqrt[3]{4}\left[\cos \frac{10\pi }{9}+i\sin \frac{10\pi }{9}\right]\\ z_3& =\sqrt[3]{4}\left[\cos \left(\frac{4\pi }{9}+\frac{4\pi }{3}\right)+i\sin \left(\frac{4\pi }{9}+\frac{4\pi }{3}\right)\right]& & k=2\\ & =\sqrt[3]{4}\left[\cos \frac{16\pi }{9}+i\sin \frac{16\pi }{9}\right]\end{array}$$

  
  ::z1=43[cos(49+03+)+isin(49+03)]k=0=43[cos49+isin49+49]z2=43[cos(49+23)]k=1=43[49+23]k=43[Cos109+isin109+isin109]z3=43[49+43]+isin(49+43)]k=2=43[cos169+isin169]]]

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  In standard form: $\begin{array}{r}{z}_1=0.276+1.563i,z_2=-1.492-0.543i,z_3=1.216-1.02i\end{array}$ .
  ::标准格式:z1=0.276+1.563i,z21.492-0.543i,z3=1.216-1.02i。

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  3. Calculate $\begin{array}{r}{\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)}^{1/3}\end{array}$
  ::3. 计算(cos4+isin4)1/3

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  Using DeMoivres Theorem for fractional powers, we get:
  ::使用DeMoivres定理器进行分数功率,我们得到:

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  $$\begin{array}{r}{\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)}^{1/3}\\ =\cos \left(\frac{1}{3}\times \frac{\pi }{4}\right)+i\sin \left(\frac{1}{3}\times \frac{\pi }{4}\right)\\ =\left(\cos \frac{\pi }{12}+i\sin \frac{\pi }{12}\right)\end{array}$$

  
  :cos4+isin4)1/3=cos(134)+isin(134)=(cos12+isin12)12)

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked to solve  $\begin{array}{r}\sqrt{3i}\end{array}$ .
  ::之前有人叫你去解决三号案

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  Finding the two square roots of $\begin{array}{r}3i\end{array}$ involves first converting the number to polar form:
  ::找到三一的两平方根首先需要将数字转换为极形:

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  For the radius:
  ::半径 :

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  $$\begin{array}{r}r=\sqrt{x^2+y^2}\\ r=\sqrt{(0{)}^2+(3{)}^2}\\ r=\sqrt{9}=3\end{array}$$

  
  ::r=x2+y2r=(0)2+(3)2r=9=3

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  And the angle:
  ::角度 :

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  $$\begin{array}{r}\cos \theta =0\\ \theta ={90}^{\circ }\end{array}$$

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

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  $$\begin{array}{r}(3i{)}^{\frac{1}{2}}=3^{\frac{1}{2}}\left(\cos \frac{{90}^{\circ }}{2}+i\sin \frac{{90}^{\circ }}{2}\right)=\sqrt{3}(\cos {45}^{\circ }+i\sin {45}^{\circ })=\frac{\sqrt{6}}{2}\left(1+i\right)\end{array}$$

  
  :3)(一)12=312(cos902+isin902)=3(cos45454545)=62(1+1)

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  To find the other root, add $\begin{array}{r}{360}^{\circ }\end{array}$ to $\begin{array}{r}\theta \end{array}$ .
  ::要找到另一根根, 请在 __ 中添加 360 。

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  $$\begin{array}{r}(3i{)}^{\frac{1}{2}}=3^{\frac{1}{2}}\left(\cos \frac{{450}^{\circ }}{2}+i\sin \frac{{450}^{\circ }}{2}\right)=\sqrt{3}(\cos {225}^{\circ }+i\sin {225}^{\circ })=\frac{\sqrt{6}}{2}\left(-1-i\right)\end{array}$$

  
  :3)(一)12=312(cos450)2+isin450)2=3(cos225)225)2=62(-1)-(i)

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  Example 2
  ::例2

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  Find $\begin{array}{r}\sqrt[3]{27i}\end{array}$ .
  ::查找273号

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  $$\begin{array}{rlrl}& & & a=0andb=27\\ & \sqrt[3]{27i}=(0+27i{)}^{\frac{1}{3}}& & x=0andy=27\\ & \text{Polar Form}& & r=\sqrt{x^2+y^2}\theta =\frac{\pi }{2}\\ & & & r=\sqrt{(0{)}^2+(27{)}^2}\\ & & & r=27\\ & & & \sqrt[3]{27i}={\left[27\left(\cos (\frac{\pi }{2}+2\pi k)+i\sin (\frac{\pi }{2}+2\pi k)\right)\right]}^{\frac{1}{3}}\text{for }k=0,1,2\\ & & & \sqrt[3]{27i}=\sqrt[3]{27}\left[\cos \left(\frac{1}{3}\right)\left(\frac{\pi }{2}+2\pi k\right)+i\sin \left(\frac{1}{3}\right)\left(\frac{\pi }{2}+2\pi k\right)\right]\text{for }k=0,1,2\\ & & & \sqrt[3]{27i}=3\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)\text{for }k=0\\ & & & \sqrt[3]{27i}=3\left(\cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6}\right)\text{for }k=1\\ & & & \sqrt[3]{27i}=3\left(\cos \frac{9\pi }{6}+i\sin \frac{9\pi }{6}\right)\text{for }k=2\\ & & & \sqrt[3]{27i}=3\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right),3\left(\frac{-\sqrt{3}}{2}+\frac{1}{2}i\right),-3i\end{array}$$

  
  ::a= 0 和 b= 2727i3= ( 0+ 277i3) = 0 和 y= y= 272727i3 = ( 0) 2+ (272r= 272727i3= [27( cos ( 2+ 2k) = k= 0, 1, 227i3= 273 [cos= ( 13) 2( 22} k) +isin ( 13) (2+2k)] k= 0, 1, 227i3= 3( cos6+isin= 6) k= 227i3= 3 (cos56+isin 56) = 127i3= 3( cos96+isin96) k= 227i3= 3( 32+12i) 3( 32+12i) 3( 32+12i) =- 3i

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  Example 3
  ::例3

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  Find the principal root of $\begin{array}{r}(1+i{)}^{\frac{1}{5}}\end{array}$ . Remember the principal root is the positive root i.e. $\begin{array}{r}\sqrt{9}=\pm 3\end{array}$ so the principal root is +3.
  ::查找 (1+i) 15 的主根。 记住主根是正根, 即 93, 所以主根是 +3 。

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  $$\begin{array}{rlrlrl}& r=\sqrt{x^2+y^2}& & \theta ={\tan }^{-1}\left(\frac{1}{1}\right)=\frac{\sqrt{2}}{2}& & \text{Polar Form}=\sqrt{2}cis\frac{\pi }{4}\\ & r=\sqrt{(1{)}^2+(1{)}^2}\\ & r=\sqrt{2}\end{array}$$

  
  ::rr= rx2+y2+y2}11(11)=22 Pollar form=2cis_4r=(1)2+(1)2r=2

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  $$\begin{array}{rl}(1+i{)}^{\frac{1}{5}}& ={\left[\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\right]}^{\frac{1}{5}}\\ (1+i{)}^{\frac{1}{5}}& ={\sqrt{2}}^{\frac{1}{5}}\left[\cos \left(\frac{1}{5}\right)\left(\frac{\pi }{4}\right)+i\sin \left(\frac{1}{5}\right)\left(\frac{\pi }{4}\right)\right]\\ (1+i{)}^{\frac{1}{5}}& =\sqrt[10]{2}\left(\cos \frac{\pi }{20}+i\sin \frac{\pi }{20}\right)\end{array}$$

  
  :1+1)15=[2(cos4+isin4)]15(1+1)15=215[cos(15)(4)+isin(15)(4)](1+1)15=210(cos20+isin20)20]

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  In standard form $\begin{array}{r}(1+i{)}^{\frac{1}{5}}=(1.06+1.06i)\end{array}$ and this is the principal root of $\begin{array}{r}(1+i{)}^{\frac{1}{5}}\end{array}$ .
  ::在标准表格(1+1)15=(1.06+1.06i)中,这是1+115的主要根。

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  Example 4
  ::例4

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  Find the fourth roots of $\begin{array}{r}81i\end{array}$ .
  ::找出81i的第四根根

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  $\begin{array}{r}81i\end{array}$ in polar form is:
  ::极形81i为:

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  $$\begin{array}{rl}& r=\sqrt{0^2+{81}^2}=81,\tan \theta =\frac{81}{0}=und\to \theta =\frac{\pi }{2}81\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)\\ & {\left[81\left(\cos \left(\frac{\pi }{2}+2\pi k\right)+i\sin \left(\frac{\pi }{2}+2\pi k\right)\right)\right]}^{\frac{1}{4}}\\ & 3\left(\cos \left(\frac{\frac{\pi }{2}+2\pi k}{4}\right)+i\sin \left(\frac{\frac{\pi }{2}+2\pi k}{4}\right)\right)\\ & 3\left(\cos \left(\frac{\pi }{8}+\frac{\pi k}{2}\right)+i\sin \left(\frac{\pi }{8}+\frac{\pi k}{2}\right)\right)\\ & z_1=3\left(\cos \left(\frac{\pi }{8}+\frac{0\pi }{2}\right)+i\sin \left(\frac{\pi }{8}+\frac{0\pi }{2}\right)\right)=3\cos \frac{\pi }{8}+3i\sin \frac{\pi }{8}=2.77+1.15i\\ & z_2=3\left(\cos \left(\frac{\pi }{8}+\frac{\pi }{2}\right)+i\sin \left(\frac{\pi }{8}+\frac{\pi }{2}\right)\right)=3\cos \frac{5\pi }{8}+3i\sin \frac{5\pi }{8}=-1.15+2.77i\\ & z_3=3\left(\cos \left(\frac{\pi }{8}+\frac{2\pi }{2}\right)+i\sin \left(\frac{\pi }{8}+\frac{2\pi }{2}\right)\right)=3\cos \frac{9\pi }{8}+3i\sin \frac{9\pi }{8}=-2.77-1.15i\\ & z_4=3\left(\cos \left(\frac{\pi }{8}+\frac{3\pi }{2}\right)+i\sin \left(\frac{\pi }{8}+\frac{3\pi }{2}\right)\right)=3\cos \frac{13\pi }{8}+3i\sin \frac{13\pi }{8}=1.15-2.77i\end{array}$$

  
  ::-2+2k) 143(cos()2+2+22)+西尼(2+2)3(2+24)3(2+22+24)3(2+2+22+24)2)3(2+2+22)2)3(2+2+2)2)3(2+2+2)3(2+2)3(88)810=281(82)810=281(2)8(8-8+2);8(8)8+1(15)2)3(3)+(5)8+3(5)8)8(8+3)8(5)8(15)15(15)+2(9)(2(8-8)2(8-8+2)(2(7-8)(2)(2(8)) (3(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)(8)

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  Review
  ::回顾

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  Find the cube roots of each complex number. Write your answers in standard form.
  ::查找每个复数的立方根。 以标准格式写入您的答复 。

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  1. $\begin{array}{r}8(\cos 2\pi +i\sin 2\pi )\end{array}$
     ::8(cos222222)
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  2. $\begin{array}{r}3(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4})\end{array}$
     ::3(cos%4+isin%4)
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  3. $\begin{array}{r}2(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4})\end{array}$
     ::2(cos34+isin34)
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  4. $\begin{array}{r}(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3})\end{array}$
     :cos%3+isin%3) (cos%3+isin%3)
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  5. $\begin{array}{r}(3+4i)\end{array}$
     :3+4i)
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  6. $\begin{array}{r}(2+2i)\end{array}$
     :2+2i)

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  Find the principal fifth roots of each complex number. Write your answers in standard form.
  ::查找每个复杂数字的主要五根根。 以标准格式写入您的答复 。

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  7. $\begin{array}{r}2(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6})\end{array}$
     ::2(cos6+isin6)
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  8. $\begin{array}{r}4(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2})\end{array}$
     ::4 (cos2+isin2)
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  9. $\begin{array}{r}32(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4})\end{array}$
     ::32(cos%4+isin%4) (cos%4+isin%4))
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  10. $\begin{array}{r}2(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3})\end{array}$
      ::2(cos%3+isin%3)
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  11. $\begin{array}{r}32i\end{array}$
      ::32i 32i
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  12. $\begin{array}{r}(1+\sqrt{5}i)\end{array}$
      :1+5i)
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  13. Find the sixth roots of -64 and plot them on the complex plane.
      ::找出64号的第六根 把它们埋在复杂的平面上
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  14. How many solutions could the equation $\begin{array}{r}{x}^6+64=0\end{array}$ have? Explain.
      ::方程式 x6+64=0 能有多少解决方案? 请解释 。
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  15. Solve $\begin{array}{r}{x}^6+64=0\end{array}$ . Use your answer to #13 to help you.
      ::解析 x6+64=0。 使用您对 # 13 的回复来帮助您 。

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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 。