4.9 产品和引引理论

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  are found in real world calculations involving: quantum mechanics, signal analysis, fluid dynamics, control theory, and many other fields.
  ::存在于真实世界的计算中,涉及:量子力学、信号分析、流体动力学、控制理论和许多其他领域。

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  In electrical engineering, complex numbers are used for calculations involving impedance (the resistance to electric flow in a circuit).
  ::在电气工程中,复杂数字用于计算阻碍因素(对电路电流的阻力)。

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  Electrical engineers are familiar with the formula:
  ::电气工程师熟悉公式:

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  $\begin{align*} V = V_0 e^{j \omega t} = V_0 \left (\cos \omega t + j \sin\omega t \right )\end{align*}$
  ::V=V0ejt=V0(cost+jsint)

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  by comparing it to the similar expression below, which is explored in this lesson, can you identify the variable j ?
  ::将它与下文的类似表达方式进行比较(这个教训对此作了探讨),你能否辨别变量j?

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  $\begin{align*}r_2 (cos \theta _2 + i sin \theta _2)\end{align*}$
  ::r2(cos%2+isin%2)

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  Product and Quotient Theorems
  ::产品和引引理论

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  The Product Theorem
  ::产品定理

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  Since complex numbers can be transformed to polar form , the multiplication of complex numbers can also be done in polar form. Suppose we know z ₁ = r ₁ ( θ ₁ + i θ ₁ ) z ₂ = r ₂ (cos θ ₂ + i sin θ ₂ )
  ::由于复数可以转换成极形, 复数的倍增也可以以极形形式进行。 假设我们知道 z1 = r1 ( 1 + i 1) z2 = r2( cos 2 + i sin 2)

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  To multiply the two complex numbers in polar form:
  ::以极形乘以两个复杂数字:

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  $\begin{align*}z_1 \cdot z_2 = r_1 (\mbox{cos}\ \theta_1 + i\ \mbox{sin}\ \theta_1) \cdot r_2(\mbox{cos}\ \theta_2 + i\ \mbox{sin}\ \theta_2)\end{align*}$
  ::z2=r1 (cos +1+i sin #1+i sin #1+r2(cos +2+i sin #2)

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  $\begin{align*}= r_1 r_2(\mbox{cos}\ \theta_1 + i\ \mbox{sin}\ \theta_1)(\mbox{cos}\ \theta_2 + i\ \mbox{sin}\ \theta_2)\end{align*}$
  ::=r1r2(cos +1+i sin #1)(cos +2+i sin #2)

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  $\begin{align*}= r_1 r_2 \cdot (\mbox{cos}\ \theta_1\ \mbox{cos}\ \theta_2 + i\ \mbox{cos}\ \theta_1\ \mbox{sin}\ \theta_2 + i\ \mbox{sin}\ \theta_1\ \mbox{cos}\ \theta_2 + i^2\ \mbox{sin}\ \theta_1\ \mbox{sin}\ \theta_2)\end{align*}$
  ::=r1r2(cos #1 cos #2+i cos #2+i cos #1 sin #2+i sin #1 cos #2+i2 sin #2+i2 sin #1 cos #1 sin#2)

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  $\begin{align*}= r_1 r_2 (\mbox{cos}\ \theta_1\ \mbox{cos}\ \theta_2\ + i\ \mbox{cos}\ \theta_1\ \mbox{sin}\ \theta_2 + i\ \mbox{sin}\ \theta_1\ \mbox{cos}\ \theta_2\ - \mbox{sin}\ \theta_1\ \mbox{sin}\ \theta_2)\end{align*}$
  ::=r1r2(cos%1 cos%2 +i coms%1 sin%2 +i spin%2 +i sin%1 scos%2 -sin%1 sin sin%1 sin%2)

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  $\begin{align*}= r_1 r_2\ (\mbox{cos}\ \theta_1\ \mbox{cos}\ \theta_2 - \mbox{sin}\ \theta_1\ \mbox{sin}\ \theta_2 + i\ \mbox{cos}\ \theta_1\ \mbox{sin}\ \theta_2 + i\ \mbox{sin}\ \theta_1\ \mbox{cos}\ \theta_2)\end{align*}$
  ::=r1r2 (cos =1 cos cos =2-sin sin =1 sin spin =2+i cos =1 sin =2+i cos spin =2+i cos 1 sin =2+i sin =2+i sin =2+i sin =2)

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  $\begin{align*}= r_1 r_2\ ([\mbox{cos}\ \theta_1\ \mbox{cos}\ \theta_2 - \mbox{sin}\ \theta_1\ \mbox{sin}\ \theta_2] + i[\mbox{cos}\ \theta_1\ \mbox{sin}\ \theta_2 + \mbox{sin}\ \theta_1\ \mbox{cos}\ \theta_2])\end{align*}$
  ::=r1r2([cos +1 sin =2]+i[cos +1 sin =2])

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  (Use i ² = -1, gather like terms, factor out $\begin{align*}i\end{align*}$ , substitute the angle sum formulas for both sine and cosine)
  :使用 I2 = - 1, 收集类似条件, 系数输出 i, 替换正弦和余弦的角和公式)

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  $\begin{align*}z_1 \cdot z_2 = r_1r_2\ \mbox{cis}\ [(\theta_1 + \theta_2)]\end{align*}$
  ::z1z2=r1r2cis[(12]]

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  This last equation states that the product of two complex numbers in polar form can be obtained by multiplying the polar r values of each of and then multiplying that value by cis of the sum of each of the two angles of the individual complex numbers. This is more concise than the rectangular form for multiplication of complex numbers.
  ::最后一个方程式指出,两个以极化形态的复杂数字的产物可以通过乘以每个复杂数字的极性 r 值获得,然后再乘以该数值,再乘以单个复杂数字两个角度的相和。这比复数倍增的矩形形式更简明。

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  The Quotient Theorem
  ::引引神论

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  Dividing complex numbers in polar form can be shown using a similar proof that was used to show multiplication of complex numbers. Here we omit the proof and give the result. For z ₁ = r ₁ (cos θ ₁ + i sin θ ₁ ) and z ₂ = r ₂ (cos θ ₂ + i sin θ ₂ ), then $\begin{align*}\frac{z_1} {z_2} = \frac{r_1} {r_2} \times \mbox{cis}\ [\theta_1 - \theta_2]\end{align*}$
  ::以极形表示的相异复杂数字可以用类似证据显示,用来显示复杂数字的乘数。在这里,我们省略了证据并给出结果。对于 z1 = r1 (cos + r1 + i sin _1) 和 z2 = r2(cos + i sin _2), 然后z1z2= r1r2xcis [1 _2]

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked to identify the variable j in the following formula:
  ::早些时候,有人要求您在以下公式中确定变量j:

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  $\begin{align*} V = V_0 e^{j \omega t} = V_0 \left (\cos \omega t + j \sin\omega t \right )\end{align*}$
  ::V=V0ejt=V0(cost+jsint)

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  In electrical calculations, the letter I is commonly used to denote current , therefore are identified with a j .
  ::在电算中,通常用字母I表示当前,因此与j相同。

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  Note the similar usage of i in $\begin{align*}r (cos \theta + i sin \theta)\end{align*}$ .
  ::请注意r(cosisin)中i 的类似用法。

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  Example 2
  ::例2

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  Multiply $\begin{align*}z_1 \cdot z_2\end{align*}$  where $\begin{align*}z_1 = 2 + 2i\end{align*}$ and $\begin{align*}z_2 = 1 - \sqrt{3}i\end{align*}$ .
  ::乘以 z1z2, 其中z1=2+2i和z2=1-3i。

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  For z ₁ ,
  ::z1, 用于 z1 。

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  $\begin{align*}r_1 = \sqrt{2^2 + 2^2}\end{align*}$
  ::r1=22+22

  $\begin{align*}= \sqrt{8}\end{align*}$

  $\begin{align*}= 2\sqrt{2}\end{align*}$

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  and
  ::和

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  $\begin{align*}\mbox{tan}\ \theta_1 = \frac{2} {2}\end{align*}$
  ::棕色 #% 1=22

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  $\begin{align*}\mbox{tan}\ \theta_1 = 1\end{align*}$
  ::皮肤 #% 1=1

  $\begin{align*}\theta_1 = \frac{\pi} {4}\end{align*}$

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  Note that $\begin{align*}\theta_1\end{align*}$ is in the first quadrant since a , and b > 0.
  ::请注意,% 1 位于自 a 以来的第一个象限, b > 0 。

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  For $\begin{align*}z_2\end{align*}$ ,
  ::对于z2, z2,

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  $\begin{align*}r_2 = \sqrt{1^2 + (-\sqrt{3})^2}\end{align*}$
  ::r2=12+(-3)2

  $\begin{align*}= \sqrt{1 + 3}\end{align*}$

  $\begin{align*}= \sqrt{4}\end{align*}$

  $\begin{align*}= 2\end{align*}$

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  and
  ::和

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  $\begin{align*}\mbox{tan}\ \theta_2 = \frac{-\sqrt{3}} {1},\end{align*}$
  ::棕色231,

  $\begin{align*}\theta_2 = \frac{5\pi} {3}\end{align*}$

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  Now we can use the formula $\begin{align*}z_1 \cdot z_2 = r_1 \cdot r_2 cis \left( \theta_1 +\theta_2 \right)\end{align*}$
  ::现在我们可以使用公式 z1z2=r1r2cis(12)

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  Substituting gives:
  ::替代给付 :

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  $\begin{align*}z_1 \cdot z_2 = 2\sqrt{2} \times 2\ \mbox{cis}\ \left [\frac{\pi} {4} + \frac{5\pi} {3}\right ]\end{align*}$
  ::z2=22x2 cis [_4+5×3]

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  $\begin{align*}= 4\sqrt{2}\ \mbox{cis}\ \left [\frac{23\pi} {12}\right ]\end{align*}$
  ::=42 cis [23-12]

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  So we have
  ::因此,我们有

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  $\begin{align*}z_1 \cdot z_2 = 4\sqrt{2} \left (\mbox{cos}\ \frac{23\pi} {12} + i\ \mbox{sin}\ \frac{23 \pi} {12}\right )\end{align*}$
  ::z2=42(cos 2312+i sin 2312)

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  Re-writing in approximate decimal form:
  ::以小数点的粗略写法重写:

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  5.656 (0.966 – 0.259i)
  ::5.656(0.966 - 0.259i)

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  5.46 - 1.46i
  ::5.46 - 1.46i

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  If the problem was done using only rectangular units then
  ::如果只使用矩形单位来解决问题,那么

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  $\begin{align*}z_1 \times z_2 = (2 + 2i)(1 - \sqrt{3}i) \ \mbox{or}\end{align*}$
  ::z2=( 2+2i)( 1-3i) 或

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  $\begin{align*}= 2 - 2\sqrt{3}i + 2i - 2\sqrt{3}i^2\end{align*}$
  ::=2 - 23i+2i - 23i2

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  Gathering like terms and using i ² = -1
  ::收集类似术语并使用 i2 = - 1

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  $\begin{align*}= (2 + 2\sqrt{3}) - (2\sqrt{3} + 2)i\end{align*}$
  ::=(2+23)-(23+2)i

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  or
  ::或

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  $\begin{align*}5.46 - 1.46i\end{align*}$
  ::5.46 - 1.46i

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  Example 3
  ::例3

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  Using polar multiplication, find the product $\begin{align*}(6 - 2\sqrt{3}i)(4 + 4\sqrt{3}i)\end{align*}$ .
  ::使用极化乘法,找到产品(6-23i)(4+43i)。

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  Let $\begin{align*}z_1 = 6 - 2\sqrt{3}i\end{align*}$ and $\begin{align*}z_2 = 4 + 4\sqrt{3}i\end{align*}$
  ::Letz1=6-23i和z2=4+43i

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  $\begin{align*}r_1 = \sqrt{(6)^2 - (2\sqrt{3})^2}\end{align*}$ and $\begin{align*}r_2 = \sqrt{(4)^2 + (4\sqrt{3})^2}\end{align*}$
  ::r1=(6)2-(23)2和r2=(4)2+(43)2

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  $\begin{align*}r_1 = \sqrt{36 + 12} = \sqrt{48} = 4\sqrt{3}\end{align*}$ and $\begin{align*}r_2 = \sqrt{16 + 48} = \sqrt{64} = 8\end{align*}$
  ::r1=36+12=48=43和r2=16+48=64=8

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  For θ ₁ , first find $\begin{align*}\mbox{tan}\ \theta_{ref} = \left |\frac{y} {x}\right |\end{align*}$
  ::=============================================================================================================================================== =============================================================================================================================================================================================================================================================================================================================================================================

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  $\begin{align*}\mbox{tan}\ \theta_{ref} = \frac{(2\sqrt{3})} {6}\end{align*}$
  ::tan tan \ ref=( 23)6

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  $\begin{align*}\mbox{tan}\ \theta_{ref} = \frac{\sqrt{3}} {3}\end{align*}$
  ::现 现 现 现 现 现 现 现 现 现 现 现 现 现 现 现 现 现 现 现

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  $\begin{align*}\theta_{ref} = \frac{\pi} {6}.\end{align*}$
  ::6号,6号,6号

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  Since x > 0 and y < 0 we know that θ ₁ is in the in the 4 ^th quadrant:
  ::由于 x > 0 和 y < 0, 我们知道第 4 象限内有 1 :

  $\begin{align*}\theta_1 = \frac{11\pi} {6}\end{align*}$

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  For θ ₂ ,
  ::来二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二,二

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  $\begin{align*}\mbox{tan}\ \theta_{ref} = \frac{(4\sqrt{3})} {4}\end{align*}$
  ::tan ref=(43)4

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  $\begin{align*}\mbox{tan}\ \theta_{ref} = \sqrt{3},\end{align*}$
  ::皮肤=3, 皮肤=3, 皮肤=3, 皮肤=3, 皮肤=3, 皮肤=3, 皮肤=3, 皮肤=3, 皮肤=3, 皮肤=3, 皮肤=3, 皮肤=3, 皮肤=3, 皮肤=3, 皮肤=3, 皮肤=3, 皮肤=3, 皮肤=3, 皮肤=3, 皮肤=3, 皮肤=3, 皮肤=3, 皮肤=3, 皮肤=3, 皮肤=3, 皮肤=3, 皮肤=3

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  $\begin{align*}\theta_{ref} = \frac{\pi} {3}\end{align*}$
  ::瑞夫3号

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  Since θ ₂ is in the first quadrant,
  ::因为2是在第一个象限,

  $\begin{align*}\theta_2 = \frac{\pi} {3}\end{align*}$

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  Using polar multiplication,
  ::利用两极乘法,

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  $\begin{align*}z_1 \times z_2 = 4\sqrt{3} \times 8\left (\mbox{cis}\ \left [\frac{11\pi} {6} + \frac{\pi} {3}\right ]\right )\end{align*}$
  ::z2=43x8(cis[1163])

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  $\begin{align*}z_1 \times z_2 = 32\sqrt{3} \left (\mbox{cis}\ \left[\frac{13\pi} {6}\right ]\right )\end{align*}$
  ::z1xz2=323(cis [13_6])

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  subtracting 2π from the augment:
  ::从增量中减去 2 :

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  $\begin{align*}z_1 \times z_2 = 32\sqrt{3} \left (\mbox{cis}\ \left [\frac{\pi} {6}\right ]\right )\end{align*}$
  ::z1xz2=323(cis[%6])

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  or in expanded form: $\begin{align*}32\sqrt{3} \left (\mbox{cos}\ \left [\frac{\pi} {6}\right ] + i\ \mbox{sin}\ \left [\frac{\pi} {6}\right ]\right )\end{align*}$
  ::或以扩展形式:323(cos[_6]+i sin [_6])

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  In decimal form this becomes: 55.426(0.866 + 0.500i) or 48 + 27.713i
  ::以小数计,以下为55.426(0.866+0.500i)或48+27.713i。

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  Check:
  ::检查 :

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  $\begin{align*}(6 - 2\sqrt{3}i)(4 + 4\sqrt{3}i) = 24 + 24\sqrt{3}i - 8\sqrt{3}i - 24i^2\end{align*}$
  :6-23i)(4+43i)=24+243i-83i-24i2)

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  $\begin{align*}= 24 + 16\sqrt{3}i + 24\end{align*}$
  ::=24+163i+24

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  $\begin{align*}= 48 + 27.713i\end{align*}$
  ::=48+27.713i

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  Example 4
  ::例4

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  Using polar division, find the quotient of $\begin{align*}\frac{z_1} {z_2}\end{align*}$ given that $\begin{align*}z_1 = 5 - 5i\end{align*}$ and $\begin{align*}z_2 = -2\sqrt{3} - 2i\end{align*}$ .
  ::使用极分,找到z1z2的商数,因为z1=5-5i和z223-2i。

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  For $\begin{align*}z_1 : r_1 = \sqrt{5^2 + (-5)^2}\end{align*}$ or $\begin{align*}5\sqrt{2}\end{align*}$ and $\begin{align*}\mbox{tan}\ \theta_1 = \frac{-5} {5}\end{align*}$ , so $\begin{align*}\theta_1 = \frac{7\pi} {4}\end{align*}$ (4th quadrant)
  ::z1:r1=52+(-5)2或52和棕色 155, 所以1=74( 第4象限)

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  For $\begin{align*}z_2 : r_2 = \sqrt{(-2\sqrt{3})^2 + (-2)^2}\end{align*}$ or $\begin{align*}\sqrt{16} = 4\end{align*}$ and $\begin{align*}\mbox{tan}\ \theta_2 = \frac{-2} {(-2\sqrt{3})}\end{align*}$ , so $\begin{align*}\theta_2 = \frac{7\pi} {6}\end{align*}$ (3 ^rd quadrant)
  ::z2:r2=(-232)2+(-2)2或16=4和tan 22(-23), 所以2=76(第3象限)

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  Using the formula, $\begin{align*}\frac{z_1} {z_2} = \frac{r_1} {r_2} \times \mbox{cis}\ [\theta_1 - \theta_2]\end{align*}$ or
  ::使用公式, z1z2=r1r2xcis [12] 或

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  $\begin{align*}= \frac{5\sqrt{2}} {4} \times \mbox{cis}\ \left [\frac{7\pi} {4} - \frac{7\pi} {6}\right ]\end{align*}$
  ::=524xcis [7.4-7-7-6]

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  $\begin{align*}= \frac{5\sqrt{2}} {4} \times \mbox{cis}\ \left [\frac{7\pi} {12}\right ]\end{align*}$
  ::=524xcis [7=12]

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  $\begin{align*}= \frac{5\sqrt{2}} {4} \left [\mbox{cos}\ \frac{7\pi} {12} + i\ \mbox{sin}\ \frac{7\pi} {12}\right ]\end{align*}$
  ::=524 [cos 712+i sin 712]

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  $\begin{align*}= 1.768[-0.259 + (0.966)i]\end{align*}$
  ::=1.768[-0.259+(0.966)]

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  $\begin{align*}= -0.458 + 1.708i\end{align*}$
  ::0.458+1.708i

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  Check by using the complex conjugate to do the division in rectangular form:
  ::使用复合二次曲线来进行矩形形式的分割 :

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  $\begin{align*}\frac{5 - 5i} {-2\sqrt{3} - 2i} \cdot \frac{-2\sqrt{3} + 2i} {-2 \sqrt{3} + 2i} = \frac{-10\sqrt{3} + 10i + 10\sqrt{3}i - 10i^2} {(-2\sqrt{3})^2 - (2i)^2}\end{align*}$
  ::5-5i-23-2i23+2i-23+2i-23+2i103+10i+103i+103i+103i+10i+10i+10i2(2-23)-2-(2i)2

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  $\begin{align*}= \frac{-10\sqrt{3} + 10i + 10\sqrt{3}i+ 10} {12 + 4}\end{align*}$
  ::+103+10i+103i+1012+4

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  $\begin{align*}= \frac{(-10\sqrt{3} + 10) + (10 + 10\sqrt{3})i} {16}\end{align*}$
  ::=(-103+10)+(10+103)i16

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  $\begin{align*}= \frac{(-17.3 + 10) + (10 + 17.3)i} {16}\end{align*}$
  ::=(-17.3+10)+(10+17.3)i16

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  $\begin{align*}= \frac{(-7.3) + (27.3)i} {16}\ \mbox{or}\end{align*}$
  ::=(-7.3)+(27.3)i16或

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  $\begin{align*}-0.456 + 1.706i\end{align*}$
  ::- -0.456+1.706i

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  The two radically different approaches yield the same answer. The small difference between the two answers is a result of decimal rounding.
  ::两种截然不同的方法产生相同的答案。 两种答案之间的小差别是小数点四舍五入的结果。

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  Example 5
  ::例5

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  Find the product: $\begin{align*}\left( 7 \left( \frac{\pi}{6} \right) \right) \bullet \left( 5 \left( \frac{-\pi}{4} \right) \right)\end{align*}$ .
  ::寻找产品: (7(6)) * (5(4)。

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  This one is easier than it looks: Recall $\begin{align*}z_1 \cdot z_2 = r_1 \cdot r_2\end{align*}$ $\begin{align*}cis (\theta_1 + \theta_2)\end{align*}$ .
  ::这比看起来容易: 回忆z1z2=r1r2cis(12) 。

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  $\begin{align*}r_1 \cdot r_2 \to 7 \cdot 5 = 35\end{align*}$ ....... By substitution and multplication
  ::r1r275=35...。

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  $\begin{align*}\theta_1 + \theta_2 \to \left( \frac{\pi}{6} \right) + \left( \frac{-\pi}{4} \right)\end{align*}$ ..... Substitute
  ::=============================================================================================================================================*====================================================================================================================================================================================================

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  $\begin{align*}\left( \frac{2\pi}{12} \right) + \left( \frac{-3\pi}{12} \right)\end{align*}$ ..... Find common denominators
  :212)+(- 312)...。 查找共同的分母

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  $\begin{align*}\left( \frac{-\pi}{12} \right)\end{align*}$ ..... Simplify
  :12). 简化

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  $\begin{align*}\therefore 35 cis \left( \frac{-\pi}{12} \right)\end{align*}$ is the product
  ::======================================================================================================================产品===================== ===============================================================================================

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  Example 6
  ::例6

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  Find the quotient: $\begin{align*}\frac{1 + 2i}{2 - i}\end{align*}$ .
  ::查找商数:1+2i2-i。

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  First, find the quotient by polar multiplication:
  ::首先,通过极乘法找到商数:

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  $\begin{align*}r_1 = \sqrt{(1)^2 + (2)^2} = \sqrt{5} \qquad r_2 = \sqrt{(2)^2 + (-1)^2} = \sqrt{5}\end{align*}$
  ::r1=(1)2+(2)2=5r2=(2)2+(-1)2=5

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  $\begin{align*}\mbox{tan}\ \theta_1 = \frac{2} {1}\end{align*}$
  ::棕色 #% 1=21

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  $\begin{align*}\mbox{tan}\ \theta_1 = 2\end{align*}$
  ::皮肤 #% 1=2

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  $\begin{align*}\theta_{ref} = 1.107\ \mbox{radians}\end{align*}$
  ::ref=1.107 弧度

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  since the angle is in the 1 ^st quadrant
  ::因为角度位于第1象限中

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  θ ₁ = 1.107 radians
  ::1 = 1.107 弧度

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  for $\begin{align*}\theta_2\end{align*}$ ,
  ::=2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,

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  $\begin{align*}\mbox{tan}\ \theta_2 = \frac{-1} {2}\end{align*}$
  ::皮肤 =================================================================================================================12

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  $\begin{align*}\mbox{tan}\ \theta_{ref} = \frac{1} {2}\end{align*}$
  ::皮肤 = 12

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  $\begin{align*}\theta_{ref} = 0.464\ \mbox{radians}\end{align*}$
  ::ref=0.464弧度

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  since θ ₂ is in the 4 ^th quadrant, between 4.712 $\begin{align*}\left (\mbox{or}\ \frac{3\pi} {2}\right )\end{align*}$ and 6.282 radians (or 2 π )
  ::2位于第4象限,在4.712(或32)至6.282弧度(或2)之间。

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  θ ₂ = 5.820 radians
  ::2 = 5.820 弧度

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  Finally, using the division formula,
  ::最后,使用分裂公式,

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  $\begin{align*}\frac{z_1} {z_2} = \frac{\sqrt{5}} {\sqrt{5}} [\mbox{cis}\ (1.107 - 5.820)]\end{align*}$
  ::z1z2=55[cis (1.107-5.820)]

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  $\begin{align*}\frac{z_1} {z_2} = [\mbox{cis}\ (-4.713)]\end{align*}$
  ::z1z2 = [cis (- 4.713)]

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  $\begin{align*}\frac{z_1} {z_2} = [\mbox{cos}\ (-4.713) + i\ \mbox{sin}\ (-4.713)]\end{align*}$
  ::z1z2 = [cos (- 4.713)+i sin (- 4.713)]

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  $\begin{align*}\frac{z_1} {z_2} = [\mbox{cos}\ (1.570) + i\ \mbox{sin}\ (1.570)]\end{align*}$
  ::z2 = [cos (1.570)+i sin (1.570)]

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  If we assume that $\begin{align*}\frac{\pi} {2} = 1.570\end{align*}$ , then
  ::如果我们假设2=1.570,那么

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  $\begin{align*}\approx \frac{z_1} {z_2} = \left [\mbox{cos}\ \left (\frac{\pi} {2}\right ) + i\ \mbox{sin}\ \left (\frac{\pi} {2}\right )\right ]\end{align*}$
  ::z1z2 = [cos (%2)+i sin (%2) = [cos (%2)+i sin (%2)]

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  $\begin{align*}\frac{z_1} {z_2} = 0 + 1i = i\end{align*}$
  ::z1z2=0+1i=i

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  Review
  ::回顾

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  1. Find the product using polar form: $\begin{align*}(2 + 2i)(\sqrt{3} - i)\end{align*}$
     ::用极形(2+2i)(3-i)查找产品
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  2. $\begin{align*}2\end{align*}$ $\begin{align*} cis (40) \bullet\end{align*}$ $\begin{align*}4 \end{align*}$ $\begin{align*}cis (20)\end{align*}$
     ::2Cis(40) 4Cis(20)
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  3. Multiply: $\begin{align*}2 \left (\mbox{cos}\ \frac{\pi} {8} + i\ \mbox{sin}\ \frac{\pi} {8}\right ) \bullet 2 \left (\mbox{cos}\ \frac{\pi} {10} + i\ \mbox{sin}\ \frac{\pi} {10}\right )\end{align*}$
     ::乘数: 2(cos 8+i sin 8) 2(cos 10+i sin 10)
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  4. $\begin{align*}\frac{2 cis (80)}{6 cis (200)}\end{align*}$
     ::Cis( 806Cis( 200) )
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  5. Divide: 3cis(130 ^o ) ÷ 4cis(270 ^o )
     ::除号: 3Cis(130o) 4Cis(270o)

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  If $\begin{align*} z_1 = 5 \left( \frac{-\pi}{4} \right)\end{align*}$ and $\begin{align*}z_2 = 5 \left( \frac{-\pi}{4} \right)\end{align*}$ find:
  ::如果z1=5(4)和z2=5(4)发现:

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  6. $\begin{align*}z_1 \cdot z_2\end{align*}$
     ::z2 z1z2 z1z2
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  7. $\begin{align*}\left( \frac{z_1}{z_2} \right)\end{align*}$
     :z1z2)
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  8. $\begin{align*}\left( \frac{z_2}{z_1} \right)\end{align*}$
     :z2z1)

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  If $\begin{align*}z_1 = 5 \left( \frac{\pi}{6} \right)\end{align*}$ and $\begin{align*} z_2 = 5 \left(\frac{\pi}{6} \right)\end{align*}$ find:
  ::如果z1=5(6)和z2=5(6)发现:

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  9. $\begin{align*}z_1 z_2\end{align*}$
     ::z1z2 z1z2
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  10. $\begin{align*}\left( \frac{z_1}{z_2} \right)\end{align*}$
      :z1z2)
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  11. $\begin{align*}\left( \frac{z_2}{z_1} \right)\end{align*}$
      :z2z1)
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  12. $\begin{align*}(z_1)^2\end{align*}$
      :z1)2
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  13. $\begin{align*}(z_2)^3\end{align*}$
      :z2)3

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  Find the products.
  ::找到产品。

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  14. Find the product using polar form: $\begin{align*}(2 + 2i)(\sqrt{3} - i)\end{align*}$
      ::使用极形查找产品: (2+2i)(3-i)
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  15. $\begin{align*}2 (cos 40^o + i sin 40^o) \bullet 4(cos 20^o + i sin 20^o)\end{align*}$
      ::2(cos40o+isin40o)%4(cos20o+isin20o)
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  16. $\begin{align*} 2 \left(cos \frac{\pi}{8} + i sin \frac{\pi}{8} \right) \bullet 2 \left(cos \frac{\pi}{10} + i sin \frac{\pi}{10} \right)\end{align*}$
      :cos%8+isin%8) =2(cos%10+isin%10)

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  Find the quotients.
  ::找到商数。

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  17. $\begin{align*} 2(cos 80^o + i sin 80^o) \div 6(cos 200^o + i sin 200^o)\end{align*}$
      ::2(cos80o+isin80o) 6(cos200o+isin200o)
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  18. $\begin{align*} 3cis(130^o) \div 4cis(270^o)\end{align*}$
      ::3Cis( 130o) 4Cis( 270o)

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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 。