节: 复杂数字的三角矩阵形式 | 6.9 复杂数字的三角测量形式

章节大纲

You have begun working with in your math class. While describing numbers in the complex plane , you realize that the plotting of a complex number is a lot like plotting a set of points on a rectangular coordinate system.
::您已经开始在数学课上工作。在描述复杂平面上的数字时, 您意识到一个复杂数字的绘图就像在矩形坐标系统中绘制一组点一样。

You also learned in math class that you can convert coordinates from a rectangular system into a polar system. As you are considering this, you plot the complex number $\begin{aligned} 2 + 3 i \end{aligned}$ . Can you somehow convert this into a type of polar plot that you've done before?
::您在数学课中也学到了可以将矩形系统坐标转换成极系统。在考虑这一点时, 您会绘制复杂的数字 2+3i 。您能以某种方式将它转换成您以前做过的极地图吗 ?

Trigonometric Form of Complex Numbers
::复杂数字的三角矩阵形式

A number in the form $\begin{aligned} a + b i \end{aligned}$ , where $\begin{aligned} a \end{aligned}$ and $\begin{aligned} b \end{aligned}$ are real numbers, and $\begin{aligned} i \end{aligned}$ is the imaginary unit, or $\begin{aligned} \sqrt{- 1} \end{aligned}$ , is called a complex number. Despite their names, complex numbers and have very real and significant applications in both mathematics and in the real world. Complex numbers are useful in pure mathematics, providing a more consistent and flexible number system that helps solve algebra and calculus problems. We will see some of these applications in problems below.
::以 a+B 格式显示的数字为 a+B, 其中a和b 是真实数字, i 是假想单位, 或-1, 称为复杂数字。尽管名称复杂, 在数学和现实世界中都有非常真实和重要的应用。复杂数字在纯数学中有用, 提供了更一致和灵活的数字系统, 有助于解决代数和计算问题。我们可以看到其中一些应用在下面的问题中。

The following diagram will introduce you to the relationship between complex numbers and .
::下图将向您介绍复杂数字和 .之间的关系。

In the figure above, the point that represents the number $\begin{aligned} x + y i \end{aligned}$ was plotted and a vector was drawn from the origin to this point. As a result, an angle in standard position , $\begin{aligned} θ \end{aligned}$ , has been formed. In addition to this, the point that represents $\begin{aligned} x + y i \end{aligned}$ is $\begin{aligned} r \end{aligned}$ units from the origin. Therefore, any point in the complex plane can be found if the angle $\begin{aligned} θ \end{aligned}$ and the $\begin{aligned} r - \end{aligned}$ value are known. The following equations relate $\begin{aligned} x, y, r \end{aligned}$ and $\begin{aligned} θ \end{aligned}$ .
::在上图中,绘制了代表 x+yi 数字的点,并绘制了从源到此点的矢量。结果,形成了标准位置的角。除此之外,代表 x+yi 的点是源的 r 单位。因此,如果知道角和 r- 值, 复合平面中的任何点都可以找到。以下方程式涉及 x,y,r 和。

$\begin{aligned} x = r \cos θ & y = r \sin θ & r^{2} = x^{2} + y^{2} & \tan θ = \frac{y}{x} \end{aligned}$

::x=rcosy=rsinr2=x2+y2tanyx

If we apply the first two equations to the point $\begin{aligned} x + y i \end{aligned}$ the result would be:
::如果我们对x+yi点应用前两个方程式,结果将是:

$\begin{aligned} x + y i = r \cos θ + r i \sin θ \to r (\cos θ + i \sin θ) \end{aligned}$

::x+yi=rcosçrisinr(cosisin)

The right side of this equation $\begin{aligned} r (\cos θ + i \sin θ) \end{aligned}$ is called the polar or trigonometric form of a complex number. A shortened version of this polar form is written as $\begin{aligned} r c i s θ \end{aligned}$ . The length $\begin{aligned} r \end{aligned}$ is called the absolute value or the modulus , and the angle $\begin{aligned} θ \end{aligned}$ is called the argument of the complex number. Therefore, the following equations define the :
::此等式 r( cosisin) 的右侧称为复数的极或三角形式。此极形式的缩写版本以 r cis 写成。 r 的长度称为绝对值或模量, 角度则称为复数的参数。因此, 以下公式定义了 :

$\begin{aligned} r^{2} = x^{2} + y^{2} & \tan θ = \frac{y}{x} & x + y i = r (\cos θ + i \sin θ) \end{aligned}$

::r2=x2+y2tanyxx+yyi=r(cosisin)

It is now time to implement these equations perform the operation of converting complex numbers in standard form to complex numbers in polar form. You will use the above equations to do this.
::现在是实施这些方程式的时候了,它可以将标准形式的复杂数字转换成极形式的复杂数字。您将使用上述方程式来做到这一点。

Let's look at a some problems that involve the trigonometric form of complex numbers.
::让我们来看看一些问题, 涉及到三角形式复杂的数字。

1. Represent the complex number $\begin{aligned} 5 + 7 i \end{aligned}$ graphically and express it in its polar form.
::1. 5+7i综合体图示5+7i,以极形表示。

Here is the graph of $\begin{aligned} 5 + 7 i \end{aligned}$ .
::这是5+7i的图。

Converting to polar from rectangular, $\begin{aligned} x = 5 \end{aligned}$ and $\begin{aligned} y = 7 \end{aligned}$ .
::从矩形、 x=5 和 Y=7 转换成极。

$\begin{aligned} r = \sqrt{5^{2} + 7^{2}} = 8.6 & \tan θ = \frac{7}{5} \\ \tan^{- 1} (\tan θ) = \tan^{- 1} \frac{7}{5} \\ θ = {54.5}^{\circ} \end{aligned}$

::=52+72=8.6tan75tan-1(tan)=tan-17554.5

So, the polar form is $\begin{aligned} 8.6 (\cos {54.5}^{\circ} + i \sin {54.5}^{\circ}) \end{aligned}$ .
::因此,极形是8.6(cos54.5) 。

Another widely used notation for the polar form of a complex number is $\begin{aligned} r ∠ θ = r (\cos θ + i \sin θ) \end{aligned}$ . Finally, there is a third way to write a complex number, in the form of $\begin{aligned} r c i s θ \end{aligned}$ , where "r" is the length of the vector in polar form, and $\begin{aligned} θ \end{aligned}$ is the angle the vector makes with the positive "x" axis. This makes a total of three ways to write the polar form of a complex number.
::复数极形式的另一种常用符号是 rr(cosisin) 。最后,有第三种方法可以写出一个复数, 以 rcis 的形式, “r” 是极形矢量的长度, 是矢量与正“ x” 轴的角。这共有三个方法可以写出复数的极形式。

$\begin{aligned} x + y i = r (\cos θ + i \sin θ) & x + y i = r c i s θ & x + y i = r ∠ θ \end{aligned}$

::x+yi=r(cosisin)x+yi=rcisix+yi=r

2. Express the following polar form of each complex number using the shorthand representations.
::2. 使用直径表示每个复杂数字的极形。

$\begin{aligned} 4.92 (\cos {214.6}^{\circ} + i \sin {214.6}^{\circ}) \end{aligned}$
::4.92(cos214.6isin214.6)

$\begin{aligned} 4.92 ∠ {214.6}^{\circ} \end{aligned}$

$\begin{aligned} 4.92 c i s {214.6}^{\circ} \end{aligned}$
::4.92 Cis 214.6

$\begin{aligned} 15.6 (\cos 37^{\circ} + i \sin 37^{\circ}) \end{aligned}$
::15.6(cos37isin37)

$\begin{aligned} 15.6 ∠ 37^{\circ} \end{aligned}$

$\begin{aligned} 15.6 c i s 37^{\circ} \end{aligned}$
::15.6 CIS 37_________________________________________________________________________________________________________________________________________________________________________________

3. Represent the complex number $\begin{aligned} - 3.12 - 4.64 i \end{aligned}$ graphically and give two notations of its polar form.
::3. 以图形形式表示复合数-3.12-4.64i,并用两点表示其极形。

From the rectangular form of $\begin{aligned} - 3.12 - 4.64 i x = - 3.12 \end{aligned}$ and $\begin{aligned} y = - 4.64 \end{aligned}$
::取自-3.12-4.64i x3.12 和 y.4.64 的矩形形式。

$\begin{aligned} r & = \sqrt{x^{2} + y^{2}} \\ r & = \sqrt{(- 3.12)^{2} + (- 4.64)^{2}} \\ r & = 5.59 \end{aligned}$

::r=x2+y2r=(-3.12)2+(-4.64)2r=5.59

$\begin{aligned} \tan θ & = \frac{y}{x} \\ \tan θ & = \frac{- 4.64}{- 3.12} \\ θ & = {56.1}^{\circ} \end{aligned}$

::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}什么? {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}什么? {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}什么? {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}什么?

This is the reference angle so now we must determine the measure of the angle in the third quadrant. $\begin{aligned} {56.1}^{\circ} + 180^{\circ} = {236.1}^{\circ} \end{aligned}$
::这是参考角度,所以我们现在必须确定第三个象限的角的测量量。 56.1180236.1

One polar notation of the point $\begin{aligned} - 3.12 - 4.64 i \end{aligned}$ is $\begin{aligned} 5.59 \end{aligned}$ $\begin{aligned} (\cos {236.1}^{\circ} + i \sin {236.1}^{\circ}) \end{aligned}$ . Another polar notation of the point is $\begin{aligned} 5.59 ∠ {236.1}^{\circ} \end{aligned}$
::点-3.12-4.64i的一个极值表示5.59(cos236.1isin236.1)。点的另一个极值表示5.59236.1。

So far we have expressed all values of theta in degrees. Polar form of a complex number can also have theta expressed in . This would be beneficial when plotting the in the polar plane.
::到目前为止,我们已经以度表示所有Theta的值。复数的极形也可以以度表示 Theta 。在极平面绘制图时, 这样做会有好处。

The answer to #2 $\begin{aligned} - 3.12 - 4.64 i \end{aligned}$ with theta expressed in radian measure would be:
::以弧度度表示的对#2-3.12-4.64i和Theta的回答是:

$\begin{aligned} \tan θ = \frac{- 4.64}{- 3.12} & \tan θ = .9788 (reference angle) \\ 0.9788 + 3.14 = 4.12 rad . \\ 5.59 (\cos 4.12 + i \sin 4.12) \end{aligned}$

::4.64-3.12tan.9788(参考角度)0.9788+3.14=4.12 rad.5.59(cos4.12+isin4.12)

Now that we have explored the polar form of complex numbers and the steps for performing these conversions, we will look at an example in circuit analysis that requires a complex number given in polar form to be expressed in standard form.
::既然我们已经探索了复杂数字的极化形式和进行这些转换的步骤,我们将在电路分析中研究一个例子,这种分析要求以标准形式表示以极化形式提供的复杂数字。

Examples
::实例

Example 1
::例1

Earlier, you were asked to convert a complex number into polar form.
::早些时候,你被要求将一个复数转换成极形。

You can now convert $\begin{aligned} 2 + 3 i \end{aligned}$ into polar form by using the equations giving the radius and angle of the number's position in the complex plane:
::现在您可以将 2+3i 转换为极形, 使用方程式给出数字在复杂平面中位置的半径和角度 :

$\begin{aligned} r & = \sqrt{x^{2} + y^{2}} \\ r & = \sqrt{(2)^{2} + (3)^{2}} \\ r & = \sqrt{13} \end{aligned}$

::rx2+y2r=(2)2+(3)2r=13

$\begin{aligned} \tan θ & = \frac{y}{x} \\ \tan θ & = \frac{3}{2} \\ θ & = {56.31}^{\circ} \end{aligned}$

::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}那又怎样? {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}

Therefore, the polar form of $\begin{aligned} 2 + 3 i \end{aligned}$ is $\begin{aligned} \sqrt{13} (\cos {56.31}^{\circ} + i \sin {56.31}^{\circ}) \end{aligned}$ .
::因此,2+3i的极形为13(cos56.31isin56.31)。

Example 2
::例2

The impedance $\begin{aligned} Z \end{aligned}$ , in ohms, in an alternating circuit is given by $\begin{aligned} Z = 4650 ∠ - {35.2}^{\circ} \end{aligned}$ . Express the value for $\begin{aligned} Z \end{aligned}$ in standard form. (In electricity, negative angles are often used.)
::Z的阻力,在ohms,交替电路由465035.2提供。用标准格式表示 Z 的值。 (在电力中,经常使用负角度。 )

The value for $\begin{aligned} Z \end{aligned}$ is given in polar form. From this notation, we know that $\begin{aligned} r = 4650 \end{aligned}$ and $\begin{aligned} θ = - {35.2}^{\circ} \end{aligned}$ Using these values, we can write:
::Z 的值以极形形式给出。从此符号中, 我们知道 r=4650 和 35.2 使用这些值, 我们可以写入 :

$\begin{aligned} Z & = 4650 (\cos (- {35.2}^{\circ}) + i \sin (- {35.2}^{\circ})) \\ x & = 4650 \cos (- {35.2}^{\circ}) \to 3800 \\ y & = 4650 \sin (- {35.2}^{\circ}) \to - 2680 \end{aligned}$

::4650(cos(- 352)+isin(-35.2)x=4650cos(-35.2) 3800y=4650sin(-35.2) 2680

Therefore the standard form is $\begin{aligned} Z = 3800 - 2680 i \end{aligned}$ ohms.
::因此,标准表格是3800 - 2680iohms。

Example 3
::例3

Express the following complex numbers in their polar form.
::以极形表示下列复杂数字。

$\begin{aligned} 4 + 3 i \end{aligned}$
::4+3i 4+3i

$\begin{aligned} - 2 + 9 i \end{aligned}$
::-2+9i

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

$\begin{aligned} - 5 - 2 i \end{aligned}$
::- 5-2i

$\begin{aligned} 4 + 3 i \to x = 4, y = 3 \end{aligned}$
$\begin{aligned} r = \sqrt{4^{2} + 3^{2}} = 5, \tan θ = \frac{3}{4} \to θ = {36.87}^{\circ} \to 5 (\cos {36.87}^{\circ} + i \sin {36.87}^{\circ}) \end{aligned}$

::4+3ix=4,y=3r=42+32=5,tann3436.875(cos36.87isin36.87)

$\begin{aligned} - 2 + 9 i \to x = - 2, y = 9 \end{aligned}$
$\begin{aligned} r = \sqrt{(- 2)^{2} + 9^{2}} = \sqrt{85} \approx 9.22, \tan θ = - \frac{9}{2} \to θ = {102.53}^{\circ} \to 9.22 (\cos {102.53}^{\circ} + i \sin {102.53}^{\circ}) \end{aligned}$

::-2+9ix2,y=9r=(-2)2+92=859.22,tan92102.539.22(cos102.53isin102.53)

$\begin{aligned} 7 - i \to x = 7, y = - 1 \end{aligned}$
$\begin{aligned} r = \sqrt{7^{2} + 1^{2}} = \sqrt{50} \approx 7.07, \tan θ = - \frac{1}{7} \to θ = {351.87}^{\circ} \to 7.07 (\cos {351.87}^{\circ} + i \sin {351.87}^{\circ}) \end{aligned}$

::7-ix=7,y1r=72+12=507.07,tan17351.877.07(cos351.8787351.87)

$\begin{aligned} - 5 - 2 i \to x = - 5, y = - 2 \end{aligned}$
$\begin{aligned} r = \sqrt{(- 5)^{2} + (- 2)^{2}} = \sqrt{29} \approx 5.39, \tan θ = \frac{2}{5} \to θ = {201.8}^{\circ} \to 5.39 (\cos {201.8}^{\circ} + i \sin {201.8}^{\circ}) \end{aligned}$

::- 5-2ix5,y2r=(-5)2+(-2)2=29539,tan25201.85.39(cos201.8_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Example 4
::例4

Express the complex number $\begin{aligned} 6 - 8 i \end{aligned}$ graphically and write it in its polar form.
::以图形形式表示6-8i号的复合体,并以极形形式写成。

$\begin{aligned} 6 - 8 i \end{aligned}$
::6-8i 6-8i

$\begin{aligned} 6 - 8 i \\ x & = 6 and y = - 8 & \tan θ = \frac{y}{x} \\ r & = \sqrt{x^{2} + y^{2}} & \tan θ = \frac{- 8}{6} \\ r & = \sqrt{(6)^{2} + (- 8)^{2}} & θ = - {53.1}^{\circ} \\ r & = 10 \end{aligned}$

::6-8ix=6 y8tanyxr=x2+y2tan86r=(6)2+(-8)253.1_r=10

Since $\begin{aligned} θ \end{aligned}$ is in the fourth quadrant then $\begin{aligned} θ = - {53.1}^{\circ} + 360^{\circ} = {306.9}^{\circ} \end{aligned}$ Expressed in polar form $\begin{aligned} 6 - 8 i \end{aligned}$ is $\begin{aligned} 10 (\cos {306.9}^{\circ} + i \sin {306.9}^{\circ}) \end{aligned}$ or $\begin{aligned} 10 ∠ {306.9}^{\circ} \end{aligned}$
::以极度表6-8i表示为10(cos306.9)或10306.9

Review
::回顾

Plot each of the following points in the complex plane.
::在复杂的平面上绘制以下各点的分布图。

$\begin{aligned} 1 + i \end{aligned}$
::1个+一

$\begin{aligned} 2 - 3 i \end{aligned}$
::2-3i 2-3i

$\begin{aligned} - 2 - i \end{aligned}$
::--2-i

$\begin{aligned} i \end{aligned}$
:一) i

$\begin{aligned} 4 - i \end{aligned}$
::4 - i 4 - i

Find the trigonometric form of the complex numbers where $\begin{aligned} 0 \leq θ < 2 π \end{aligned}$ .
::在 02处找到复数数的三角形式。

$\begin{aligned} 8 - 6 i \end{aligned}$
::8 - 6i

$\begin{aligned} 5 + 12 i \end{aligned}$
::5+12i 5+12i

$\begin{aligned} 2 - 2 i \end{aligned}$
::2-2-2i

$\begin{aligned} 3 + 3 i \end{aligned}$
::3+3i 3+3i

$\begin{aligned} 2 + 3 i \end{aligned}$
::2+3i 2+3i

$\begin{aligned} 5 - 6 i \end{aligned}$
::5-66i

Write each complex number in standard form.
::以标准格式填写每个复数。

$\begin{aligned} 4 (\cos 30^{\circ} + i \sin 30^{\circ}) \end{aligned}$
::4 (cos30isin30)

$\begin{aligned} 3 (\cos \frac{π}{4} + i \sin \frac{π}{4}) \end{aligned}$
::3(cos%4+isin%4)

$\begin{aligned} 2 (\cos \frac{7 π}{6} + i \sin \frac{7 π}{6}) \end{aligned}$
::2(cos76+isin76)

$\begin{aligned} 2 (\cos \frac{π}{12} + i \sin \frac{π}{12}) \end{aligned}$
::2(cos12+isin12)

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。

章节大纲

Trigonometric Form of Complex Numbers ::复杂数字的三角矩阵形式

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)

Trigonometric Form of Complex Numbers
::复杂数字的三角矩阵形式

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Review
::回顾

Review (Answers)
::回顾(答复)