介绍复数

闲聊灌水
21

Typing Chinese is really a pain. The IM often stops responding; moreover, the Discourse math plugin seems a little presumptuous.

故 $T = z_1^{-1} z_2^{-1}$ 。

故 T = z_1^{-1} z_2^{-1} 。

故 $T = z_1^{-1} z_2^{-1}$。

故 $T = z_1^{-1} z_2^{-1}$。

Well, if I do not insert a space before Chinese punctuation marks, the formula surrounded by two dollar signs ($) will not be parsed correctly.

I cannot fix it, because I cannot control the plugin.

I do not know what to do to fix my IM. However, that does not really bother me, since at least the Chinese IM in one of my VMs functions normally.

Most users of Appinn Forum do not type formulae, so the “feature” does not bother them.

Few people bother to read something on math on an application forum.

I cannot see any rule which forbids the usage of English.

Taking these things into consideration, I decide to present the remaining materials in English. My English is poor, though.

22

Powers

Definition. Suppose that z is a complex number, and that n is a nonnegative integer. Define

z^n = \begin{cases} 1, & \quad \text{if $n = 0$;} \\ z \cdot z^{n-1}, & \quad \text{else.} \end{cases}

Example. Take z = (1,2). Then

z^2 = (1 \cdot 1 - 2 \cdot 2, 1 \cdot 2 + 2 \cdot 1) = (-3,4)

Example. Call \mathrm{i} = (0,1). Then

\mathrm{i}^0 = (1,0), \quad \mathrm{i}^1 = (0,1), \quad \mathrm{i}^2 = (-1,0), \quad \mathrm{i}^3 = (0,-1), \quad \mathrm{i}^4 = (1,0).

The verification is left as an exercise for readers. As a challenge, readers are invited to write a formula for \mathrm{i}^n, where n is a nonnegative integer.

Propositions. Suppose that z, w are complex numbers. Suppose that m, n are nonnegative integers. The following equalities hold:

  1. z^m z^n = z^{m+n};
  2. (z^m)^n = z^{mn};
  3. z^m w^m = (zw)^m.
You could see the source code here. 
## Powers

**Definition.** Suppose that $z$ is a complex number, and that $n$ is a nonnegative integer. Define
$$
z^n = \begin{cases}
    1, & \quad \text{if $n = 0$;} \\
    z \cdot z^{n-1}, & \quad \text{else.}
\end{cases}
$$

**Example.** Take $z = (1,2)$. Then
$$
z^2 = (1 \cdot 1 - 2 \cdot 2, 1 \cdot 2 + 2 \cdot 1) = (-3,4)
$$

**Example.** Call $\mathrm{i} = (0,1)$. Then
$$
\mathrm{i}^0 = (1,0), \quad \mathrm{i}^1 = (0,1), \quad \mathrm{i}^2 = (-1,0), \quad \mathrm{i}^3 = (0,-1), \quad \mathrm{i}^4 = (1,0).
$$
The verification is left as an exercise for readers. As a challenge, readers are invited to write a formula for $\mathrm{i}^n$, where $n$ is a nonnegative integer.

> **Propositions.** Suppose that $z$, $w$ are complex numbers. Suppose that $m$, $n$ are nonnegative integers. The following equalities hold:
> 
> 0. $z^m z^n = z^{m+n}$;
> 1. $(z^m)^n = z^{mn}$;
> 2. $z^m w^m = (zw)^m$.
1 个赞
23

You are good at English.

24

Thanks. It seems that reading English every day paid off.

25

Definition. Suppose that z is a nonzero complex number, and that n is a negative integer. Define

z^n = (z^{-1})^{-n}.

Question. Why do we not define z^n if z = 0 and n < 0?

Example. Take z = (1,2). Clearly, z \neq 0. Then

z^{-1} = \left(\frac{1}{1^2 + 2^2}, \frac{-2}{1^2 + 2^2}\right) = \left(\frac15, -\frac25\right).

Hence

z^{-2} = \left( \frac15 \cdot \frac15 - \frac{-2}5 \cdot \frac{-2}5, \frac15 \cdot \frac{-2}5 + \frac{-2}5 \cdot \frac15 \right) = \left(-\frac{3}{25}, -\frac{4}{25} \right),

which is exactly (z^2)^{-1}. The reader should verify it.

Example. Recall that \mathrm{i} = (0,1). Then

\mathrm{i}^{-1} = \left( \frac{0}{0^2 + 1^2}, \frac{-1}{0^2 + 1^2} \right) = (0, -1) = \mathrm{i}^3.

As a challenge, readers are invited to write a formula for \mathrm{i}^n, where n is an integer.

Propositions. Suppose that z, w are nonzero complex numbers. Suppose that m, n are integers. The following equalities hold:

  1. z^m z^n = z^{m+n};
  2. (z^m)^n = z^{mn};
  3. z^m w^m = (zw)^m.
You could see the source code here. 
**Definition.** Suppose that $z$ is a **nonzero** complex number, and that $n$ is a negative integer. Define
$$
z^n = (z^{-1})^{-n}.
$$

**Question.** Why do we not define $z^n$ if $z = 0$ and $n < 0$?

**Example.** Take $z = (1,2)$. Clearly, $z \neq 0$. Then
$$
z^{-1} = \left(\frac{1}{1^2 + 2^2}, \frac{-2}{1^2 + 2^2}\right) = \left(\frac15, -\frac25\right).
$$
Hence
$$
z^{-2} = \left( \frac15 \cdot \frac15 - \frac{-2}5 \cdot \frac{-2}5, \frac15 \cdot \frac{-2}5 + \frac{-2}5 \cdot \frac15 \right) = \left(-\frac{3}{25}, -\frac{4}{25} \right),
$$
which is exactly $(z^2)^{-1}$. The reader should verify it.

**Example.** Recall that $\mathrm{i} = (0,1)$. Then
$$
\mathrm{i}^{-1} = \left( \frac{0}{0^2 + 1^2}, \frac{-1}{0^2 + 1^2} \right) = (0, -1) = \mathrm{i}^3.
$$
As a challenge, readers are invited to write a formula for $\mathrm{i}^n$, where $n$ is **an integer**.

> **Propositions.** Suppose that $z$, $w$ are **nonzero** complex numbers. Suppose that $m$, $n$ are **integers**. The following equalities hold:
> 
> 0. $z^m z^n = z^{m+n}$;
> 1. $(z^m)^n = z^{mn}$;
> 2. $z^m w^m = (zw)^m$.
26

As for now, complex numbers are ordered pairs (x,y), where x, y are real numbers. We will write complex numbers in a new way.

Recall that we write 0 = (0,0) and 1 = (1,0). We continue in the same fashion; that is, we identify (x,0) with x.

Suppose that a and b are real numbers. Here are some reasons for which the identification makes sense.

  • If a = b, then (a,0) = (b,0), and vice versa.
  • (a,0) + (b,0) = (a+b, 0).
  • (a,0) \cdot (b,0) = (ab, 0).
  • No real number equals (x,y) if y \neq 0 (that is, the set of all real numbers and the set of all complex numbers (x,y) where y \neq 0 do not intersect).

Suppose that a is a real number. Then

a(x,y) = (a,0)(x,y) = (ax-0y, ay+0x)=(ax,ay).

Recall that \mathrm{i} = (0,1). Hence

\begin{aligned} x + \mathrm{i}y = {} & (x,0) + (0,1)(y,0) \\ = {} & (x,0) + (y,0)(0,1) \\ = {} & (x,0) + (0,y) \\ = {} & (x,y). \end{aligned}

From now on, we will write (x,y) as x + \mathrm{i} y (and vice versa, if necessary). Of course, x + y \mathrm{i} also works.

From now on, we will say that (a,0) is a real number (if a is a real number).

From now on, we will say that every real number is also a complex number.

Recall that \mathrm{i}^2 = (-1, 0) = -1. Suppose that x_1, y_1, x_2, y_2 are real numbers. By laws of addition and multiplication,

\begin{aligned} & \begin{aligned} (x_1 + \mathrm{i} y_1) + (x_2 + \mathrm{i} y_2) = {} & ((x_1 + \mathrm{i} y_1) + x_2) + \mathrm{i} y_2 \\ = {} & (x_1 + (\mathrm{i} y_1 + x_2)) + \mathrm{i} y_2 \\ = {} & (x_1 + (x_2 + \mathrm{i} y_1)) + \mathrm{i} y_2 \\ = {} & ((x_1 + x_2) + \mathrm{i} y_1) + \mathrm{i} y_2 \\ = {} & (x_1 + x_2) + (\mathrm{i} y_1 + \mathrm{i} y_2) \\ = {} & (x_1 + x_2) + \mathrm{i} (y_1 + y_2), \end{aligned} \\ & \begin{aligned} (x_1 + \mathrm{i} y_1) (x_2 + \mathrm{i} y_2) = {} & x_1 (x_2 + \mathrm{i} y_2) + (\mathrm{i} y_1) (x_2 + \mathrm{i} y_2) \\ = {} & (x_1 x_2 + x_1 (\mathrm{i} y_2)) + ((\mathrm{i} y_1) x_2 + (\mathrm{i} y_1) (\mathrm{i} y_2)) \\ = {} & (x_1 x_2 + (x_1 \mathrm{i}) y_2) + (\mathrm{i} (y_1 x_2) + ((\mathrm{i} y_1) \mathrm{i}) y_2) \\ = {} & (x_1 x_2 + (\mathrm{i} x_1) y_2) + (\mathrm{i} (y_1 x_2) + (\mathrm{i} (y_1 \mathrm{i})) y_2) \\ = {} & (x_1 x_2 + \mathrm{i} (x_1 y_2)) + (\mathrm{i} (y_1 x_2) + (\mathrm{i} (\mathrm{i} y_1)) y_2) \\ = {} & (x_1 x_2 + \mathrm{i} (x_1 y_2)) + (\mathrm{i} (y_1 x_2) + ((\mathrm{i} \mathrm{i}) y_1) y_2) \\ = {} & (x_1 x_2 + \mathrm{i} (x_1 y_2)) + (\mathrm{i} (y_1 x_2) + ((\mathrm{i} \mathrm{i}) y_1) y_2) \\ = {} & (x_1 x_2 + \mathrm{i} (x_1 y_2)) + (\mathrm{i} (y_1 x_2) + (\mathrm{i} \mathrm{i}) (y_1 y_2)) \\ = {} & (x_1 x_2 + \mathrm{i} (x_1 y_2)) + (\mathrm{i} (y_1 x_2) + (-1) (y_1 y_2)) \\ = {} & (x_1 x_2 + \mathrm{i} (x_1 y_2)) + (\mathrm{i} (y_1 x_2) - y_1 y_2) \\ = {} & x_1 x_2 + (\mathrm{i} (x_1 y_2) + (\mathrm{i} (y_1 x_2) - y_1 y_2)) \\ = {} & x_1 x_2 + ((\mathrm{i} (x_1 y_2) + \mathrm{i} (y_1 x_2)) - y_1 y_2) \\ = {} & x_1 x_2 + (\mathrm{i} (x_1 y_2 + y_1 x_2) - y_1 y_2) \\ = {} & x_1 x_2 + (-y_1 y_2 + \mathrm{i} (x_1 y_2 + y_1 x_2)) \\ = {} & (x_1 x_2 - y_1 y_2) + \mathrm{i} (x_1 y_2 + y_1 x_2). \end{aligned} \end{aligned}

Hence, one can always easily compute the sum or product of two complex numbers, if they remember that \mathrm{i}^2 = -1 and the usual rules of arithmetic (commutativity, associativity, the distributive law, etc.).

Postscript.

  • 0 + b\mathrm{i} can be written as b\mathrm{i}.
  • x + 0\mathrm{i} can be written as x.
  • 1\mathrm{i} and (-1)\mathrm{i} can be written as \mathrm{i} and -\mathrm{i}, respectively.
You could see the source code here. 
As for now, complex numbers are **ordered pairs** $(x,y)$, where $x$, $y$ are real numbers. We will write complex numbers in a new way.

Recall that we write $0 = (0,0)$ and $1 = (1,0)$. We continue in the same fashion; that is, we identify $(x,0)$ with $x$.

Suppose that $a$ and $b$ are real numbers. Here are some reasons for which the identification makes sense.
- If $a = b$, then $(a,0) = (b,0)$, and vice versa.
- $(a,0) + (b,0) = (a+b, 0)$.
- $(a,0) \cdot (b,0) = (ab, 0)$.
- No real number equals $(x,y)$ if $y \neq 0$ (that is, the set of all real numbers and the set of all complex numbers $(x,y)$ where $y \neq 0$ do not intersect).

Suppose that $a$ is a real number. Then
$$
a(x,y) = (a,0)(x,y) = (ax-0y, ay+0x)=(ax,ay).
$$
Recall that $\mathrm{i} = (0,1)$. Hence
$$
\begin{aligned}
x + \mathrm{i}y
= {} & (x,0) + (0,1)(y,0) \\
= {} & (x,0) + (y,0)(0,1) \\
= {} & (x,0) + (0,y) \\
= {} & (x,y).
\end{aligned}
$$

> From now on, we will write $(x,y)$ as $x + \mathrm{i} y$ (and vice versa, if necessary). Of course, $x + y \mathrm{i}$ also works.

> From now on, we will say that $(a,0)$ is a real number (if $a$ is a real number).

> From now on, we will say that every real number is also a complex number.

Recall that $\mathrm{i}^2 = (-1, 0) = -1$. Suppose that $x_1$, $y_1$, $x_2$, $y_2$ are real numbers. By laws of addition and multiplication,
$$
\begin{aligned}
& \begin{aligned}
(x_1 + \mathrm{i} y_1) + (x_2 + \mathrm{i} y_2)
= {} & ((x_1 + \mathrm{i} y_1) + x_2) + \mathrm{i} y_2 \\
= {} & (x_1 + (\mathrm{i} y_1 + x_2)) + \mathrm{i} y_2 \\
= {} & (x_1 + (x_2 + \mathrm{i} y_1)) + \mathrm{i} y_2 \\
= {} & ((x_1 + x_2) + \mathrm{i} y_1) + \mathrm{i} y_2 \\
= {} & (x_1 + x_2) + (\mathrm{i} y_1 + \mathrm{i} y_2) \\
= {} & (x_1 + x_2) + \mathrm{i} (y_1 + y_2),
\end{aligned} \\
& \begin{aligned}
(x_1 + \mathrm{i} y_1) (x_2 + \mathrm{i} y_2)
= {} & x_1 (x_2 + \mathrm{i} y_2) + (\mathrm{i} y_1) (x_2 + \mathrm{i} y_2) \\
= {} & (x_1 x_2 + x_1 (\mathrm{i} y_2)) + ((\mathrm{i} y_1) x_2 + (\mathrm{i} y_1) (\mathrm{i} y_2)) \\
= {} & (x_1 x_2 + (x_1 \mathrm{i}) y_2) + (\mathrm{i} (y_1 x_2) + ((\mathrm{i} y_1) \mathrm{i}) y_2) \\
= {} & (x_1 x_2 + (\mathrm{i} x_1) y_2) + (\mathrm{i} (y_1 x_2) + (\mathrm{i} (y_1 \mathrm{i})) y_2) \\
= {} & (x_1 x_2 + \mathrm{i} (x_1 y_2)) + (\mathrm{i} (y_1 x_2) + (\mathrm{i} (\mathrm{i} y_1)) y_2) \\
= {} & (x_1 x_2 + \mathrm{i} (x_1 y_2)) + (\mathrm{i} (y_1 x_2) + ((\mathrm{i} \mathrm{i}) y_1) y_2) \\
= {} & (x_1 x_2 + \mathrm{i} (x_1 y_2)) + (\mathrm{i} (y_1 x_2) + ((\mathrm{i} \mathrm{i}) y_1) y_2) \\
= {} & (x_1 x_2 + \mathrm{i} (x_1 y_2)) + (\mathrm{i} (y_1 x_2) + (\mathrm{i} \mathrm{i}) (y_1 y_2)) \\
= {} & (x_1 x_2 + \mathrm{i} (x_1 y_2)) + (\mathrm{i} (y_1 x_2) + (-1) (y_1 y_2)) \\
= {} & (x_1 x_2 + \mathrm{i} (x_1 y_2)) + (\mathrm{i} (y_1 x_2) - y_1 y_2) \\
= {} & x_1 x_2 + (\mathrm{i} (x_1 y_2) + (\mathrm{i} (y_1 x_2) - y_1 y_2)) \\
= {} & x_1 x_2 + ((\mathrm{i} (x_1 y_2) + \mathrm{i} (y_1 x_2)) - y_1 y_2) \\
= {} & x_1 x_2 + (\mathrm{i} (x_1 y_2 + y_1 x_2) - y_1 y_2) \\
= {} & x_1 x_2 + (-y_1 y_2 + \mathrm{i} (x_1 y_2 + y_1 x_2)) \\
= {} & (x_1 x_2 - y_1 y_2) + \mathrm{i} (x_1 y_2 + y_1 x_2).
\end{aligned}
\end{aligned}
$$
Hence, one can always easily compute the sum or product of two complex numbers, if they remember that $\mathrm{i}^2 = -1$ and the usual rules of arithmetic (commutativity, associativity, the distributive law, etc.).

**Postscript.**

- $0 + b\mathrm{i}$ can be written as $b\mathrm{i}$.
- $x + 0\mathrm{i}$ can be written as $x$.
- $1\mathrm{i}$ and $(-1)\mathrm{i}$ can be written as $\mathrm{i}$ and $-\mathrm{i}$, respectively.
27

One does not have difficulty in writing x + \mathrm{i}y with paper and a pen. However, it is not easy to type x + \mathrm{i}y. According to the custom, we write \mathrm{i} (instead of i) if we refer to the complex number (0,1).

$x + \mathrm{i}y$ produces x + \mathrm{i}y. It is longer than (x,y) ($(x,y)$) is, so I did not use the \mathrm{i} notation at the beginning of the introduction.

28

Suppose that z and w are complex numbers.

Here are some useful identities which will be used later.

\begin{aligned} & (z + w)(z - w) = z^2 - w^2, \\ & (z + w)^2 = z^2 + 2zw + w^2, \\ & (z - w)^2 = z^2 - 2zw + w^2, \\ & (z - w)(z^2 + zw + w^2) = z^3 - w^3, \\ & (z + w)(z^2 - zw + w^2) = z^3 + w^3. \end{aligned}

As an exercise, readers are invited to show that

z^2 + w^2 = (z - \mathrm{i}w)(z + \mathrm{i}w).
You could see the source code here. 
Suppose that $z$ and $w$ are complex numbers.

---

Here are some useful identities which will be used later.
$$
\begin{aligned}
& (z + w)(z - w) = z^2 - w^2, \\
& (z + w)^2 = z^2 + 2zw + w^2, \\
& (z - w)^2 = z^2 - 2zw + w^2, \\
& (z - w)(z^2 + zw + w^2) = z^3 - w^3, \\
& (z + w)(z^2 - zw + w^2) = z^3 + w^3.
\end{aligned}
$$

As an exercise, readers are invited to show that
$$
z^2 + w^2 = (z - \mathrm{i}w)(z + \mathrm{i}w).
$$
29

Conjugate

Definition. Suppose that z = (x,y) is a complex number. Denoted as \overline{z}, (x,-y) is said to be the (complex) conjugate of z.

Example. The conjugate of 1 - 2\mathrm{i} is 1 + 2\mathrm{i}.

Suppose that z and w are complex numbers.

  1. Suppose that c = \overline{z}. Then \overline{c} = z.
  2. Suppose that s = z + w. Then \overline{s} = \overline{z} + \overline{w}.
  3. Suppose that p = z \cdot w. Then \overline{p} = \overline{z} \cdot \overline{w}.
  4. \overline{z} + z is real.
  5. \mathrm{i} (\overline{z} - z) is real.
  6. \overline{z} z = 0 if z = 0.
  7. \overline{z} z > 0 unless z = 0.

Suppose that z = (x,y) and that w = (u,v).

0. By the definition of the complex conjugate, c = (x,-y). Hence \overline{c} = (x,-(-y)), which is exactly (x,y).

1. By the definition of addition,

s = z + w = (x+u, y+v).

The conjugate of s is

\begin{aligned} (x+u, -(y+v)) = {} & (x+u, -y-v) \\ = {} & (x,-y) + (u,-v) \\ = {} & \overline{z} + \overline{w}. \end{aligned}

2. By the definition of multiplication,

p = z w = (xu - yv, xv + yu).

The conjugate of p is

\begin{aligned} & (xu - yv, -(xv + yu)) \\ = {} & (xu - yv, -xv-yu) \\ = {} & (xu - (-y)(-v), x(-v) + (-y)u) \\ = {} & (x,-y) (u,-v) \\ = {} & \overline{z} \cdot \overline{w}. \end{aligned}

3. \overline{z} + z = (x,-y) + (x,y) = (2x,0) = 2x, which is real.

4. \mathrm{i}(\overline{z} - z) = (0,1)((x,-y) - (x,y)) = (0,1)(0,-2y) = (2y,0) = 2y, which is real.

5. Somewhat too obvious.

6. It is known that z_1^2 + z_2^2 = (z_1 - \mathrm{i}z_2)(z_1 + \mathrm{i}z_2). Taking z_1 = x and z_2 = y yields

x^2 + y^2 = (x - \mathrm{i}y) (x + \mathrm{i}y) = \overline{z} z.

Since z \neq 0, x^2 + y^2 > 0.

You could see the source code here. 
## Conjugate

**Definition.** Suppose that $z = (x,y)$ is a complex number. Denoted as $\overline{z}$, $(x,-y)$ is said to be the *(complex) conjugate* of $z$.

**Example.** The conjugate of $1 - 2\mathrm{i}$ is $1 + 2\mathrm{i}$.

> Suppose that $z$ and $w$ are complex numbers.
>
> 0. Suppose that $c = \overline{z}$. Then $\overline{c} = z$.
> 1. Suppose that $s = z + w$. Then $\overline{s} = \overline{z} + \overline{w}$.
> 2. Suppose that $p = z \cdot w$. Then $\overline{p} = \overline{z} \cdot \overline{w}$.
> 3. $\overline{z} + z$ is real.
> 4. $\mathrm{i} (\overline{z} - z)$ is real.
> 5. $\overline{z} z = 0$ if $z = 0$.
> 6. $\overline{z} z > 0$ unless $z = 0$.

---

Suppose that $z = (x,y)$ and that $w = (u,v)$.

0\. By the definition of the complex conjugate, $c = (x,-y)$. Hence $\overline{c} = (x,-(-y))$, which is exactly $(x,y)$.

1\. By the definition of addition,
$$
s = z + w = (x+u, y+v).
$$
The conjugate of $s$ is
$$
\begin{aligned}
    (x+u, -(y+v))
    = {} & (x+u, -y-v) \\
    = {} & (x,-y) + (u,-v) \\
    = {} & \overline{z} + \overline{w}.
\end{aligned}
$$

2\. By the definition of multiplication,
$$
p = z w = (xu - yv, xv + yu).
$$
The conjugate of $p$ is
$$
\begin{aligned}
    & (xu - yv, -(xv + yu)) \\
    = {} & (xu - yv, -xv-yu) \\
    = {} & (xu - (-y)(-v), x(-v) + (-y)u) \\
    = {} & (x,-y) (u,-v) \\
    = {} & \overline{z} \cdot \overline{w}.
\end{aligned}
$$

3\. $\overline{z} + z = (x,-y) + (x,y) = (2x,0) = 2x$, which is real.

4\. $\mathrm{i}(\overline{z} - z) = (0,1)((x,-y) - (x,y)) = (0,1)(0,-2y) = (2y,0) = 2y$, which is real.

5\. Somewhat too obvious.

6\. It is known that $z_1^2 + z_2^2 = (z_1 - \mathrm{i}z_2)(z_1 + \mathrm{i}z_2)$. Taking $z_1 = x$ and $z_2 = y$ yields
$$
x^2 + y^2 = (x - \mathrm{i}y) (x + \mathrm{i}y) = \overline{z} z.
$$
Since $z \neq 0$, $x^2 + y^2 > 0$.
30

Suppose that b is a nonnegative (real) number. There exists a unique nonnegative number r such that r^2 = b, which is often denoted as \sqrt{b}. Moreover, if b > 0, then \sqrt{b} > 0.

This is an important proposition. We will take it for granted, since a rigorous proof requires much deeper mathematics.

Suppose that a and b are nonnegative numbers. If a > b, then \sqrt{a} > \sqrt{b}, and vice versa.

This is easy. Suppose a > b \geq 0. Then \sqrt{a} + \sqrt{b} \geq \sqrt{a} > 0. Hence {1 \over \sqrt{a} + \sqrt{b}} > 0. Using algebraic identities, we have

\sqrt{a} - \sqrt{b} = {(\sqrt{a})^2 - (\sqrt{b})^2 \over \sqrt{a} + \sqrt{b}} = {a - b \over \sqrt{a} + \sqrt{b}} > 0.

Conversely, suppose that \sqrt{a} > \sqrt{b}. Then

a - b = (\sqrt{a})^2 - (\sqrt{b})^2 = (\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b}) > 0.
You could see the source code here. 
> Suppose that $b$ is a nonnegative (real) number. There exists a unique nonnegative number $r$ such that $r^2 = b$, which is often denoted as $\sqrt{b}$. Moreover, if $b > 0$, then $\sqrt{b} > 0$.

This is an important proposition. We will take it for granted, since a rigorous proof requires much deeper mathematics.

> Suppose that $a$ and $b$ are nonnegative numbers. If $a > b$, then $\sqrt{a} > \sqrt{b}$, and vice versa.

This is easy. Suppose $a > b \geq 0$. Then $\sqrt{a} + \sqrt{b} \geq \sqrt{a} > 0$. Hence ${1 \over \sqrt{a} + \sqrt{b}} > 0$. Using algebraic identities, we have
$$
\sqrt{a} - \sqrt{b} = {(\sqrt{a})^2 - (\sqrt{b})^2 \over \sqrt{a} + \sqrt{b}} = {a - b \over \sqrt{a} + \sqrt{b}} > 0.
$$

Conversely, suppose that $\sqrt{a} > \sqrt{b}$. Then
$$
a - b = (\sqrt{a})^2 - (\sqrt{b})^2 = (\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b}) > 0.
$$
31

It is the winter solstice today. I decide to eat some dumplings :dumpling:.

32

Somewhat busy and tired. I had a nucleic acid test. The hospital is “a little” far away from the place in which I live.

I promise that I will finish the introduction.

33

I had not saw you introduce 復平面。

and if you show us how the 復數 calculate in the 復平面，I think it is better to know what is 復數.

34

I agree with you.

Since every point in the two-dimensional orthogonal coordinate system uniquely corresponds to some ordered pair, and every ordered pair is a complex number (or uniquely corresponds to some complex number, which depends upon the definition of complex numbers), complex numbers can be visualized as points in the 2D OCS. The OCS is consequently said to be the complex plane.

Addition and subtraction of CNs are essentially the same as those of 2D vectors. To be precise, the set of all CNs is a vector space over the set of all real numbers with respect to usual arithmetic operations.

Doing multiplication or division geometrically requires a good command of the polar coordinate system and two basic trigonometric functions, sine and cosine. To be specific, multiplication can be thought of as a combination of two actions (precisely, transformations): rotation followed by scaling. Division is the inverse of multiplication.

I think that I will introduce the geometrical representation of complex numbers after a proper introduction to the 2D OCS and some basic trigonometry.

I hope that I managed to make myself understood.

35

Absolute value

Definition. Suppose that z is a complex number. Define the absolute value (or the norm) of z to be

|z| = \sqrt{\overline{z} z}.

Example. Suppose that z = 1 + 2\mathrm{i}. Then \overline{z} z = 1^2 + 2^2. Hence |z| = \sqrt{5}, which is approximately 2.236\,067\,977\,5.

Definition. Suppose that z = (x, y) is a complex number. Define the real and imaginary parts of z to be x and y, respectively. We occasionally write

\Re z = x \quad \text{and} \quad \Im z = y.

The following result immediately follows from the definitions.

Propositions. Suppose that z is a complex number. Then

|z| = \sqrt{(\Re z)^2 + (\Im z)^2}.
You could see the source code here. 
## Absolute value

**Definition.** Suppose that $z$ is a complex number. Define *the absolute value* (or *the norm*) of $z$ to be
$$
|z| = \sqrt{\overline{z} z}.
$$

**Example.** Suppose that $z = 1 + 2\mathrm{i}$. Then $\overline{z} z = 1^2 + 2^2$. Hence $|z| = \sqrt{5}$, which is approximately $2.236\,067\,977\,5$.

**Definition.** Suppose that $z = (x, y)$ is a complex number. Define *the real* and *imaginary parts* of $z$ to be $x$ and $y$, respectively. We occasionally write
$$
\Re z = x \quad \text{and} \quad \Im z = y.
$$

The following result immediately follows from the definitions.

> **Propositions.** Suppose that $z$ is a complex number. Then
> $$
> |z| = \sqrt{(\Re z)^2 + (\Im z)^2}.
> $$
36

Suppose that z and w are complex numbers.

  1. |z| = 0 if z = 0.
  2. |z| > 0 unless z = 0.
  3. |zw| = |z| \cdot |w|.
  4. |\Re z| \leq |z| and |\Im z| \leq |z|.
  5. |z + w| \leq |z| + |w|.
  6. ||z| - |w|| \leq |z - w|.
  7. |z| \leq |\Re z| + |\Im z|.

0. It is obvious.

1. Suppose that z \neq 0 and that |z| \leq 0. By definition, |z| = \sqrt{\overline{z} z} \geq 0. Hence |z| = 0, which means that

\overline{z} = \frac{\overline{z} z}{z} = \frac{|z|^2}{z} = 0,

which in turn implies that z = \overline{0} = 0. A contradiction.

2. Note that

\begin{aligned} |zw|^2 = {} & \overline{zw} \cdot (zw) = (\overline{z} \cdot \overline{w})(w z) \\ = {} & ((\overline{z} \cdot \overline{w})w)z = (\overline{z} (\overline{w} w)) z \\ = {} & \overline{z} ((|w| \cdot |w|) z) = \overline{z} (z (|w| \cdot |w|)) \\ = {} & (\overline{z} z) (|w| \cdot |w|) = (|z| \cdot |z|) (|w| \cdot |w|) \\ = {} & |z| (|z| (|w| \cdot |w|)) = |z| ((|z| \cdot |w|) |w|) \\ = {} & |z| ((|w| \cdot |z|) |w|) = |z| (|w| (|z| \cdot |w|)) \\ = {} & (|z| \cdot |w|) (|z| \cdot |w|) = (|z| \cdot |w|)^2. \end{aligned}

3. Note that

|\Re z| = \sqrt{(\Re z)^2} \leq \sqrt{(\Re z)^2 + (\Im z)^2} = |z|.

The same skill also applies to the other inequality.

4. Note that

\begin{aligned} & (|z| + |w|)^2 - |z + w|^2 \\ = {} & (|z|^2 + 2 |z| |w| + |w|^2) - (\overline{z} + \overline{w}) (z + w) \\ = {} & (\overline{z}z + \overline{w}w + 2|\overline{z} w|) - (\overline{z}z + \overline{w} w + (\overline{z} w + z \overline{w})) \\ = {} & 2(|\overline{z} w| - \Re {(\overline{z} w)}) \\ \geq {} & 2(|\overline{z} w| - |\Re {(\overline{z} w)}|) \\ \geq {} & 0. \end{aligned}

5. Note that

|z| - |w| = |z| - (|w| + |z - w|) + |z - w| \leq |z - w|

and that |w| - |z| \leq |w - z| = |z - w|.

6. Note that

|z| = |\Re z + \mathrm{i} \Im z| \leq |\Re z| + |\mathrm{i} \Im z| = |\Re z| + |\mathrm{i}| \cdot |\Im z|.
You could see the source code here. 
> Suppose that $z$ and $w$ are complex numbers.
>
> 0. $|z| = 0$ if $z = 0$.
> 1. $|z| > 0$ unless $z = 0$.
> 2. $|zw| = |z| \cdot |w|$.
> 3. $|\Re z| \leq |z|$ and $|\Im z| \leq |z|$.
> 4. $|z + w| \leq |z| + |w|$.
> 5. $||z| - |w|| \leq |z - w|$.
> 6. $|z| \leq |\Re z| + |\Im z|$.

---

0\. It is obvious.

1\. Suppose that $z \neq 0$ and that $|z| \leq 0$. By definition, $|z| = \sqrt{\overline{z} z} \geq 0$. Hence $|z| = 0$, which means that
$$
\overline{z} = \frac{\overline{z} z}{z} = \frac{|z|^2}{z} = 0,
$$
which in turn implies that $z = \overline{0} = 0$. A contradiction.

2\. Note that
$$
\begin{aligned}
    |zw|^2
    = {} & \overline{zw} \cdot (zw) = (\overline{z} \cdot \overline{w})(w z) \\
    = {} & ((\overline{z} \cdot \overline{w})w)z = (\overline{z} (\overline{w} w)) z \\
    = {} & \overline{z} ((|w| \cdot |w|) z) = \overline{z} (z (|w| \cdot |w|)) \\
    = {} & (\overline{z} z) (|w| \cdot |w|) = (|z| \cdot |z|) (|w| \cdot |w|) \\
    = {} & |z| (|z| (|w| \cdot |w|)) = |z| ((|z| \cdot |w|) |w|) \\
    = {} & |z| ((|w| \cdot |z|) |w|) = |z| (|w| (|z| \cdot |w|)) \\
    = {} & (|z| \cdot |w|) (|z| \cdot |w|) = (|z| \cdot |w|)^2.
\end{aligned}
$$

3\. Note that
$$
|\Re z| = \sqrt{(\Re z)^2} \leq \sqrt{(\Re z)^2 + (\Im z)^2} = |z|.
$$
The same skill also applies to the other inequality.

4\. Note that
$$
\begin{aligned}
    & (|z| + |w|)^2 - |z + w|^2 \\
    = {} & (|z|^2 + 2 |z| |w| + |w|^2) - (\overline{z} + \overline{w}) (z + w) \\
    = {} & (\overline{z}z + \overline{w}w + 2|\overline{z} w|)
    - (\overline{z}z + \overline{w} w + (\overline{z} w + z \overline{w}))             \\
    = {} & 2(|\overline{z} w| - \Re {(\overline{z} w)})          \\
    \geq {} & 2(|\overline{z} w| - |\Re {(\overline{z} w)}|)        \\
    \geq {} & 0.
\end{aligned}
$$

5\. Note that
$$
|z| - |w| = |z| - (|w| + |z - w|) + |z - w| \leq |z - w|
$$
and that $|w| - |z| \leq |w - z| = |z - w|$.

6\. Note that
$$
|z| = |\Re z + \mathrm{i} \Im z| \leq |\Re z| + |\mathrm{i} \Im z| = |\Re z| + |\mathrm{i}| \cdot |\Im z|.
$$
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37

Well, the introduction is almost over. You have seen the most essential things of complex numbers. Since the polar form of a complex number requires trigonometry, I decided to skip it. This explains why you did not see the term “argument.”

38

I decide to conclude the introduction with a not-so-hard problem.

Suppose that z, w are complex numbers. Show that

|z + w|^2 + |z - w|^2 = 2|z|^2 + 2|w|^2.

Try giving a geometrical interpretation of the equation, if it is possible.

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39

过年真是学习的好时机

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40

首先，絕對值對於偶次幂来說，意義不大。

實部：
$$(z^2+w^2)*2=2z^2+2w^2$$
因此實部可以是任何實數

虛部：
$$(-z^2-w^2)*2=-2z^2-2w^2$$
虛部也可以是任何虛數