Complex Numbers 复数

本页译自 LibreTexts · A First Course in Linear Algebra (Kuttler) 第 6 章「Complex Numbers」。公式经 MathJax 渲染，自定义宏已注入；正文与译文段段对照。

6.1 Complex Numbers and Their Operations 复数及其运算

Outcomes.

1. Understand the geometric significance of a complex number as a point in the plane.
2. Prove algebraic properties of addition and multiplication of complex numbers, and apply these properties. Understand the action of taking the conjugate of a complex number.
3. Understand the absolute value of a complex number and how to find it as well as its geometric significance.

1. 理解复数作为平面上一点所具有的几何意义。
2. 证明复数加法与乘法的代数性质，并应用这些性质；理解取共轭这一运算。
3. 理解复数的绝对值（模）及其求法，以及它的几何意义。

Although very powerful, the real numbers are inadequate to solve equations such as $x^2+1=0$, and this is where complex numbers come in. We define the number $i$ as the imaginary number such that $i^2 = -1$, and define complex numbers as those of the form $z = a + bi$ where $a$ and $b$ are real numbers. We call this the standard form, or Cartesian form, of the complex number $z$. Then, we refer to $a$ as the real part of $z$, and $b$ as the imaginary part of $z$. It turns out that such numbers not only solve the above equation, but in fact also solve any polynomial of degree at least 1 with complex coefficients. This property, called the Fundamental Theorem of Algebra, is sometimes referred to by saying $\mathbb{C}$ is algebraically closed. Gauss is usually credited with giving a proof of this theorem in 1797 but many others worked on it and the first completely correct proof was due to Argand in 1806.

Just as a real number can be considered as a point on the line, a complex number $z = a + bi$ can be considered as a point $\left( a,b\right)$ in the plane whose $x$ coordinate is $a$ and whose $y$ coordinate is $b.$ For example, in the following picture, the point $z = 3+2i$ can be represented as the point in the plane with coordinates $\left( 3,2\right) .$

Addition of complex numbers is defined as follows. $\left( a+bi\right) +\left( c+di\right) =\left( a+c\right) +\left( b+d\right)i $

This addition obeys all the usual properties as the following theorem indicates.

Theorem 1: Properties of Addition of Complex Numbers 复数加法的性质

Let $z,w,$ and $v$ be complex numbers. Then the following properties hold.

• Commutative Law for Addition $z+w=w+z$
• Additive Identity $z+0=z$
• Existence of Additive Inverse $\begin{array}{l} \mbox{For each } z\in \mathbb{C}, \mbox{ there exists } -z\in \mathbb{C} \mbox{ such that } z+\left( -z\right) =0. \\ \mbox{In fact if } z=a+bi, \mbox{ then } -z=-a-bi. \end{array}$
• Associative Law for Addition $\left( z+w\right) +v= z +\left( w+v\right) $

• 加法交换律 $z+w=w+z$
• 加法单位元 $z+0=z$
• 加法逆元的存在性 $\begin{array}{l} \mbox{对每个 } z\in \mathbb{C}, \mbox{ 存在 } -z\in \mathbb{C} \mbox{ 使得 } z+\left( -z\right) =0. \\ \mbox{事实上若 } z=a+bi, \mbox{ 则 } -z=-a-bi. \end{array}$
• 加法结合律 $\left( z+w\right) +v= z +\left( w+v\right) $

Proof. The proof of this theorem is left as an exercise for the reader.

Now, multiplication of complex numbers is defined the way you would expect, recalling that $i^{2} = -1$.

$$\begin{aligned} \left( a+bi\right) \left( c+di\right) &=ac+adi+bci+i^{2}bd \\ &=\left( ac-bd\right) +\left( ad + bc \right)i \end{aligned}$$

Consider the following examples.

Example 1: Multiplication of Complex Numbers 复数的乘法

• $(2-3i)(-3+4i) = 6+17i$
• $(4-7i)(6-2i) = 10-50i$
• $(-3+6i)(5-i) = -9+33i$

• $(2-3i)(-3+4i) = 6+17i$
• $(4-7i)(6-2i) = 10-50i$
• $(-3+6i)(5-i) = -9+33i$

The following are important properties of multiplication of complex numbers.

Theorem 2: Properties of Multiplication of Complex Numbers 复数乘法的性质

Let $z,w$ and $v$ be complex numbers. Then, the following properties of multiplication hold.

• Commutative Law for Multiplication $zw=wz$
• Associative Law for Multiplication $\left( zw\right) v=z\left( wv\right)$
• Multiplicative Identity $1z=z$
• Existence of Multiplicative Inverse $\mbox{For each } z\neq 0, \mbox{ there exists } z^{-1} \mbox{ such that } zz^{-1}=1$
• Distributive Law $z\left( w+v\right) =zw+zv$

• 乘法交换律 $zw=wz$
• 乘法结合律 $\left( zw\right) v=z\left( wv\right)$
• 乘法单位元 $1z=z$
• 乘法逆元的存在性 $\mbox{对每个 } z\neq 0, \mbox{ 存在 } z^{-1} \mbox{ 使得 } zz^{-1}=1$
• 分配律 $z\left( w+v\right) =zw+zv$

You may wish to verify some of these statements. The real numbers also satisfy the above axioms, and in general any mathematical structure which satisfies these axioms is called a field. There are many other fields, in particular even finite ones particularly useful for cryptography, and the reason for specifying these axioms is that linear algebra is all about fields and we can do just about anything in this subject using any field. Although here, the fields of most interest will be the familiar field of real numbers, denoted as $\mathbb{R}$, and the field of complex numbers, denoted as $\mathbb{C}$.

An important construction regarding complex numbers is the complex conjugate denoted by a horizontal line above the number, $\overline{z}$. It is defined as follows.

Definition 1: Conjugate of a Complex Number 复数的共轭

Let $z = a+bi$ be a complex number. Then the conjugate of $z$, written $\overline{z}$ is given by $\overline{a+bi}= a-bi$

Geometrically, the action of the conjugate is to reflect a given complex number across the $x$ axis. Algebraically, it changes the sign on the imaginary part of the complex number. Therefore, for a real number $a$, $\overline{a} = a$.

Example 2: Conjugate of a Complex Number 复数的共轭（例）

• If $z=3+4i$, then $\overline{z}=3-4i$, i.e., $\overline{3+4i}=3-4i$.
• $\overline{-2+5i}= -2-5i$.
• $\overline{i}= -i$.
• $\overline{7}= 7$.

• 若 $z=3+4i$，则 $\overline{z}=3-4i$，即 $\overline{3+4i}=3-4i$。
• $\overline{-2+5i}= -2-5i$。
• $\overline{i}= -i$。
• $\overline{7}= 7$。

Consider the following computation.

$$\begin{aligned} \left( \overline{a+bi}\right) \left( a+bi\right) &= \left( a-bi\right) \left( a+bi\right) \\ &= a^{2}+b^{2}-\left( ab-ab\right)i =a^{2}+b^{2}\end{aligned}$$

Notice that there is no imaginary part in the product, thus multiplying a complex number by its conjugate results in a real number.

Theorem 3: Properties of the Conjugate 共轭的性质

Let $z$ and $w$ be complex numbers. Then, the following properties of the conjugate hold.

• $\overline{z\pm w} = \overline{z} \pm \overline{w}$.
• $\overline{(zw)} = \overline{z}~ \overline{w}$.
• $\overline{(\overline{z})}=z$.
• $\overline{\left(\frac{z}{w}\right)} = \frac{\overline{z}}{\overline{w}}$.
• $z$ is real if and only if $\overline{z}=z$.

• $\overline{z\pm w} = \overline{z} \pm \overline{w}$。
• $\overline{(zw)} = \overline{z}~ \overline{w}$。
• $\overline{(\overline{z})}=z$。
• $\overline{\left(\frac{z}{w}\right)} = \frac{\overline{z}}{\overline{w}}$。
• $z$ 为实数当且仅当 $\overline{z}=z$。

Division of complex numbers is defined as follows. Let $z=a+bi$ and $w=c+di$ be complex numbers such that $c,d$ are not both zero. Then the quotient $z$ divided by $w$ is

$$\begin{aligned} \frac{z}{w} &= \frac{a+bi}{c+di} \\ &= \frac{a+bi}{c+di}\times \frac{c-di}{c-di} \\ &= \frac{(ac+bd)+(bc-ad)i}{c^2+d^2} \\ & = \frac{ac+bd}{c^2+d^2} +\frac{bc-ad}{c^2+d^2}i.\end{aligned}$$

In other words, the quotient $\frac{z}{w}$ is obtained by multiplying both top and bottom of $\frac{z}{w}$ by $\overline{w}$ and then simplifying the expression.

Example 3: Division of Complex Numbers 复数的除法

$$\frac{1}{i} = \frac{1}{i}\times \frac{-i}{-i} =\frac{-i}{-i^2}=-i$$

$$\frac{2-i}{3+4i} = \frac{2-i}{3+4i}\times \frac{3-4i}{3-4i} =\frac{(6-4)+(-3-8)i}{3^2+4^2} =\frac{2-11i}{25} =\frac{2}{25} - \frac{11}{25}i$$

$$\frac{1-2i}{-2+5i} = \frac{1-2i}{-2+5i}\times \frac{-2-5i}{-2-5i} =\frac{(-2-10) + (4-5)i}{2^2+5^2} =-\frac{12}{29}-\frac{1}{29}i$$

Interestingly every nonzero complex number $a+bi$ has a unique multiplicative inverse. In other words, for a nonzero complex number $z$, there exists a number $z^{-1}$ (or $\frac{1}{z}$) so that $zz^{-1} = 1$. Note that $z=a+bi$ is nonzero exactly when $a^{2}+b^{2}\neq 0$, and its inverse can be written in standard form as defined now.

Definition 2: Inverse of a Complex Number 复数的逆元

Let $z = a+bi$ be a complex number. Then the multiplicative inverse of $z$, written $z^{-1}$ exists if and only if $a^{2}+b^{2}\neq 0$ and is given by

$$z^{-1} = \frac{1}{a+bi} = \frac{1}{a+bi}\times \frac{a-bi}{a-bi}=\frac{a-bi}{a^{2}+b^{2}}=\frac{a}{a^{2}+b^{2}}-i\frac{b}{ a^{2}+b^{2}}$$

Note that we may write $z^{-1}$ as $\frac{1}{z}$. Both notations represent the multiplicative inverse of the complex number $z$. Consider now an example.

Example 4: Inverse of a Complex Number 复数的逆元（例）

Consider the complex number $z = 2 + 6i$. Then $z^{-1}$ is defined, and

$$\begin{aligned} \frac{1}{z} &= \frac{1}{2+6i} \\ &= \frac{1}{2+6i}\times \frac{2-6i}{2-6i} \\ &= \frac{2-6i}{2^2+6^2} \\ &= \frac{2-6i}{40} \\ &= \frac{1}{20} - \frac{3}{20}i \end{aligned}$$

You can always check your answer by computing $zz^{-1}$.

Another important construction of complex numbers is that of the absolute value, also called the modulus. Consider the following definition.

Definition 3: Absolute Value 绝对值（模）

The absolute value, or modulus, of a complex number, denoted $\left\| z \right\|$ is defined as follows. $\left\| a+bi\right\| = \sqrt{a^{2}+b^{2}}$

Thus, if $z$ is the complex number $z=a+bi$, it follows that $\left\| z\right\| =\left( z\overline{z}\right) ^{1/2}$

Also from the definition, if $z=a+bi$ and $w=c+di$ are two complex numbers, then $\left\vert zw\right\vert =\left\vert z\right\vert \left\vert w\right\vert .$ Take a moment to verify this.

The triangle inequality is an important property of the absolute value of complex numbers. There are two useful versions which we present here, although the first one is officially called the triangle inequality.

Proposition 1: Triangle Inequality 三角不等式

Let $z,w$ be complex numbers.

The following two inequalities hold for any complex numbers $z,w$: $\begin{array}{l} \left\| z+w\right\| \leq \left\| z\right\| +\left\| w\right\| \\ \left\| \left\| z\right\| -\left\| w\right\| \right\| \leq \left\| z-w\right\| \end{array}$ The first one is called the Triangle Inequality.

Proof. Let $z=a+bi$ and $w=c+di$. First note that $z \overline{w}=\left( a+bi\right) \left( c-di\right) =ac+bd+\left( bc-ad\right)i$ and so $\left\vert ac+bd\right\vert \leq \left\vert z\overline{w}\right\vert =\left\vert z\right\vert \left\vert w\right\vert .$

Then, $\left\vert z+w\right\vert ^{2}=\left( a+c+i\left( b+d\right) \right) \left( a+c-i\left( b+d\right) \right)$ $=\left( a+c\right) ^{2}+\left( b+d\right) ^{2}=a^{2}+c^{2}+2ac+2bd+b^{2}+d^{2}$ $\leq \left\vert z\right\vert ^{2}+\left\vert w\right\vert ^{2}+2\left\vert z\right\vert \left\vert w\right\vert =\left( \left\vert z\right\vert +\left\vert w\right\vert \right) ^{2}$

Taking the square root, we have that $\left\vert z+w\right\vert \leq \left\vert z\right\vert +\left\vert w\right\vert$ so this verifies the triangle inequality.

To get the second inequality, write $z=z-w+w,$ $w=w-z+z$ and so by the first form of the inequality we get both: $\left\vert z\right\vert \leq \left\vert z-w\right\vert +\left\vert w\right\vert ,$ $\left\vert w\right\vert \leq \left\vert z-w\right\vert +\left\vert z\right\vert$

Hence, both $\left\vert z\right\vert -\left\vert w\right\vert$ and $\left\vert w\right\vert -\left\vert z\right\vert$ are no larger than $\left\vert z-w\right\vert$. This proves the second version because $\left\vert \left\vert z\right\vert -\left\vert w\right\vert \right\vert$ is one of $\left\vert z\right\vert -\left\vert w\right\vert$ or $\left\vert w\right\vert -\left\vert z\right\vert$.

With this definition, it is important to note the following. You may wish to take the time to verify this remark.

Let $z=a+bi$ and $w=c+di.$ Then

$$\left\| z-w\right\| =\sqrt{\left( a-c\right) ^{2}+\left( b-d\right) ^{2}}.$$

Thus the distance between the point in the plane determined by the ordered pair $\left( a,b\right)$ and the ordered pair $\left( c,d\right)$ equals $\left\| z-w\right\|$ where $z$ and $w$ are as just described.

For example, consider the distance between $\left( 2,5\right)$ and $\left( 1,8\right) .$ Letting $z=2+5i$ and $w=1+8i,$ $z-w=1-3i$, $\left( z-w\right) \left( \overline{z-w}\right) =\left( 1-3i\right) \left( 1+3i\right) = 10$ so $\left\vert z-w\right\vert =\sqrt{10}$.

Recall that we refer to $z=a+bi$ as the standard form of the complex number. In the next section, we examine another form in which we can express the complex number.

6.2 Polar Form 极坐标形式

Outcomes.

1. Convert a complex number from standard form to polar form, and from polar form to standard form.

1. 将复数从标准形式化为极坐标形式，以及从极坐标形式化回标准形式。

In the previous section, we identified a complex number $z=a+bi$ with a point $\left( a, b\right)$ in the coordinate plane. There is another form in which we can express the same number, called the polar form. The polar form is the focus of this section. It will turn out to be very useful if not crucial for certain calculations as we shall soon see.

Suppose $z=a+bi$ is a complex number, and let $r=\sqrt{a^{2}+b^{2}} = \|z\|$. Recall that $r$ is the modulus of $z$. Note first that $\left( \frac{a}{r} \right) ^{2}+\left( \frac{b}{r}\right) ^{2}= \frac{a^2+b^2}{r^2}=1$ and so $\left( \frac{a}{r},\frac{b}{r}\right)$ is a point on the unit circle. Therefore, there exists an angle $\theta$ (in radians) such that $\cos \theta =\frac{a}{r},$ $\sin \theta =\frac{b}{r}$. In other words $\theta$ is an angle such that $a = r\cos \theta$ and $b=r \sin \theta$, that is $\theta = \cos^{-1}(a/r)$ and $\theta = \sin^{-1}(b/r)$. We call this angle $\theta$ the argument of $z$.

We often speak of the principal argument of $z$. This is the unique angle $\theta \in (-\pi, \pi]$ such that $\cos \theta =\frac{a}{r},$ $\sin \theta =\frac{b}{r}$.

The polar form of the complex number $z=a+bi = r \left( \cos \theta +i\sin \theta \right)$ is for convenience written as: $z = r e^{i \theta}$ where $\theta$ is the argument of $z$.

Definition 1: Polar Form of a Complex Number 复数的极坐标形式

Let $z = a + bi$ be a complex number. Then the polar form of $z$ is written as $z = re^{i\theta}$ where $r = \sqrt{a^2 + b^2}$ and $\theta$ is the argument of $z$.

When given $z = re^{i\theta}$, the identity $e^{i\theta} = \cos\theta + i \sin\theta$ will convert $z$ back to standard form. Here we think of $e^{i \theta}$ as a short cut for $\cos \theta +i\sin \theta$. This is all we will need in this course, but in reality $e^{i \theta}$ can be considered as the complex equivalent of the exponential function where this turns out to be a true equality.

Thus we can convert any complex number in the standard (Cartesian) form $z = a+bi$ into its polar form. Consider the following example.

Example 1: Standard to Polar Form 由标准形式化为极坐标形式

Let $z = 2 + 2i$ be a complex number. Write $z$ in the polar form $z = re^{i \theta}$

Solution. First, find $r$. By the above discussion, $r=\sqrt{ a^{2}+b^{2}} = \|z\|$. Therefore, $r = \sqrt{2^{2} + 2^{2}} = \sqrt{8} =2\sqrt{2} $

Now, to find $\theta$, we plot the point $\left( 2, 2 \right)$ and find the angle from the positive $x$ axis to the line between this point and the origin. In this case, $\theta = 45^{\circ} = \frac{\pi}{4}$. That is we found the unique angle $\theta$ such that $\theta = \cos^{-1}(1/\sqrt{2})$ and $\theta = \sin^{-1}(1/\sqrt{2})$.

Note that in polar form, we always express angles in radians, not degrees.

Hence, we can write $z$ as $z = 2\sqrt{2} e^{i\frac{\pi}{4}}$

Notice that the standard and polar forms are completely equivalent. That is not only can we transform a complex number from standard form to its polar form, we can also take a complex number in polar form and convert it back to standard form.

Example 3: Polar to Standard Form 由极坐标形式化为标准形式

Let $z = 2 e^{ 2\pi i/3}$. Write $z$ in the standard form $z = a+bi $

Solution. Let $z = 2 e^{2\pi i/3}$ be the polar form of a complex number. Recall that $e^{i\theta} = \cos \theta + i \sin \theta$. Therefore using standard values of $\sin$ and $\cos$ we get: $\begin{aligned} z = 2 e^{i 2\pi/3} &= 2 (\cos (2\pi/3)+i\sin (2\pi/3)) \\ &= 2 \left ( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right) \\ &=-1 + \sqrt{3}i \end{aligned}$ which is the standard form of this complex number.

You can always verify your answer by converting it back to polar form and ensuring you reach the original answer.

6.3 Roots of Complex Numbers 复数的根

Outcomes.

1. Understand De Moivre's theorem and be able to use it to find the roots of a complex number.

1. 理解棣莫弗定理，并能用它来求复数的根。

A fundamental identity is the formula of De Moivre with which we begin this section.

Theorem 1: De Moivre's Theorem 棣莫弗定理

For any positive integer $n$, we have $\left( e^{i \theta} \right)^n = e^{i n \theta} $

Thus for any real number $r>0$ and any positive integer $n$, we have:

$$\left( r\left( \cos \theta+i\sin \theta\right) \right) ^{n}=r^{n}\left( \cos n \theta +i\sin n\theta\right)$$

Proof. The proof is by induction on $n$. It is clear the formula holds if $n=1.$ Suppose it is true for $n.$ Then, consider $n+1$.

$$\left( r\left( \cos \theta+i\sin \theta\right) \right) ^{n+1}=\left( r\left( \cos \theta+i\sin \theta\right) \right) ^{n}\left( r\left( \cos \theta+i\sin \theta\right) \right)$$

which by induction equals

$$\begin{aligned} &=r^{n+1}\left( \cos n\theta+i\sin n\theta\right) \left( \cos \theta+i\sin \theta\right) \\ &= r^{n+1}\left( \left( \cos n\theta\cos \theta-\sin n\theta\sin \theta\right) +i\left( \sin n\theta\cos \theta+\cos n\theta\sin \theta\right) \right)\\ &=r^{n+1}\left( \cos \left( n+1\right) \theta+i\sin \left( n+1\right) \theta\right)\end{aligned}$$

by the formulas for the cosine and sine of the sum of two angles.

The process used in the previous proof, called mathematical induction is very powerful in Mathematics and Computer Science and explored in more detail in the Appendix.

Now, consider a corollary of Theorem 1.

Corollary 1: Roots of Complex Numbers 复数的根

Let $z$ be a non zero complex number. Then there are always exactly $k$ many $k^{th}$ roots of $z$ in $\mathbb{C}$.

Proof. Let $z=a+bi$ and let $z=\left\vert z\right\vert \left( \cos \theta+i\sin \theta\right)$ be the polar form of the complex number. By De Moivre's theorem, a complex number $w= r e^{i \alpha} = r\left( \cos \alpha +i\sin \alpha \right)$ is a $k^{th}$ root of $z$ if and only if $w^k = (r e^{i \alpha})^k = r^k e^{ik\alpha} = r^{k}\left( \cos k\alpha +i\sin k\alpha \right) =\left\vert z\right\vert \left( \cos \theta+i\sin \theta\right)$

This requires $r^{k}=\left\vert z\right\vert$ and so $r=\left\vert z\right\vert ^{1/k}$. Also, both $\cos \left( k\alpha \right) =\cos \theta$ and $\sin \left( k\alpha \right) =\sin \theta.$ This can only happen if $k\alpha =\theta+2 \ell \pi $ for $\ell$ an integer. Thus $\alpha = \frac{\theta+2 \ell \pi }{k},$ $\ell = 0, 1, 2, \cdots, k-1$ and so the $k^{th}$ roots of $z$ are of the form $\left\vert z\right\vert ^{1/k}\left( \cos \left( \frac{\theta+2 \ell \pi }{k}\right) +i\sin \left( \frac{\theta+2 \ell \pi }{k}\right) \right) ,$ $\ell = 0, 1, 2, \cdots, k-1$

Since the cosine and sine are periodic of period $2\pi ,$ there are exactly $k$ distinct numbers which result from this formula.

The procedure for finding the $k^{th}$ roots of $z \in \mathbb{C}$ is as follows.

Procedure 1: Finding Roots of a Complex Number 求复数的根

Let $w$ be a complex number. We wish to find the $n^{th}$ roots of $w$, that is all $z$ such that $z^n = w$.

There are $n$ distinct $n^{th}$ roots and they can be found as follows:.

1. Express both $z$ and $w$ in polar form $z=re^{i\theta}, w=se^{i\phi}$. Then $z^n = w$ becomes: $(re^{i\theta})^n = r^n e^{i n \theta} = se^{i\phi}$. We need to solve for $r$ and $\theta$.
2. Solve the following two equations: $r^n = s$ and $e^{i n \theta} = e^{i \phi}.$
3. The solutions to $r^n = s$ are given by $r = \sqrt[n]{s}$.
4. The solutions to $e^{i n \theta} = e^{i \phi}$ are given by: $n\theta = \phi + 2\pi \ell,$ for $\ell = 0,1,2, \cdots, n-1$ or $\theta = \frac{\phi}{n} + \frac{2}{n} \pi \ell,$ for $\ell = 0,1,2, \cdots, n-1$
5. Using the solutions $r, \theta$ to the equations given in 上述方程, construct the $n^{th}$ roots of the form $z = re^{i\theta}$.

1. 把 $z$ 与 $w$ 都写成极坐标形式 $z=re^{i\theta}, w=se^{i\phi}$。于是 $z^n = w$ 化为：$(re^{i\theta})^n = r^n e^{i n \theta} = se^{i\phi}$。我们需要求解 $r$ 与 $\theta$。
2. 解下面两个方程：$r^n = s$ 与 $e^{i n \theta} = e^{i \phi}$。
3. 方程 $r^n = s$ 的解为 $r = \sqrt[n]{s}$。
4. 方程 $e^{i n \theta} = e^{i \phi}$ 的解为：$n\theta = \phi + 2\pi \ell,$ 其中 $\ell = 0,1,2, \cdots, n-1$；或者 $\theta = \frac{\phi}{n} + \frac{2}{n} \pi \ell,$ 其中 $\ell = 0,1,2, \cdots, n-1$。
5. 利用上述方程关于 $r, \theta$ 的解，构造出形如 $z = re^{i\theta}$ 的 $n$ 次根。

Notice that once the roots are obtained in the final step, they can then be converted to standard form if necessary. Let's consider an example of this concept. Note that according to Corollary 1, there are exactly $3$ cube roots of a complex number.

Example 1: Finding Cube Roots 求立方根

Find the three cube roots of $i.$ In other words find all $z$ such that $z^3 = i$.

Solution. First, convert each number to polar form: $z = re^{i\theta}$ and $i = 1 e^{i \pi/2}$. The equation now becomes $(re^{i\theta})^3 = r^3 e^{3i\theta} = 1 e^{i \pi/2}$.

Therefore, the two equations that we need to solve are $r^3 = 1$ and $3i\theta = i \pi/2$. Given that $r \in \mathbb{R}$ and $r^3 = 1$ it follows that $r=1$.

Solving the second equation is as follows. First divide by $i$. Then, since the argument of $i$ is not unique we write $3\theta = \pi/2 + 2\pi\ell$ for $\ell = 0,1,2$.

$$\begin{aligned} 3\theta &= \pi/2 + 2\pi\ell \\ \mbox{for} \\ \ell = 0,1,2 \\ \theta &= \pi/6 + \frac{2}{3} \pi\ell \\ \mbox{for} \\ \ell = 0,1,2 \end{aligned}$$

For $\ell = 0$: $\theta = \pi/6 + \frac{2}{3} \pi (0) = \pi/6$

For $\ell = 1$: $\theta = \pi/6 + \frac{2}{3} \pi(1) = \frac{5}{6} \pi$

For $\ell = 2$: $\theta = \pi/6 + \frac{2}{3} \pi(2) = \frac{3}{2} \pi$

Therefore, the three roots are given by $1e^{i \pi/6}, 1e^{i \frac{5}{6}\pi}, 1e^{i \frac{3}{2}\pi}$

Written in standard form, these roots are, respectively, $\frac{\sqrt{3}}{2} + i \frac{1}{2}, -\frac{\sqrt{3}}{2} + i \frac{1}{2}, -i$

The ability to find $k^{th}$ roots can also be used to factor some polynomials.

Example 2: Solving a Polynomial Equation 解多项式方程

Factor the polynomial $x^{3}-27.$

Solution. First find the cube roots of 27. By the above procedure , these cube roots are $3,3\left( \displaystyle \frac{-1}{2}+i\displaystyle\frac{\sqrt{3}}{2}\right) ,$ and $3\left( \displaystyle\frac{-1}{2}-i\displaystyle\frac{\sqrt{3}}{2}\right).$

You may wish to verify this using the above steps.

Therefore, $x^{3}-27 = \left( x-3\right) \left( x-3\left( \frac{-1}{2}+i\frac{\sqrt{3}}{2}\right) \right) \left( x-3\left( \frac{-1}{2}-i\frac{\sqrt{3}}{2}\right) \right)$

Note also $\left( x-3\left( \frac{-1}{2}+i\frac{\sqrt{3}}{2}\right) \right) \left( x-3\left( \frac{-1}{2}-i\frac{\sqrt{3}}{2}\right) \right) = x^{2}+3x+9$ and so $x^{3}-27=\left( x-3\right) \left( x^{2}+3x+9\right)$ where the quadratic polynomial $x^{2}+3x+9$ cannot be factored without using complex numbers.

Note that even though the polynomial $x^{3}-27$ has all real coefficients, it has some complex zeros, $3\left( \frac{-1}{2}+i\frac{\sqrt{3}}{2}\right) ,$ and $3\left( \frac{-1}{2}-i\frac{\sqrt{3}}{2}\right)$. These zeros are complex conjugates of each other. It is always the case that if a polynomial has real coefficients and a complex root, it will also have a root equal to the complex conjugate.

6.4 The Quadratic Formula 二次公式

Outcomes.

1. Use the Quadratic Formula to find the complex roots of a quadratic equation.

1. 使用二次公式求二次方程的复根。

The roots (or solutions) of a quadratic equation $ax^{2}+bx+c=0$ where $a,b,c$ are real numbers are obtained by solving the familiar quadratic formula given by $x= \frac{-b\pm \sqrt{b^{2}-4ac}}{2a} $

When working with real numbers, we cannot solve this formula if $b^{2}-4ac<0.$ However, complex numbers allow us to find square roots of negative numbers, and the quadratic formula remains valid for finding roots of the corresponding quadratic equation. In this case there are exactly two distinct (complex) square roots of $b^{2}-4ac$, which are $i\sqrt{4ac-b^{2}}$ and $-i\sqrt{4ac-b^{2}}$.

Here is an example.

Example 1: Solutions to Quadratic Equation 二次方程的解

Find the solutions to $x^{2}+2x+5=0$.

Solution. In terms of the quadratic equation above, $a=1$, $b=2$, and $c=5$. Therefore, we can use the quadratic formula with these values, which becomes

$$x= \frac{-b\pm \sqrt{b^{2}-4ac}}{2a} = \frac{-2 \pm \sqrt{\left(2\right)^{2} - 4 (1)(5)}}{2(1)}$$

Solving this equation, we see that the solutions are given by $x=\frac{-2i\pm \sqrt{4-20}}{2}=\frac{-2\pm 4i}{2}=-1\pm 2i$

We can verify that these are solutions of the original equation. We will show $x = -1+2i$ and leave $x = -1-2i$ as an exercise.

$$\begin{aligned} x^{2}+2x+5 &= (-1+2i)^2 + 2(-1+2i) + 5 \\ &= 1 - 4i - 4 -2 + 4i + 5 \\ &= 0\end{aligned}$$

Hence $x = -1+2i$ is a solution.

What if the coefficients of the quadratic equation are actually complex numbers? Does the formula hold even in this case? The answer is yes. This is a hint on how to do Exercise 6.E.26 below, a special case of the fundamental theorem of algebra, and an ingredient in the proof of some versions of this theorem.

Consider the following example.

Example 2: Solutions to Quadratic Equation 二次方程的解（复系数）

Find the solutions to $x^{2}-2ix-5=0$.

Solution. In terms of the quadratic equation above, $a=1$, $b=-2i$, and $c=-5$. Therefore, we can use the quadratic formula with these values, which becomes $x= \frac{-b\pm \sqrt{b^{2}-4ac}}{2a} = \frac{2i \pm \sqrt{\left(-2i\right)^{2} - 4 (1)(-5)}}{2(1)}$

Solving this equation, we see that the solutions are given by $x=\frac{2i\pm \sqrt{-4+20}}{2}=\frac{2i\pm 4}{2}=i\pm 2$

We can verify that these are solutions of the original equation. We will show $x = i + 2$ and leave $x = i-2$ as an exercise.

$$\begin{aligned} x^{2}-2ix-5 &= (i+2)^2 - 2i (i+2) - 5 \\ &= -1 + 4i + 4 + 2 - 4i - 5 \\ &= 0\end{aligned}$$

Hence $x = i+2$ is a solution.

We conclude this section by stating an essential theorem.

Theorem 1: The Fundamental Theorem of Algebra 代数基本定理

Any polynomial of degree at least $1$ with complex coefficients has a root which is a complex number.

6.E Exercises 习题

Exercise 1 习题 1

Let $z = 2+7i$ and let $w = 3-8i$. Compute the following.

1. $z+w$
2. $z-2w$
3. $zw$
4. $\frac{w}{z}$

1. $z+w$
2. $z-2w$
3. $zw$
4. $\frac{w}{z}$

Answer

1. $z+w=5-i$
2. $z-2w=-4+23i$
3. $zw=62+5i$
4. $\frac{w}{z}=-\frac{50}{53}-\frac{37}{53}i$

1. $z+w=5-i$
2. $z-2w=-4+23i$
3. $zw=62+5i$
4. $\frac{w}{z}=-\frac{50}{53}-\frac{37}{53}i$

Exercise 2 习题 2

Let $z = 1-4i$. Compute the following.

1. $\overline{z}$
2. $z^{-1}$
3. $\|z\|$

1. $\overline{z}$
2. $z^{-1}$
3. $\|z\|$

Exercise 3 习题 3

Let $z = 3+5i$ and $w = 2-i$. Compute the following.

1. $\overline{zw}$
2. $\|zw\|$
3. $z^{-1}w$

1. $\overline{zw}$
2. $\|zw\|$
3. $z^{-1}w$

Exercise 4 习题 4

If $z$ is a complex number, show there exists a complex number $w$ with $\|w\| = 1$ and $wz = \|z\|$.

Answer

If $z=0$, let $w=1$. If $z\neq 0$, let $w=\frac{\overline{z}}{\|z\|}$

Exercise 5 习题 5

If $z, w$ are complex numbers prove $\overline{zw} = \overline{z}\overline{w}$ and then show by induction that $\overline{z_1\cdots z_m} = \overline{z_1}\cdots\overline{z_m}$. Also verify that $\overline{\sum_{k=1}^m z_k}=\sum_{k=1}^m\overline{z_k}$. In words this says the conjugate of a product equals the product of the conjugates and the conjugate of a sum equals the sum of the conjugates.

Answer

$$\overline{(a+bi)(c+di)} = \overline{ac-bd+(ad+bc)i} = (ac-bd)-(ad+bc)i = (a-bi)(c-di) = ac-bd-(ad+bc)i$$ which is the same thing. Thus it holds for a product of two complex numbers. Now suppose you have that it is true for the product of $n$ complex numbers. Then $$\overline{z_1\cdots z_{n+1}} = \overline{z_1\cdots z_n}\,\overline{z_{n+1}}$$ and now, by induction this equals $$\overline{z_1}\cdots\overline{z_n}\,\overline{z_{n+1}}.$$ As to sums, this is even easier: $$\overline{\sum_{j=1}^n(x_j+iy_j)} = \overline{\sum_{j=1}^n x_j + i\sum_{j=1}^n y_j} = \sum_{j=1}^n x_j - i\sum_{j=1}^n y_j = \sum_{j=1}^n (x_j-iy_j) = \sum_{j=1}^n \overline{(x_j+iy_j)}.$$

Exercise 6 习题 6

Suppose $p(x) = a_nx^n +a_{n-1}x^{n-1} +\cdots +a_1x+a_0$ where all the $a_k$ are real numbers. Suppose also that $p(z) = 0$ for some $z \in \mathbb{C}$. Show it follows that $p(\overline{z}) = 0$ also.

Answer

If $p(z)=0$, then you have $\begin{aligned}\overline{p(z)}&=0=\overline{a_nz^n+a_{n-1}z^{n-1}+\cdots +a_1z+a_0} \\ &=\overline{a_nz^n}+\overline{a_{n-1}z^{n-1}}+\cdots +\overline{a_1z}+\overline{a_0} \\ &=\overline{a_n}\,\overline{z}^n+\overline{a_{n-1}}\,\overline{z}^{n-1}+\cdots +\overline{a_1}\,\overline{z}+\overline{a_0} \\ &=a_n\overline{z}^n+a_{n-1}\overline{z}^{n-1}+\cdots +a_1\overline{z}+a_0 \\ &=p(\overline{z})\end{aligned}$

Exercise 7 习题 7

I claim that $1=-1$. Here is why. $-1=i^2=\sqrt{-1}\sqrt{-1}=\sqrt{(-1)^2}=\sqrt{1}=1$ This is clearly a remarkable result but is there something wrong with it? If so, what is wrong?

Answer

The problem is that there is no single $\sqrt{-1}$.

Exercise 8 习题 8

Let $z = 3+3i$ be a complex number written in standard form. Convert $z$ to polar form, and write it in the form $z = re^{i\theta}$.

Exercise 9 习题 9

Let $z = 2i$ be a complex number written in standard form. Convert $z$ to polar form, and write it in the form $z = re^{i\theta}$.

Exercise 10 习题 10

Let $z = 4e^{\frac{2\pi}{3}i}$ be a complex number written in polar form. Convert $z$ to standard form, and write it in the form $z = a+bi$.

Exercise 11 习题 11

Let $z = -1e^{\frac{\pi}{6}i}$ be a complex number written in polar form. Convert $z$ to standard form, and write it in the form $z = a+bi$.

Exercise 12 习题 12

If $z$ and $w$ are two complex numbers and the polar form of $z$ involves the angle $\theta$ while the polar form of $w$ involves the angle $\phi$, show that in the polar form for $zw$ the angle involved is $\theta +\phi$.

Answer

You have $z = \|z\|(\cos \theta +i\sin \theta)$ and $w = \|w\|(\cos \phi +i\sin \phi)$. Then when you multiply these, you get $\begin{aligned} &\|z\|\,\|w\| (\cos\theta +i\sin\theta )(\cos \phi+i\sin \phi) \\ =&\|z\|\,\|w\| (\cos\theta\cos \phi-\sin\theta\sin \phi+i(\cos\theta\sin \phi+\cos \phi\sin\theta )) \\ =&\|z\|\,\|w\| (\cos (\theta +\phi)+i\sin (\theta+\phi))\end{aligned}$

Exercise 13 习题 13

Give the complete solution to $x^4+16=0$.

Answer

Solution is: $(1-i)\sqrt{2},$ $-(1+i)\sqrt{2},$ $-(1-i)\sqrt{2},$ $(1+i)\sqrt{2}$

Exercise 14 习题 14

Find the complex cube roots of $8$.

Answer

The cube roots are the solutions to $z^3 - 8 = 0$, Solution is: $-1 + i\sqrt{3},$ $-1 + i\sqrt{3},$ $2$

Exercise 15 习题 15

Find the four fourth roots of $-16$.

Answer

The fourth roots are the solutions to $z^4 + 16 = 0$, Solution is: $(1-i)\sqrt{2},$ $-(1+i)\sqrt{2},$ $-(1-i)\sqrt{2},$ $(1+i)\sqrt{2}$

Exercise 16 习题 16

De Moivre's theorem says $\left(r(\cos t +i\sin t)\right)^n = r^n (\cos nt +i\sin nt)$ for $n$ a positive integer. Does this formula continue to hold for all integers n, even negative integers? Explain.

Answer

Yes, it holds for all integers. First of all, it clearly holds if $n = 0$. Suppose now that n is a negative integer. Then $-n > 0$ and so $\left(r(\cos t+i\sin t)\right)^n=\frac{1}{\left(r(\cos t+i\sin t)\right)^{-n}}=\frac{1}{r^{-n}(\cos (-nt)+i\sin (-nt))}$ $\begin{aligned}&=\frac{r^n}{(\cos (nt)-i\sin (nt))}=\frac{r^n(\cos (nt)+i\sin (nt))}{(\cos (nt)-i\sin (nt))(\cos (nt)+i\sin (nt))} \\ &=r^n(\cos (nt)+i\sin (nt))\end{aligned}$ because $(\cos (nt)-i\sin (nt))(\cos (nt)+i\sin (nt))=1$.

Exercise 17 习题 17

Factor $x^3 +8$ as a product of linear factors. Hint: Use the result of Exercise 14.

Answer

Solution is: $i\sqrt{3}+1,$ $1-i\sqrt{3},$ $-2$ and so this polynomial equals $(x+2)\left(x-\left(i\sqrt{3}+1\right)\right)\left(x-\left(1-i\sqrt{3}\right)\right)$

Exercise 18 习题 18

Write $x^3 +27$ in the form $(x+3)(x^2 +ax+b)$ where $x^2 +ax +b$ cannot be factored any more using only real numbers.

Answer

$$x^3+27=(x+3)(x^2-3x+9)$$

Exercise 19 习题 19

Completely factor $x^4 +16$ as a product of linear factors. Hint: Use the result of Exercise 15.

Answer

Solution is: $(1-i)\sqrt{2},$ $-(1+i)\sqrt{2},$ $-(1-i)\sqrt{2},$ $(1+i)\sqrt{2}.$ These are just the fourth roots of $-16$. Then to factor, you get $\left(x-\left((1-i)\sqrt{2}\right)\right)\left(x-\left(-(1+i)\sqrt{2}\right)\right)$ $\left(x-\left(-(1-i)\sqrt{2}\right)\right)\left(x-\left((1+i)\sqrt{2}\right)\right)$

Exercise 20 习题 20

Factor $x^4 + 16$ as the product of two quadratic polynomials each of which cannot be factored further without using complex numbers.

Answer

$x^4+16=\left(x^2-2\sqrt{2}x+4\right)\left(x^2+2\sqrt{2}x+4\right)$. You can use the information in the preceding problem. Note that $(x-z) (x-\overline{z})$ has real coefficients.

Exercise 21 习题 21

If $n$ is an integer, is it always true that $(\cos \theta -i\sin \theta)^n = \cos(n\theta)-i\sin(n\theta)$? Explain.

Answer

Yes, this is true. $\begin{aligned}(\cos\theta -i\sin\theta)^n&=(\cos(-\theta)+i\sin(-\theta ))^n \\ &=\cos (-n\theta )+i\sin(-n\theta ) \\ &=\cos (n\theta )-i\sin (n\theta )\end{aligned}$

Exercise 22 习题 22

Suppose $p(x) = a_nx^n +a_{n-1}x^{n-1} +\cdots +a_1x+a_0$ is a polynomial and it has $n$ zeros, $z_1, z_2,\cdots ,z_n$ listed according to multiplicity. ($z$ is a root of multiplicity $m$ if the polynomial $f (x) = (x-z)^m$ divides $p(x)$ but $(x-z) f (x)$ does not.) Show that $p(x)=a_n(x-z_1)(x-z_2)\cdots (x-z_n)$

Answer

$p(x) = (x-z_1)q(x)+r(x)$ where $r(x)$ is a nonzero constant or equal to $0$. However, $r(z_1) = 0$ and so $r(x) = 0$. Now do to $q(x)$ what was done to $p(x)$ and continue until the degree of the resulting $q(x)$ equals $0$. Then you have the above factorization.

Exercise 23 习题 23

Show that $1+i,$ $2+i$ are the only two roots to $p(x) = x^2 -(3+2i)x+ (1+3i)$ Hence complex zeros do not necessarily come in conjugate pairs if the coefficients of the equation are not real.

Answer

$(x-(1+i)) (x-(2+i)) = x^2 -(3+2i)x+1+3i$

Exercise 24 习题 24

Give the solutions to the following quadratic equations having real coefficients.

1. $x^2-2x+2=0$
2. $3x^2+x+3=0$
3. $x^2-6x+13=0$
4. $x^2+4x+9=0$
5. $4x^2+4x+5=0$

1. $x^2-2x+2=0$
2. $3x^2+x+3=0$
3. $x^2-6x+13=0$
4. $x^2+4x+9=0$
5. $4x^2+4x+5=0$

Answer

1. Solution is: $1+i,$ $1-i$
2. Solution is: $\frac{1}{6}i\sqrt{35}-\frac{1}{6},$ $-\frac{1}{6}i\sqrt{35}-\frac{1}{6}$
3. Solution is: $3+2i,$ $3-2i$
4. Solution is: $i\sqrt{5}-2,$ $-i\sqrt{5}-2$
5. Solution is: $-\frac{1}{2}+i,$ $-\frac{1}{2}-i$

1. 解为：$1+i,$ $1-i$
2. 解为：$\frac{1}{6}i\sqrt{35}-\frac{1}{6},$ $-\frac{1}{6}i\sqrt{35}-\frac{1}{6}$
3. 解为：$3+2i,$ $3-2i$
4. 解为：$i\sqrt{5}-2,$ $-i\sqrt{5}-2$
5. 解为：$-\frac{1}{2}+i,$ $-\frac{1}{2}-i$

Exercise 25 习题 25

Give the solutions to the following quadratic equations having complex coefficients.

1. $x^2+2x+1+i=0$
2. $4x^2+4ix-5=0$
3. $4x^2+(4+4i)x+1+2i=0$
4. $x^2-4ix-5=0$
5. $3x^2+(1-i)x+3i=0$

1. $x^2+2x+1+i=0$
2. $4x^2+4ix-5=0$
3. $4x^2+(4+4i)x+1+2i=0$
4. $x^2-4ix-5=0$
5. $3x^2+(1-i)x+3i=0$

Answer

1. Solution is: $x=-1+\frac{1}{2}\sqrt{2}-\frac{1}{2}i\sqrt{2},$ $x=-1-\frac{1}{2}\sqrt{2}+\frac{1}{2}i\sqrt{2}$
2. Solution is: $x=1-\frac{1}{2}i,$ $x=-1-\frac{1}{2}i$
3. Solution is: $x=-\frac{1}{2},$ $x=-\frac{1}{2}-i$
4. Solution is: $x=-1+2i,$ $x=1+2i$
5. Solution is: $x=-\frac{1}{6}+\frac{1}{6}\sqrt{19}+\left(\frac{1}{6}-\frac{1}{6}\sqrt{19}\right)i,$ $x=-\frac{1}{6}-\frac{1}{6}\sqrt{19}+\left(\frac{1}{6}+\frac{1}{6}\sqrt{19}\right)i$

1. 解为：$x=-1+\frac{1}{2}\sqrt{2}-\frac{1}{2}i\sqrt{2},$ $x=-1-\frac{1}{2}\sqrt{2}+\frac{1}{2}i\sqrt{2}$
2. 解为：$x=1-\frac{1}{2}i,$ $x=-1-\frac{1}{2}i$
3. 解为：$x=-\frac{1}{2},$ $x=-\frac{1}{2}-i$
4. 解为：$x=-1+2i,$ $x=1+2i$
5. 解为：$x=-\frac{1}{6}+\frac{1}{6}\sqrt{19}+\left(\frac{1}{6}-\frac{1}{6}\sqrt{19}\right)i,$ $x=-\frac{1}{6}-\frac{1}{6}\sqrt{19}+\left(\frac{1}{6}+\frac{1}{6}\sqrt{19}\right)i$

Exercise 26 习题 26

Prove the fundamental theorem of algebra for quadratic polynomials having coefficients in $\mathbb{C}$. That is, show that an equation of the form $ax^2 + bx + c = 0$ where $a, b, c$ are complex numbers, $a\neq 0$ has a complex solution. Hint: Consider the fact, noted earlier that the expressions given from the quadratic formula do in fact serve as solutions.