4.7 复杂数字的产品和引号

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  复杂数字的产品和数字

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  Working with can be decidedly strange. It is natural to wonder why we put so much time and effort into learning how to properly manipulate numbers that don't even exist! There are, however, many real and rather fun and interesting applications of complex numbers in the "real world."
  ::工作总是会非常奇怪。很自然,我们为什么花这么多时间和精力来学习如何正确操控根本不存在的数字呢?然而,在“现实世界”中,复杂的数字有许多真实的、相当有趣和有趣的应用。

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  Products and Quotients of Complex Numbers
  ::复杂数字的产品和数字

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  In this lesson, we will explore the multiplication and division of complex numbers. In general, operations on complex numbers behave just like operations on regular polynomials: combine like terms, FOIL binomials, distribute, and so on. The differences are generally apparent when the products and quotients of complex numbers are simplified.
  ::在这个教训中,我们将探索复杂数字的乘法和分法。 一般来说,复杂数字的操作就像常规多式操作一样:将条件、FOIL的二进制、分布等等结合起来。 当复杂数字的产品和商数简化时,差异一般都很明显。

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  Recall the unusual case of $\begin{align*}i^2\end{align*}$ :  $\begin{align*}i = \sqrt{-1} \therefore i^2 = (\sqrt{-1})^2 = -1\end{align*}$
  ::回顾i2的异常情况:i2:i_1_i2=(1_1)2_1

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  Since $\begin{align*}i^2 = -1\end{align*}$ , multiplying complex numbers often results in imaginary terms becoming real terms.
  ::由于 i2\% 1, 乘以复杂数字往往导致假想值变成实际值。

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  Multiplying Complex Numbers
  ::乘数复合体数

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  When multiplying complex numbers in rectangular form , recall the method for multiplying two binomials (sometimes called FOIL): ( m + n )( x + y ) = mx + my + nx + ny . We use the same procedure for multiplying complex numbers:
  ::当以矩形形式乘以复杂数字时,请记得乘以两个二元数的方法(有时称为FOIL): (m+ n)(x + y) = mx + my + nx + ny。我们使用相同的程序来乘以复杂的数字:

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  ( a + bi )( c + di ) = ac + adi + bci + bdi ²
  :a) + bi(c) + di) = ac + adi + bci + bdi2

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  But, unlike the algebraic expression, the above expression contains the number, i .
  ::但是,与代数表达式不同,上述表达式包含数字,即数字。

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  Recall that i ² = -1, so bdi ² = bd (-1) = - bd and
  ::回顾i2 = -1, 所以 bdi2 = bd(-1) = - bd和

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  adi + bci can be combined and then factored as ( ad + bc ) i . Thus we have the general result,
  ::adi + bci + bci 可合并,然后作为(ad + bc)i 计算。 因此,我们得出了一般结果,

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  ( a + bi )( c + di ) = ( ac - bd ) + ( ad + bc ) i
  :a) + b(c) + di) = (ac - bd) + (ad+ bc)i

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  Dividing Complex Numbers
  ::Dividing 复杂数字

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  To divide two complex numbers is similar to dividing two irrational numbers. Recall that in that problem, the procedure was to find the irrational conjugate of the denominator and then multiply both the numerator and the denominator by that conjugate, for example:
  ::将两个复杂数字分开类似于将两个非理性数字分开。 回顾在这个问题中,程序是找到分母的非理性共和点,然后乘以分子和分母,例如:

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  Divide: $\begin{align*}\frac{3} {1 + \sqrt{2}}\end{align*}$
  ::差额: 31%2

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  First find the irrational conjugate of the denominator: $\begin{align*}1 - \sqrt{2}\end{align*}$ , then multiply both the numerator and the denominator by this value:
  ::首先找到分母的非理性共解值 : 1\\\ 2, 然后将分子和分母乘以此值 :

  $\begin{align*}\frac{3}{1 + \sqrt{2}} \times \frac{1 - \sqrt{2}} {1 - \sqrt{2}} = \frac{3 - 3\sqrt{2}} {1 - 2}\end{align*}$

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  this reduces to
  ::降为

  $\begin{align*}= \frac{3 - 3\sqrt{2}} {-1}\end{align*}$

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  or
  ::或

  $\begin{align*}= -3 + 3\sqrt{2}\end{align*}$

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  The process is very much the same with complex numbers. With complex numbers, since you are interested in eliminating from the denominator, you find the complex conjugate of the denominator and multiply BOTH the numerator AND the denominator by it.
  ::过程与复杂的数字非常相同。 复杂的数字, 因为您想要从分母中消除, 您就会发现分母的复杂组合, 并乘以分子和分母的倍数。

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  You find the complex conjugate in the same way you found the conjugate of irrational numbers, change the sign of the imaginary part. For instance, the complex conjugate of 4 + 3i is 4 – 3i
  ::您发现复杂的共产体的方式与您找到非理性数字的共产体的方式相同, 改变想象部分的符号。 例如, 4+3i的复杂共产体是 4 - 3i

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  A complex number multiplied by its complex conjugate will yield a real number. By recalling ( a + b )( a - b ) = a ² - b ² simplifying a complex number and its conjugate can be easy, for example:
  ::复数乘以其复杂的共产体将产生一个实际数字。通过回顾(a+b)(a-b) = a2-b2 简化一个复杂数字,其共产体很容易,例如:

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  The conjugate of 4 + 3i is found by retaining the real part (4) and reversing only the sign of the imaginary part (that is, 3i becomes -3i)
  ::4+3i的结合通过保留真实部分(4)和仅颠倒假想部分的标志(即3i变成-3i)而找到。

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  (4 + 3 i )(4 - 3 i ) = 16 - 12i + 12i - 9i ² . Notice that -12 i and 12 i cancel. Also recall that i ² = -1
  :4 + 3i)(4 + 3i)(4 + 3i)(4 + 3i) = 16 - 12i + 12i - 9i2. 通知 - 12i 和 12i 取消。

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  That gives us:
  ::这使我们:

  16 + 9 = 25

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  Therefore: (4 + 3i)(4 – 3i) = 25
  ::因此: (4 + 3i)(4 - 3i) = 25

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  The product of this complex number and its conjugate is 25.
  ::这一复杂数字的产物及其共鸣为25。

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  When multiplying complex numbers sometimes intuition about the nature of the product can mislead.
  ::当复杂的数字成倍增加时,有时对产品性质的直觉会误导。

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  For example in (a + b)(a + b), where all of the terms are real numbers, no terms of each of the four products will cancel. Some of the terms may be combined. However in (1 + i)(1 - i), where some terms are real numbers and some terms are , this is no longer true. Two of these terms cancel: the first product yields 1 while the last product yields i ² or -1, and those terms cancel!
  ::例如,在(a)+b(a)+(a)+(b)+(b)中,如果所有条件都是实际数字,则四个产品中每个产品都不得取消任何条件。有些条件可以合并。然而,在(a)+i(1-1-i)中,有些条件是实际数字,有些条件是真实数字,有些条件是真实数字,而有些条件是真实数字)中,则不再如此。其中两个条件取消:第一个产品生产1,而最后一个产品生产i2或-1,这些条件取消!

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  Examples
  ::实例

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  Example 1
  ::例1

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  Multiply: $\begin{align*}(6 + 3i) (2 - 3i)\end{align*}$ .
  ::乘数: (6+3i)(2-3i)

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  $\begin{align*}(6 + 3i)(2 - 3i) = 12 - 18i + 6i - 9i^2\end{align*}$
  :6+3i)(2-3i)=12-18i+6i-9i2

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  $\begin{align*}= 21 - 12i\end{align*}$
  ::=21-12i

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  (Note when combining like terms: -9 i ² reduces to -9(-1) or 9)
  :注意将类似术语合并时注意: -9i2 减至 -9(-1)或9)

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  Example 2
  ::例2

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  Multiply: $\begin{align*}(5 - 7i) (5 + 7i)\end{align*}$ .
  ::乘以: (5-7i)(5+7i).

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  $\begin{align*}(5 - 7i)(5 + 7i) = 25 + 35i - 35i - 49i^2\end{align*}$
  :5-7i)(5+7i)=25+35i-35i-49i2)

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  or
  ::或

  $\begin{align*}= 25 + 0 - 49(-1)\end{align*}$

  $\begin{align*}= 74\end{align*}$

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  Example 3
  ::例3

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  Multiply:  $\begin{align*}(6+7i )(2+3i)\end{align*}$ .
  ::乘数: (6+7i)(2+3i)

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  To multiply $\begin{align*}(6 + 7i ) \cdot (2 + 3i)\end{align*}$ , treat each complex number as a binomial and apply the FOIL method to find the product.
  ::要乘以(6+7i)(2+3i),将每个复合数作为二元数处理,并采用FOIL方法寻找产品。

  $\begin{align*}6 \cdot 2 = 12\end{align*}$	multiply the First terms
  $\begin{align*}6 \cdot 3i = 18i\end{align*}$	multiply the Outer terms
  $\begin{align*}7i \cdot 2 = 14i\end{align*}$	multiply the Inner terms
  $\begin{align*}7i \cdot 3i = -21\end{align*}$	multiply the Last terms and simplify the $\begin{align*}i^2\end{align*}$

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  $\begin{align*}12 +18i + 14i - 21 \to 12 + 32i -21 \to -9 + 32i\end{align*}$ Combine like terms and simplify
  ::12+18i+14i-2112+32i-219+32i

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  $\begin{align*}\therefore (6 + 7i) \cdot (2 + 3i) = -9 + 32i\end{align*}$
  :6+7i)(2+3i)9+32i

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  Example 4
  ::例4

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  Divide : $\begin{align*}\frac{6 - 3i} {4 + 3i}\end{align*}$ .
  ::除以:6-3i4+3i。

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  First, observe that the complex conjugate of the denominator is 4 – 3i.
  ::第一,注意分母的复杂结合是4-3i。

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  Multiply both the numerator and the denominator by 4 – 3i: $\begin{align*}\frac{6 - 3i} {4 + 3i} \times \frac{4 - 3i} {4 - 3i} = \frac{24 - 18i + 12i + 9i^2} {16 - 12i + 12i - 9i^2}\end{align*}$
  ::乘以 4-3i: 6-3i4+3ix4-3i4-3i4-3i4-3i4-3i=24-18i+12i+9i216-12i+12i-9i2

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  $\begin{align*}= \frac{15 - 6i} {25}\end{align*}$
  ::=15-625岁

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  $\begin{align*}\frac{6 - 3i} {4 + 3i} = \frac{15 - 6i} {25}\end{align*}$
  ::6-3i4+3i=15-625

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  Example 5
  ::例5

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  Divide:  $\begin{align*}\frac{5 + 2i}{7 + 4i}\end{align*}$ .
  ::除以5+2i7+4i。

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  To divide  $\begin{align*}\frac{5 + 2i}{7 + 4i}\end{align*}$ , first determine the conjugate of the denominator $\begin{align*}(7 + 4i)\end{align*}$ the conjugate is the same two terms with the opposite sign between them, in this case $\begin{align*}(7 - 4i)\end{align*}$ .
  ::要划分 5+2i7+4i,首先确定分母(7+4i)的共性(7+4i),共性是相同的两个条件,两者的符号是相反的(7+4i)。

  $\begin{align*}\left(\frac{5 +2i}{7 + 4i} \right) \bullet \left(\frac{7 - 4i}{7 - 4i} \right)\end{align*}$	Multiply the numerator and denominator by the conjugate
  $\begin{align*}\frac{35 + 14i - 20i - 8i^2}{49 - 28i + 28i - 16i^2}\end{align*}$	FOIL
  $\begin{align*}\frac{35 - 6i + 8}{49 + 16}\end{align*}$	Simplify
  $\begin{align*}\frac{43 - 6i}{65}\end{align*}$	Simplify

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  $\begin{align*}\therefore \frac{5 + 2i}{7 + 4i} = \frac{43 - 6i}{65}\end{align*}$
  ::5+2i7+4i=43-6i65

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  Example 6
  ::例6

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  Divide:  $\begin{align*}\frac{4 + i}{5 - i}\end{align*}$ .
  ::除以:4+i5-i。

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  To divide $\begin{align*}\frac{4 + i}{5 - i}\end{align*}$ , first identify the conjugate of the denominator: $\begin{align*}(5 + i)\end{align*}$
  ::要划分 4+i5-i,首先确定分母的组合: (5+i)

  $\begin{align*}\left(\frac{4 + i}{5 - i} \right) \bullet \left(\frac{5 + i}{5 + i} \right)\end{align*}$	Multiply the numerator and denominator by the conjugate
  $\begin{align*}\frac{20 + 4i + 5i + i^2}{25 + 5i - 5i - i^2}\end{align*}$	FOIL
  $\begin{align*}\frac{35 + 9i - 1}{25 + 1}\end{align*}$	Simplify
  $\begin{align*}\frac{34 + 9i}{26}\end{align*}$	Simplify

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  $\begin{align*}\therefore \frac{4 + i}{5 - i} = \frac{34 + 9i}{26}\end{align*}$
  ::+4+i5-i=34+926

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  Review
  ::回顾

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  1. $\begin{align*}(-6i) + (-4i)\end{align*}$
     :-6i)+(-4i)
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  2. $\begin{align*}(12i + 1) + (6i + 11)\end{align*}$
     :12i+1)+(6i+11)

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  Multiply the complex numbers.
  ::乘以复杂的数字。

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  3. $\begin{align*}-7i(7i + 5)\end{align*}$
     ::- 7i(7i+5)
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  4. $\begin{align*}(2i + 3)^2\end{align*}$
     :2i+3)2
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  5. $\begin{align*}(-2i)(-11i)(-12i)\end{align*}$
     :-2i)-(-1)-(-1)-(-12i)
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  6. $\begin{align*}(8i - 5)(11i^3 - 9)\end{align*}$
     :8i-5)(11i3-9)
  7. $\begin{align*}(9\sqrt{-9}) \cdot (14\sqrt{-4})\end{align*}$
  8. $\begin{align*}(10\sqrt{-1}) \cdot (24 \sqrt{-144})\end{align*}$
  9. $\begin{align*}(18 \sqrt{-49}) \cdot (5 \sqrt{-144})\end{align*}$

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  Divide the complex numbers.
  ::将复数除以。

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  10. $\begin{align*}\frac{11i - 2}{i + 1}\end{align*}$
      ::11i-2i+1 11i-2i+1
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  11. $\begin{align*}\frac{10i + 10}{-8i - 10}\end{align*}$
      ::10i+10-8i-10
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  12. $\begin{align*}\frac{7i + 12}{-12i - 8}\end{align*}$
      ::7i+12-12i-8
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  13. $\begin{align*}\frac{10i + 5}{-i - 7}\end{align*}$
      ::10i+5-i-7
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  14. $\begin{align*}\frac{4i^2 + 4}{-6i^2 - 1}\end{align*}$
      ::4i2+4-6i2-1
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  15. $\begin{align*}\frac{5i^2 - 5}{-8i^2 + 3}\end{align*}$
      ::5i2-5-8i2+3 5i2-5-8i2+3
  16. $\begin{align*}\frac{-4\sqrt{-60}}{-2\sqrt{-20}}\end{align*}$
  17. $\begin{align*}\frac {-100\sqrt{-154}}{10\sqrt{-22}}\end{align*}$
  18. $\begin{align*}\frac {48 \sqrt{-170}}{8 \sqrt{-17}}\end{align*}$

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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 。