8.6 复杂数字

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  复数数

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  Impedance is the measure of the opposition that a circuit presents to a current when a voltage is applied.  Electrical engineers regularly work with the formula   $\begin{array}{r}V=IZ\end{array}$ , where V is voltage (in volts), I is current (in amps), and Z is the impedance (in ohms) ¹ . If a current is 6 amps and goes through a part of the circuit with an impedance of  $\begin{array}{r}70+120i\end{array}$  ohms, what is an expression for the voltage of the circuit? 
  ::电路在电压应用时对电流的反作用度是电路对电流的测量。电动工程师经常使用V=IZ的公式,V是电压(伏),I是电流(安普斯),Z是阻力(奥姆斯)。1. 如果电流是6安普斯,经过电路的一部分,阻力为70+120安普斯,电路电压的表达方式是什么?

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  $\begin{array}{r}70+120i\end{array}$  is an example of a complex number . We'll discuss numbers like this in this section.
  ::70+120i是一个复杂数字的例子。我们将在本节讨论这样的数字。

  

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  Complex Numbers 
  ::复数数

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  So far we have solved quadratic equations in this chapter, and our solutions have been real numbers, such as 2, -5, $\begin{array}{r}\sqrt{11}\end{array}$ , and $\begin{array}{r}\frac{1}{3}\end{array}$ . However, let's consider a quadratic equation where we do not get a real number solution, say,  $\begin{array}{r}{x}^2+25=0\end{array}$ . We can solve this by the square root method.
  ::到目前为止,我们已经解决了本章中的四方方程式,我们的解决办法是实际数字,如2、5、11和13。然而,让我们考虑一个四方方方程式,在这个方程式中,我们没有真正的数字解决方案,例如,x2+25=0。我们可以用平方根方法解决这个问题。

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  $$\begin{array}{r}{x}^2+25=0\\ \underset{\_}{-25-25}\\ x^2=\text{-}25\\ \sqrt{x^2}=\sqrt{\text{-}25}\\ x=\sqrt{\text{-}25}\end{array}$$

  
  ::x2+25=0 - 25- 25_ x2=- 25x2=- 25x2=- 25x2=- 25x=- 25x=- 25

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  At this point, we are stuck. We cannot take the square root of a negative number in the real numbers. But we would like  a way to describe the solutions of this equation. That is where complex numbers come in. 
  ::此时此刻,我们被卡住了。我们不能从实际数字负数的平方根中找到正方根。 但我们想用一种方式来描述这个等式的解决方案。 这就是复杂数字出现的地方。

   

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  We can at least simplify -25 if we break it up into a positive factor and a factor of -1. 
  ::如果我们将其分成积极因素和1因素,我们至少可以简化-25。

  $$\begin{array}{r}\sqrt{\text{-}25}=\sqrt{25\cdot \text{-}1}=\sqrt{25}\cdot \sqrt{\text{-}1}=5\sqrt{\text{-}1}\end{array}$$

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  What remains is to assign $\begin{array}{r}\sqrt{\text{-}1}\end{array}$ a name, $\begin{array}{r}i\end{array}$ .
  ::剩下的是指定 - 1 名称, i 。

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  $$\begin{array}{r}\sqrt{\text{-}25}=5\sqrt{\text{-}1}=5i\end{array}$$

  
  ::-25=5-1=5i

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  $\begin{array}{r}5i\end{array}$  is an example of a complex number, which we define below.
  ::5i 是一个复杂数字的例子,下文对此作了说明。

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     Complex Numbers
  ::复数数

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  A  complex number  is a number of the form   $\begin{array}{r}a+bi,\end{array}$  where   $\begin{array}{r}a\end{array}$   and   $\begin{array}{r}b\end{array}$   are real numbers. We call  $\begin{array}{r}a\end{array}$  the real part of the complex number and   $\begin{array}{r}b\end{array}$  the imaginary part .
  ::复数是 a+Bi 格式的数项,其中a和b为真实数字。我们称之为复数的真实部分,b为假想部分。

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  Note: 
  ::注:

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  • If  $\begin{array}{r}b=0\end{array}$  , then   $\begin{array}{r}a\end{array}$  remains  and $\begin{array}{r}a\end{array}$  is a real number.  As you can see in the Venn Diagram below, all real numbers are complex numbers; that is, the set of real numbers is a subset of the set of complex numbers. 
    ::如果 b=0,那么遗骨和a 是一个真实数字。从下面的Venn 图表中可以看到,所有真实数字都是复杂的数字;也就是说,真实数字是一组复杂数字的一个子集。
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  • If  $\begin{array}{r}a=0\end{array}$  , then the number is only   $\begin{array}{r}bi\end{array}$   and is called a pure .
    ::如果 a=0,则数字为双数,称为纯数。
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  • Two complex numbers are equal if their real parts are equal and their imaginary parts are equal. 
    ::两个复杂的数字是相等的,如果它们的实际部分是相等的,而它们的假想部分是相等的。

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  The Venn Diagram below shows the relationships between the sets of numbers we will use. The term "imaginary" was used by mathematicians when these numbers were first described, because they believed them to be fake. We will avoid using the term "imaginary number" unless the number is a pure imaginary number. Complex numbers are not fake; in fact, they have many real-world applications. 
  ::下面的文恩图表显示了我们将要使用的一组数字之间的关系。 数学家在首次描述这些数字时使用了“ 想象” 一词, 因为他们相信这些数字是假的。 除非数字是一个纯粹的假想数字, 否则我们将避免使用“ 想象数字 ” 。 复杂数字不是伪造的; 事实上, 它们有许多真实世界应用 。

  

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  by Mathispower4u demonstrates how to identify and simplify complex numbers. 
  ::由 Mathispower4u 演示如何识别和简化复杂数字。

   

   

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  Example 1
  ::例1

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  Rewrite $\begin{array}{r}\sqrt{\text{-}49}\end{array}$  and $\begin{array}{r}\sqrt{\text{-}162}\end{array}$ in terms of $\begin{array}{r}i\end{array}$ and simplify each  radical.
  ::重写 -49 和 -162 以i 重写 -49 和 -162, 并简化每个激进。

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  Solution:  Write -49 as the product of a positive number and -1. 
  ::解决方案: 将 - 49 写成正数和 - 1 的产物。

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  $$\begin{array}{r}\sqrt{\text{-}49}=\sqrt{49}\cdot \sqrt{\text{-}1}=7i\end{array}$$

  
  ::-49=49-1=7i

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  Again, write -162 as the square root of a positive number and -1.
  ::重写 -162 作为正数的平方根和 -1 。

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  $$\begin{array}{r}\sqrt{\text{-}162}=\sqrt{162}\cdot \sqrt{\text{-}1}=i\sqrt{162}=i\sqrt{81\cdot 2}=9i\sqrt{2}\end{array}$$

  
  ::-162=162-1=i162=i812=9i2

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     WARNING
  ::警告

  When there is an  i and a root, we put the  i in front of the root to avoid confusion over whether the  i is inside the root or not, e.g.,  $\begin{array}{r}\sqrt{2}i\end{array}$  or  $\begin{array}{r}\sqrt{2i}\end{array}$ .  

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  by Tyler Wallace demonstrates how to identify and simplify complex numbers. 
  ::泰勒·华莱士教授如何识别和简化复杂数字。

   

   

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  Powers of  $\begin{array}{r}i\end{array}$  
  ::i 权力 i 权力

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  $\begin{array}{r}i\end{array}$  has some interesting properties, one of which is its  powers. Let's consider: 
  ::我有一些有趣的属性, 其中之一是它的力量。

  $\begin{array}{r}{i}^2\end{array}$	$\begin{array}{r}{i}^2=i\cdot i=\sqrt{\text{-}1}\cdot \sqrt{\text{-}1}={\sqrt{\text{-}1}}^2=\text{-}1\end{array}$
  $\begin{array}{r}{i}^3\end{array}$	$\begin{array}{r}{i}^3=i^2\cdot i=\text{-}1\cdot i=\text{-}i\end{array}$
  $\begin{array}{r}{i}^4\end{array}$	$\begin{array}{r}{i}^4=i^2\cdot i^2=\text{-}1\cdot \text{-}1=1\end{array}$
  $\begin{array}{r}{i}^5\end{array}$	$\begin{array}{r}{i}^5=i^4\cdot i=1\cdot i=i\end{array}$
  $\begin{array}{r}{i}^6\end{array}$	$\begin{array}{r}{i}^6=i^4\cdot i^2=1\cdot \text{-}1=\text{-}1\end{array}$

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  Notice there is  a pattern. T he powers of   $\begin{array}{r}i\end{array}$   repeat every 4 powers.  If we want to find   $\begin{array}{r}{i}^{19}\end{array}$ , divide 19 by 4 and determine the remainder. The remainder will tell you what power of $\begin{array}{r}i\end{array}$  the original expression is the same as.
  ::注意有一个模式。 我重复每四种权力的权力。 如果我们想要找到 i19, 将19 除以 4 和决定剩余部分。 其余部分将告诉您 i 的原表达式与 i 相同的力量 。

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  $$\begin{array}{rl}\frac{19}{4}& =4\text{R}3\\ \\ i^{19}& =i^{16}\cdot i^3=1\cdot i^3=\text{-}i\end{array}$$

  
  ::194=4 R3i19=i16i3=1i3=-i

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  Example 2
  ::例2

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  Find:
  ::查找 :

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  a.  $\begin{array}{r}{i}^{32}\end{array}$
  ::a. i32

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  b.  $\begin{array}{r}{i}^{50}\end{array}$
  ::b. 150个

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  c.  $\begin{array}{r}{i}^7\end{array}$
  ::c. c. i7

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  Solution:
  ::解决方案 :

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  a. 32 is divisible by 4, so $\begin{array}{r}{i}^{32}=1\end{array}$ .
  ::a. 32 可除以 4 , 所以 i32= 1 。

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  b.  $\begin{array}{r}50÷4=12\end{array}$ , with a remainder of 2. Therefore, $\begin{array}{r}{i}^{50}=i^2=\text{-}1\end{array}$ .
  ::b. 504=12,其余为2。 因此,i50=i2=1。

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  c.  $\begin{array}{r}7÷4=1\end{array}$ , with a remainder of 3. Therefore, $\begin{array}{r}{i}^7=i^3=\text{-}i.\end{array}$
  ::c. 74=1,其余为3。 因此,i7=i3=-i。

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  Addition and Subtraction of Complex Numbers
  ::复杂数字的增减

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  To add and subtract complex numbers, we combine real parts with real parts and imaginary parts with imaginary parts. 
  ::为了增加和减去复杂数字,我们把真实部分与真实部分结合起来,想象部分与想象部分结合起来。

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  Example 3
  ::例3

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  Simplify the complex expressions.
  ::简化复杂表达式。

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  a.  $\begin{array}{r}(6-4i)+(5+8i)\end{array}$
  ::a. (6-4i)+(5+8i)

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  b.  $\begin{array}{r}(8-3i)-(12-i)\end{array}$
  ::b. (8-3i)-(12-i)

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  Solution:  Treat  $\begin{array}{r}i\end{array}$  like the variable part of an expression when you are adding or subtracting—combine the coefficients and keep the  $\begin{array}{r}i\end{array}$ .  
  ::解答: 当您在添加或减- 连接系数并保留 i 时, 对待我就像表达式的变量部分 。

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  a.  $\begin{array}{r}(6-4i)+(5+8i)=6-4i+5+8i=11+4i\end{array}$
  ::a. (6-4i)+(5+8i)=6-4i+5+8i=11+4i

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  b. Distribute -1 to each part in the 2nd complex number,  and combine like parts.
  ::b. 将-1分配给第二个复合数字的每个部分,并合并成相同的部件。

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  $$\begin{array}{r}(8-3i)-(12-i)=8-3i-12+i=\text{-}4-2i\end{array}$$

  
  :8-3i)-(12-i)=8-3i-12+i=4-2i)

   

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  Example 4 
  ::例4

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  Simplify the  complex expression  $\begin{array}{r}9-(4+i)+(2-7i).\end{array}$
  ::简化复杂的表达式 9-( 4+i)+( 2-7i) 。

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  Solution:   $\begin{array}{r}9-(4+i)+(2-7i)=9-4-i+2-7i=7-8i\end{array}$
  ::溶液: 9-(4+一)+(2-7i)=9-4-i+2-7i=7-8i

   

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  Multiplication of Complex Numbers
  ::复杂数字的乘数

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  Multiplication of complex numbers is similar to multiplication of variable expressions. Use the distributive property and the FOIL method to multiply, and then combine like terms.  Simplify any powers of  i . 
  ::复杂数字的乘法与变量表达式的乘法相似。使用分配属性和FOIL方法进行乘法,然后将类似术语合并。简化 i 的任何权力。

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  Example 5
  ::例5

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  Multiply :
  ::乘数 :

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  a.  $\begin{array}{r}6i(1-4i)\end{array}$
  ::a. 6i(1-4i)

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  b.  $\begin{array}{r}(5-2i)(3+8i)\end{array}$
  ::b. (5-2i)(3+8i)

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  Solution:
  ::解决方案 :

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  a. Distribute the $\begin{array}{r}6i\end{array}$ to both parts inside the parentheses.
  ::a. 将6i分到括号内两部分。

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  $$\begin{array}{r}6i(1-4i)=6i-24i^2\end{array}$$

  
  ::6i( 1- 4i) =6i- 24i2

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  Substitute $\begin{array}{r}{i}^2=\text{-}1\end{array}$ and simplify further.
  ::替代i2=1,进一步简化。

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  $$\begin{array}{rl}& =6i-24(\text{-}1)\\ & =24+6i\end{array}$$

  
  ::=6i-24(-1)=24+6i

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  It is conventional to put the  real part 1st.
  ::将真正的第一部分放在第一部分是传统的。

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  b. FOIL the two numbers  together.
  ::b. 将两个数字加在一起。

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  $$\begin{array}{rl}(5-2i)(3+8i)& =15+40i-6i-16i^2\\ & =15+34i-16i^2\end{array}$$

  
  :5-2i)(3+8i)=15+40i-6i-16i2=15+34i-16i)

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  Substitute $\begin{array}{r}{i}^2=\text{-}1\end{array}$ and simplify further.
  ::替代i2=1,进一步简化。

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  $$\begin{array}{rl}& =15+34i-16(\text{-}1)\\ & =15+34i+16\\ & =31+34i\end{array}$$

  
  ::=15+34i-16(-1)=15+34i+16=31+34i

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  Example 6
  ::例6

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  If a current is 6 amps and goes through a part of the circuit with an impedance of $\begin{array}{r}70+120i\end{array}$   ohms, what is an expression for the voltage of the circuit? 
  ::如果电流是6安普,经过部分电路,阻力为70+120奥姆,电路电压的表达方式是什么?

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  Solution: As we saw in the Introduction,  $\begin{array}{r}V=IZ\end{array}$ , so $\begin{array}{r}V=6(70+120i)=6\cdot 70+6\cdot 120i=420+720i\end{array}$ .   
  ::解答:正如我们在导言中看到的那样,V=IZ,所以V=6(70+120i)=670+6120i=420+720i。

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     WARNING
  ::警告

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  When computing with radicals, be sure to rewrite even  square  roots of negative numbers as complex numbers 1st, and then compute the result. 
  ::当使用基数计算时,一定要将负数的平方根重写为复杂的数字1,然后计算结果。

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  $$\begin{array}{rl}\sqrt{\text{-}2}\cdot \sqrt{\text{-}3}& \ne \sqrt{\text{-}2\cdot \text{-}3}\ne \sqrt{6}\\ \\ \sqrt{\text{-}2}\cdot \sqrt{\text{-}3}& =i\sqrt{2}\cdot i\sqrt{3}=i^2\sqrt{6}=\text{-}\sqrt{6}\end{array}$$

  
  ::-2-3-2-3-6-2-3=i2i3=i26=-6

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  Division of Complex Numbers
  ::复杂数字司

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  Dividing complex numbers is a bit more complicated. Similar to numbers expressed in radical form, we tend not to leave complex numbers in the denominator of a fraction. To get rid of the complex number in the denominator, we multiply by what is called the complex conjugate . A complex number of  the form $\begin{array}{r}a+bi\end{array}$  has a complex conjugate of the form  $\begin{array}{r}a-bi\end{array}$ . For example, the complex conjugate of $\begin{array}{r}\text{-}6+5i\end{array}$ would be $\begin{array}{r}\text{-}6-5i\end{array}$ . Notice what happens when we multiply complex conjugates.
  ::分解复杂的数字比较复杂。 类似以激进形式表示的数字, 我们往往不会在分母的分母中留下复杂的数字。 为了摆脱分母中的复杂数字, 我们乘以所谓的复杂共产体。 a+bi 形式中的复杂数字与 a- bi 形式有复杂的共产体。 例如, 6+5i 的复杂共产体将是 - 6- 5i 。 注意当我们把复杂的共产体相乘时会发生什么 。

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  $$\begin{array}{rl}(a+bi)(a-bi)& =a^2-abi+abi-b^2{i}^2=a^2-b^2(\text{-}1)=a^2+b^2\\ \\ (\text{-}6+5i)(\text{-}6-5i)& =(\text{-}6{)}^2+30i-30i-25i^2=36-25(\text{-}1)=36+25=61\end{array}$$

  
  ::a+bi(a-bi)=a2-abi+abi-b2i2=a2-b2(-1)=a2-b2(-1)=a2+b2(-6+5i)(-6-5i)=(-6)2+30i-30i--25i2=36-25(-1)=36+25=61

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  There is no imaginary part in the result. The result is just a real number. When  dividing complex numbers, our 1st step will be to multiply the numerator and the denominator by the complex conjugate of the denominator.
  ::结果是没有虚构的部分。 结果只是一个真实的数字。 当分割复杂数字时, 我们的第一个步骤就是将分子和分母乘以分母的复杂组合。

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  Example 7
  ::例7

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  Divide  $\begin{array}{r}\frac{8-3i}{6i}\end{array}$ .
  ::8-3i6i除以8-3i6i。

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  Solution:  Our 1st step is to multiply both the numerator and denominator by the conjugate of the denominator. In this case, the conjugate of $\begin{array}{r}6i\end{array}$     is $\begin{array}{r}\text{-}6i\end{array}$  .
  ::解答:我们的第一个步骤是将分子和分母乘以分母的组合。在这种情况下,6i的组合是 - 6i。

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  $$\begin{array}{rl}\frac{8-3i}{6i}\cdot \frac{\text{-}6i}{\text{-}6i}& =\frac{\text{-}48i+18i^2}{36i^2}\\ & =\frac{\text{-}18-48i}{\text{-}36}\\ & =\frac{18}{36}+\frac{48}{36}i\\ & =\frac{1}{2}+\frac{4}{3}i\end{array}$$

  
  ::8-3i6i-6i-6i=-48i+18i236i2=-1848i-36=1836+4836i=12+43i

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  When the complex number contains fractions, write the number in standard form, keeping the real and imaginary parts separate. Reduce both fractions separately.
  ::当复数包含分数时, 请以标准格式写入数字, 将真实部分和想象部分分开。 将两个分数分开 。

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  Example 8
  ::例8

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  Divide  $\begin{array}{r}\frac{3-5i}{2+9i}\end{array}$ .
  ::除以 3-5i2+9i。

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  Solution: M ultiply the numerator  and the denominator  by the complex conjugate of the denominator, $\begin{array}{r}2-9i\end{array}$ .
  ::解答:乘以2-9i这一分母的复杂组合体的分子和分母。

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  $$\begin{array}{rl}\frac{3-5i}{2+9i}\cdot \frac{2-9i}{2-9i}& =\frac{6-27i-10i+45i^2}{4-18i+18i-81i^2}\\ & =\frac{6-37i-45}{4+81}\\ & =\frac{\text{-}39-37i}{85}\\ & =\text{-}\frac{39}{85}-\frac{37}{85}i\end{array}$$

  
  ::3-5i2+9i2-9i2-9i2-9i=6-27i-10i+45i24-18i+18i-81i2=6-37i-48i2=6-375i-454+81=-39-3785=-3985-3785i

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  Example 9
  ::例9

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  Divide  $\begin{array}{r}\frac{8+i}{6-4i}\end{array}$ .
  ::8+i6-4i除以8+i6-4i。

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  Solution: Multiply the numerator and denominator by the complex conjugate of the denominator, $\begin{array}{r}6+4i\end{array}$ .
  ::解答:乘以6+4i的分母的复杂组合,将分子和分母乘以6+4i。

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  $$\begin{array}{rl}\frac{8+i}{6-4i}\cdot \frac{6+4i}{6+4i}& =\frac{48+32i+6i+4i^2}{36+24i-24i-16i^2}\\ & =\frac{48+38i-4}{36+16}\\ & =\frac{44+38i}{52}\\ & =\frac{44}{52}+\frac{38}{52}i\\ & =\frac{11}{13}+\frac{19}{26}i\end{array}$$

  
  ::8+i6-4i6+4i6+4i6+4i6+4i=48+32i+6i+6i+4i+4i236+24i-24i-166i2=48+38i-436+16=44+38i52=4452+3852i=1113+1926i

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  by Mathispower4u demonstrates how to add, subtract, multiply, and divide complex numbers. 
  ::用 Mathispower4u 演示如何添加、减、乘和分隔复杂数字。

   

   

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  Feature: Getting Lost in a Pineapple
  ::特色:迷失在菠萝中

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  by Meredith Beaton 
  ::梅雷迪思·贝顿(签名)

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  A fractal is a never-ending pattern that repeats itself on different scales (this is called self-similarity). While the fractal itself can look extremely complex, the initial shape or item to be repeated is often extremely simple.
  ::分形是一种在不同的尺度上重复的永不终止的模式(这被称为自我相似性 ) 。 虽然分形本身看起来非常复杂,但要重复的最初形状或项目往往非常简单。

  

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  Amazing But True
  ::令人惊异但真实

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  In mathematics, a fractal is created not by shapes, but by calculating a simple equation over and over and over, thousands of times, each time using the answer for the next repetition of the equation.
  ::在数学中,分形不是通过形状形成的,而是通过一个简单的方程式, 一次又一次地, 一次又一次地, 一次地, 一次地, 一次地, 一次地, 一次地, 一次地, 一次地, 一次地, 一次地, 一次地, 一次地, 一次地, 一次地, 一次地, 一次地, 一次地, 一次地, 一次地用答案 来重复这个方程式的下一次重复。

  

   

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  In 1975, mathematician Benoit Mandelbrot published the first book on fractals, including the Mandelbrot Set of fractals. Sampling complex numbers and performing a mathematical operation on those numbers is the process that creates the Mandelbrot Set of fractals. Therefore, complex numbers are used to generate fractals. These fractals, Mandelbrot argued, could be used to explain rough phenomena in nature, such as lungs, mountaintops, coastlines, and others.
  ::1975年,数学家Benoit Mandelbrot出版了第一本关于分形的书,其中包括曼德尔布罗特分形集,对复杂数字进行取样和对这些数字进行数学操作的过程是创建曼德尔布罗特分形集的过程,因此,复杂的数字被用来产生分形。Mandelbrot认为,这些分形可以用来解释自然界的粗糙现象,例如肺部、山顶、海岸线等。

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  by IBM provides an interview with Benoit Mandelbrot. In the interview, Mandelbrot shares his love for mathematics and how it led him to his wondrous discovery of fractals. His work lives on today in many innovations in science, design, telecommunications, medicine, renewable energy, film (special effects), gaming (computer graphics), and more.    
  ::曼德尔布罗特(Benoit Mandelbrot)提供了与Benoit Mandelbrot(Benoit Mandelbrot)的访谈。 在访谈中,曼德尔布罗特分享了他对数学的热爱,以及他是如何导致他神奇地发现分形的。 他今天在科学、设计、电信、医药、可再生能源、电影(特别效果)、游戏(计算机图形)等许多创新领域工作。

   

   

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  Summary
  ::摘要

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  • To simplify a radical with -1 as a factor, rewrite the radical as a product of two radicals—a positive factor and a factor of -1.
    ::为了简化激进,以-1作为因素,将激进改写为两个激进的产物——积极因素和-1因素。
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  • To add and subtract complex numbers, combine the real parts and combine the imaginary parts.
    ::要添加和减去复数,要合并实际部分,要合并假想部分。
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  • To multiply complex numbers, apply the distributive property or the FOIL technique.
    ::乘以复杂数字,应用分配财产或FOIL技术。
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  • To divide complex numbers, multiply the numerator and the denominator by the complex conjugate of the denominator.  
    ::要分隔复杂数字, 将分子和分母乘以分母的复杂组合。

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  Review
  ::回顾

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  Simplify each expression.
  ::简化每个表达式。

  1.   $\begin{array}{r}\sqrt{\text{-}242}\end{array}$

  2.  $\begin{array}{r}6\sqrt{\text{-}45}\end{array}$

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  3.  $\begin{array}{r}\text{-}12i\sqrt{98}\end{array}$
  ::-3.1298

  4.  $\begin{array}{r}\sqrt{\text{-}32}\cdot \sqrt{\text{-}27}\end{array}$

  5.  $\begin{array}{r}\sqrt{\text{-}\frac{16}{80}}\end{array}$

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  6. a.  $\begin{array}{r}16i^{22}\end{array}$   b.  $\begin{array}{r}\text{-}9i^{65}\end{array}$   c.  $\begin{array}{r}2i^{91}\end{array}$
  ::6. a. 16i22 b. - 9i65 c. 2i91

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  7.   $\begin{array}{r}(8-i)-(3+4i)+15i\end{array}$
  ::7. (8-i)-(3+4i)+15i

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  8.   $\begin{array}{r}(0.2+1.5i)-(\text{-}0.6+i)\end{array}$
  ::8. (0.2+1.5i)-(-0.6+一)

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  9.   $\begin{array}{r}6+(18-i)-(2+12i)\end{array}$
  ::9. 6+(18-i)-(2+12i)

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  10.   $\begin{array}{r}\text{-}2i(11-4i)\end{array}$
  ::-10-2i(11-4i)

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  11.   $\begin{array}{r}(4+5i)(3+16i)\end{array}$
  ::11. (4+5i)(3+16i)

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  12.   $\begin{array}{r}4i(2-3i)(7+3i)\end{array}$
  ::12. 4i(2-3i)(7+3i)

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  13.   $\begin{array}{r}\frac{4+9i}{3i}\end{array}$
  ::13. 4+9i3i

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  14.   $\begin{array}{r}\frac{2-i}{2+i}\end{array}$
  ::14.2-i2+i

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  15.   $\begin{array}{r}\frac{14+9i}{7-20i}\end{array}$
  ::15. 14+9i7-20i

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  Explore More
  ::探索更多

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  1. The convenience of conjugates shows up many times in mathematics. Where in the "Factoring"   chapter did we use conjugates?
  ::1. 在数学中,共产主义的方便性多次出现,在 " 因素 " 一章中,我们在哪里使用共产主义?

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  2. What is the error? 

  $$\begin{array}{r}\sqrt{\text{-}6}\cdot \sqrt{\text{-}6}=\sqrt{(\text{-}6{)}^2}=\sqrt{36}=6\end{array}$$

   
  ::2. 什么是错误? -6-6=(-6)2=36=6

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  3. Consider $\begin{array}{r}2+3i\end{array}$ . Now, multiply it by $\begin{array}{r}i\end{array}$ . Plot the new complex number. Now, find the slopes of the lines that connect the origin to each of these points. What relationship do they have?
  ::3. 考虑 2+3i 。 现在, 将其乘以 i. 绘制新的复数。 现在, 找到将起源与其中每个点连接的线条的斜度。 它们有什么关系 ?

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  4. Recall from the Feature the connection between complex numbers and fractals. One example of a fractal is the Mandelbrot Set. Below is a graph of the Mandelbrot Set, the set of complex numbers, c , where the function values for  $\begin{array}{r}f(z)=z^2+c\end{array}$  do not grow without bound when you iterate and start with $\begin{array}{r}z=0\end{array}$ . For example, when $\begin{array}{r}c=1\end{array}$ , the 1st five values are 0, 1, 2, 5, and 26. Notice that 1 is not included in the shaded region below.    
  ::4. 从特征中回顾复杂数字和分形之间的联系。一个分形的例子是Mandelbrot 集。下面是Mandelbrot 集图,一组复杂数字,c,其中f(z)=z2+c的函数值不会在不附带约束的情况下生长,当您用 z=0 开始转动时。例如,当 c=1 时,前五个值为 0, 1, 2, 5 和 26. 注意 1 不包括在下面的阴影区域中 。

  

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  Find the 1st five values for $\begin{array}{r}c=\text{-}1\end{array}$  and  $\begin{array}{r}c=i\end{array}$ . 
  ::查找 c=-1 和 c=i 的值前五位。

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  5. Electrical engineers typically use $\begin{array}{r}j\end{array}$  rather than $\begin{array}{r}i\end{array}$  to represent imaginary numbers. This is to prevent confusion with the $\begin{array}{r}i\end{array}$  used as a variable to represent electrical current. If the voltage of a circuit is $\begin{array}{r}35+5j\end{array}$  volts and the current is $\begin{array}{r}3+j\end{array}$  amps, find the impedance.
  ::5. 电气工程师通常使用j而不是i来代表想象数字,这是为了防止与i作为代表电流的变量混为一谈。如果电路电压为35+5j伏特,电流为3+j安普斯,则会发现阻力。

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  Answers to Review and Explore More Problems
  ::对审查和探讨更多问题的答复

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  Please see the Appendix. 
  ::请参看附录。

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  PLIX
  ::PLIX

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  Try these interactives that reinforce the concepts explored in this section.
  ::尝试这些加强本节所探讨概念的交互作用。

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  References
  ::参考参考资料

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  1. "Electrical Impedance," last edited April 27, 2017, .
  ::1. 2017年4月27日,