Section: 定义复杂数字 | 5.8 确定复杂数字

Section outline

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

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)