Course: 5.8 确定复杂数字

Section outline

Select section General

Collapse Expand
General

Collapse all Expand all
- Select activity 新闻通告
  
  新闻通告 Forum
Select section 定义复杂数字

Collapse Expand
定义复杂数字
The coldest possible temperature, known as absolute zero is almost –460 degrees Fahrenheit. What is the square root of this number?
::最冷的温度,被称为绝对零的温度几乎 — — 460华氏度。这个数字的平方根是什么?

Complex Numbers
::复数数

Before this concept, all numbers have been real numbers. 2, -5, $\begin{aligned} \sqrt{11} \end{aligned}$ , and $\begin{aligned} \frac{1}{3} \end{aligned}$ are all examples of real numbers. With what we have previously learned, we cannot find $\begin{aligned} \sqrt{- 25} \end{aligned}$ because you cannot take the square root of a negative number. There is no real number that, when multiplied by itself, equals -25. Let’s simplify $\begin{aligned} \sqrt{- 25} \end{aligned}$ .
::在这个概念之前,所有数字都是真实数字。 2, 5, 11, 和13都是真实数字的例子。以我们以前所学的,我们无法找到25, 因为你不能从负数的平方根中找到25。没有真实数字, 当自己乘以乘以25时, 等于25。让我们简化-25。

$\begin{aligned} \sqrt{- 25} = \sqrt{25 \cdot - 1} = 5 \sqrt{- 1} \end{aligned}$

In order to take the square root of a negative number we are going to assign $\begin{aligned} \sqrt{- 1} \end{aligned}$ a variable, $\begin{aligned} i \end{aligned}$ . $\begin{aligned} i \end{aligned}$ represents an . Now, we can use $\begin{aligned} i \end{aligned}$ to take the square root of a negative number.
::为了从负数的平方根中取出一个负数的平方根, 我们要指定 - 1 一个变量, i. i 代表一个。现在我们可以用我来从负数的平方根中取出一个负数的平方根。

$\begin{aligned} \sqrt{- 25} = \sqrt{25 \cdot - 1} = 5 \sqrt{- 1} = 5 i \end{aligned}$

::-25=251=5-1=5i

All complex numbers have the form $\begin{aligned} a + b i \end{aligned}$ , where $\begin{aligned} a \end{aligned}$ and $\begin{aligned} b \end{aligned}$ are real numbers. $\begin{aligned} a \end{aligned}$ is the real part of the complex number and $\begin{aligned} b \end{aligned}$ is the imaginary part . If $\begin{aligned} b = 0 \end{aligned}$ , then $\begin{aligned} a \end{aligned}$ is left and the number is a real number. If $\begin{aligned} a = 0 \end{aligned}$ , then the number is only $\begin{aligned} b i \end{aligned}$ and called a pure imaginary number . If $\begin{aligned} b \neq 0 \end{aligned}$ and $\begin{aligned} a \neq 0 \end{aligned}$ , the number will be an imaginary number.
::所有复数都有 a+Bi 的窗体, 其中 a 和 b 是真实数字。 a 是复数的真实部分, b 是假想部分。如果 b=0, 则左键是一个数字。如果 a=0, 则数字只是一个实际数字。如果 a=0, 数字仅是双数, 称为纯假想数字。如果 b=0 和 a=0, 数字将是一个虚想数字。

Let's find $\begin{aligned} \sqrt{- 162} \end{aligned}$ .
::我们找找 -162

First pull out the $\begin{aligned} i \end{aligned}$ . Then, simplify $\begin{aligned} \sqrt{162} \end{aligned}$ .
::先拿出i 然后简化162

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

::- 1621162=i162=i812=9i2

Powers of i
::i 权力 i 权力

In addition to now being able to take the square root of a negative number, $\begin{aligned} i \end{aligned}$ also has some interesting properties. Try to find $\begin{aligned} i^{2}, i^{3}, \end{aligned}$ and $\begin{aligned} i^{4} \end{aligned}$ .
::除了现在可以选择负数的平方根外, 我还有一些有趣的属性。试着找到 i2, i3 和 i4 。

Step 1: Write out $\begin{aligned} i^{2} \end{aligned}$ and simplify. $\begin{aligned} i^{2} = i \cdot i = \sqrt{- 1} \cdot \sqrt{- 1} = {\sqrt{- 1}}^{2} = - 1 \end{aligned}$
::第1步: 写出i2 并简化 i2=ii111121

Step 2: Write out $\begin{aligned} i^{3} \end{aligned}$ and simplify. $\begin{aligned} i^{3} = i^{2} \cdot i = - 1 \cdot i = - i \end{aligned}$
::第2步:写出i3并简化。 i3=i2_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_i_

Step 3: Write out $\begin{aligned} i^{4} \end{aligned}$ and simplify. $\begin{aligned} i^{4} = i^{2} \cdot i^{2} = - 1 \cdot - 1 = 1 \end{aligned}$
::第3步:写出i4并简化。 i4=i2i2=i2i2}11=1

Step 4: Write out $\begin{aligned} i^{5} \end{aligned}$ and simplify. $\begin{aligned} i^{5} = i^{4} \cdot i = 1 \cdot i = i \end{aligned}$
::第4步:写出i5并简化。 i5=i4i=1i=i

Step 5: Write out $\begin{aligned} i^{6} \end{aligned}$ and simplify. $\begin{aligned} i^{6} = i^{4} \cdot i^{2} = 1 \cdot - 1 = - 1 \end{aligned}$
::第5步:写出i6并简化。 i6=i4i2=11111

Step 6: Do you see a pattern? Describe it and try to find $\begin{aligned} i^{19} \end{aligned}$ .
::第6步: 你看到一个模式吗? 描述它并尝试找到 i19 。

You should see that the powers of $\begin{aligned} i \end{aligned}$ repeat every 4 powers. So, all the powers that are divisible by 4 will be equal to 1. To find $\begin{aligned} i^{19} \end{aligned}$ , divide 19 by 4 and determine the remainder. That will tell you what power it is the same as.
::您应该看到我重复每四个权力的权力。因此, 所有四分之四的权力将等于 1 。要找到 i19, 将19 除以 4, 并确定剩余的权力。这将告诉您什么权力与什么权力相同。

$\begin{aligned} i^{19} = i^{16} \cdot i^{3} = 1 \cdot i^{3} = - i \end{aligned}$

::i19=i16_i3=1}i19=i16_i3=1}i19=i16_i3_i

Now, let's find the following powers of i.
::现在,让我们找到我下面的力量

$\begin{aligned} i^{32} \end{aligned}$
::i32 i32

32 is divisible by 4, so $\begin{aligned} i^{32} = 1 \end{aligned}$ .
::32 以 4 变4 表示, 所以 i32= 1 。

$\begin{aligned} i^{50} \end{aligned}$
::150 i50

$\begin{aligned} 50 \div 4 = 12 \end{aligned}$ , with a remainder of 2. Therefore, $\begin{aligned} i^{50} = i^{2} = - 1 \end{aligned}$ .
::504=12,剩余2, 因此,i50=i2=1。

$\begin{aligned} i^{7} \end{aligned}$
::i7 i7

$\begin{aligned} 7 \div 4 = 1 \end{aligned}$ , with a remainder of 3. Therefore, $\begin{aligned} i^{7} = i^{3} = - i \end{aligned}$
::74=1,其余为3,因此,i7=i3

Finally, let's simplify the following complex expressions.
::最后,让我们简化以下复杂的表达方式。

$\begin{aligned} (6 - 4 i) + (5 + 8 i) \end{aligned}$
:6-4i)+(5+8i)

$\begin{aligned} (6 - 4 i) + (5 + 8 i) = 6 - 4 i + 5 + 8 i = 11 + 4 i \end{aligned}$
:6-4i)+(5+8i)=6-4i+5+8i=11+4i

$\begin{aligned} 9 - (4 + i) + (2 - 7 i) \end{aligned}$
::9-(4+一)+(2-7i)

$\begin{aligned} 9 - (4 + i) + (2 - 7 i) = 9 - 4 - i + 2 - 7 i = 7 - 8 i \end{aligned}$
::9-(4+一)+(2-7i)=9-4-4-i+2-7i=7-8i

To add or subtract complex numbers, you need to combine like terms. Be careful with negatives and properly distributing them. Your answer should always be in standard form , which is $\begin{aligned} a + b i \end{aligned}$ .
::要添加或减去复数, 您需要将类似条件合并。注意底片并正确分配。您的回答应该总是以标准格式( a+Bi) 表示, 即 a+Bi 。

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the square root of -460 degrees.
::早些时候,你被要求找到 -460度的平方根

We're looking for $\begin{aligned} \sqrt{- 460} \end{aligned}$ .
::我们正在寻找 -460。

First we need to pull out the $\begin{aligned} i \end{aligned}$ . Then, we need to simplify $\begin{aligned} \sqrt{460} \end{aligned}$ .
::首先,我们需要拿出i。然后,我们需要简化460。

$\begin{aligned} \sqrt{- 460} = \sqrt{- 1} \cdot \sqrt{460} = i \sqrt{460} = i \sqrt{4 \cdot 115} = 2 i \sqrt{115} \end{aligned}$

::- 4601460=i460=i4115=2i115

Example 2
::例2

Simplify $\begin{aligned} \sqrt{- 49} \end{aligned}$ .
::简化-49。

Rewrite $\begin{aligned} \sqrt{- 49} \end{aligned}$ in terms of $\begin{aligned} i \end{aligned}$ and simplify the radical.
::重写 -49 以i 和简化激进。

$\begin{aligned} \sqrt{- 49} = i \sqrt{49} = 7 i \end{aligned}$

::- 49=i49=7i

Example 3
::例3

Simplify $\begin{aligned} \sqrt{- 125} \end{aligned}$ .
::简化-1125。

Rewrite $\begin{aligned} \sqrt{- 125} \end{aligned}$ in terms of $\begin{aligned} i \end{aligned}$ and simplify the radical.
::重写-125的i 并简化激进。

$\begin{aligned} \sqrt{- 125} = i \sqrt{125} = i \sqrt{25 \cdot 5} = 5 i \sqrt{5} \end{aligned}$

::-125=i125=i25=i25=5=5i5

Example 4
::例4

Simplify $\begin{aligned} i^{210} \end{aligned}$ .
::简化 i210 。

$\begin{aligned} 210 \div 4 = 52 \end{aligned}$ , with a remainder of 2. Therefore, $\begin{aligned} i^{210} = i^{2} = - 1 \end{aligned}$ .
::=210=i212=i2*1。

Example 5
::例5

Simplify $\begin{aligned} (8 - 3 i) - (12 - i) \end{aligned}$ .
::简化 (8- 3i)-( 12- i) 。

Distribute the negative and combine like terms.
::将负面的分布在一边, 并结合类似术语。

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

:8-3i)-(12-i)=8-3i-12+i4-2i)

Review
::回顾

Simplify each expression and write in standard form.
::简化每个表达式,并以标准格式写入。

$\begin{aligned} \sqrt{- 9} \end{aligned}$

$\begin{aligned} \sqrt{- 242} \end{aligned}$

$\begin{aligned} 6 \sqrt{- 45} \end{aligned}$

$\begin{aligned} - 12 i \sqrt{98} \end{aligned}$
::- 1298年

$\begin{aligned} \sqrt{- 32} \cdot \sqrt{- 27} \end{aligned}$

$\begin{aligned} 7 i \sqrt{- 126} \end{aligned}$
::7i-126 7i-126

$\begin{aligned} i^{8} \end{aligned}$
::i8 i8

$\begin{aligned} 16 i^{22} \end{aligned}$
::16i22 16i22

$\begin{aligned} - 9 i^{65} \end{aligned}$
::- 965 - 965

$\begin{aligned} i^{365} \end{aligned}$
::i365

$\begin{aligned} 2 i^{91} \end{aligned}$
::2 i91

$\begin{aligned} \sqrt{- \frac{16}{80}} \end{aligned}$

$\begin{aligned} (11 - 5 i) + (6 - 7 i) \end{aligned}$
:11-5i)+(6-7i)

$\begin{aligned} (14 + 2 i) - (20 + 9 i) \end{aligned}$
:14+2i)-(20+9i)

$\begin{aligned} (8 - i) - (3 + 4 i) + 15 i \end{aligned}$
:8-一)-(3+4i)+15i

$\begin{aligned} - 10 i - (1 - 4 i) \end{aligned}$
::--10i-(1-4i)

$\begin{aligned} (0.2 + 1.5 i) - (- 0.6 + i) \end{aligned}$
:0.2+1.5i)-(-0.6+i)

$\begin{aligned} 6 + (18 - i) - (2 + 12 i) \end{aligned}$
::6+(18-i)-(2+12i)

$\begin{aligned} - i + (19 + 22 i) - (8 - 14 i) \end{aligned}$
::-i+(19+22i)-(8)-14i)

$\begin{aligned} 18 - (4 + 6 i) + (17 - 9 i) + 24 i \end{aligned}$
::18-(4+6i)+(17-9i)+24i

Review (Answers)
::回顾(答复)

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。

Section outline

General

定义复杂数字

Complex Numbers ::复数数

Powers of i ::i 权力 i 权力

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Review ::回顾

Review (Answers) ::回顾(答复)

Complex Numbers
::复数数

Powers of i
::i 权力 i 权力

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Example 5
::例5

Review
::回顾

Review (Answers)
::回顾(答复)