Introduction to Complex Numbers | 复数入门

📚 Introduction to Complex Numbers | 复数入门

tal in A-Level Mathematics, particularly in solving quadratic equations, trigonometry, and calculus.

复数将实数系统扩展，以包含像 x² = -1 这类方程的解。复数在A-Level数学中至关重要，特别是在解二次方程、三角函数和微积分中。

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1. The Imaginary Unit i | 虚数单位 i

1. This means i is a number whose square is negative, something impossible for real numbers.

虚数单位的定义为 i² = -1。这意味着 i 是一个平方为负数的数，而这对于实数来说是不可能的。

From this definition, the powers of i follow a repeating pattern:

根据这一定义，i 的幂呈现循环模式：

• i¹ = i
• i² = -1
• i³ = i² × i = -i
• i⁴ = (i²)² = 1

ne of these four values. For example, i²⁰¹⁹ = i³ = -i because 2019 ÷ 4 leaves remainder 3.

i 的任何整数幂都可以化简为这四个值之一。例如，i²⁰¹⁹ = i³ = -i，因为 2019 ÷ 4 余 3。

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2. Definition of Complex Numbers | 复数的定义

s, and i is the imaginary unit. Here, a is called the real part, written Re(z), and b is called the imaginary part, written Im(z).

复数写作 z = a + bi 的形式，其中 a 和 b 是实数，i 是虚数单位。这里 a 被称为实部，记作 Re(z)；b 被称为虚部，记作 Im(z)。

m(z) = -2. Notice that the imaginary part is the coefficient of i, not the whole term -2i.

例如，若 z = 3 - 2i，则 Re(z) = 3，Im(z) = -2。注意虚部是 i 的系数，而不是整个项 -2i。

The set of complex numbers is denoted by ℂ. All real numbers are also complex numbers because any real number a can be written as a + 0i.

复数集用 ℂ 表示。所有实数也是复数，因为任意实数 a 都可以写成 a + 0i 的形式。

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3. The Complex Plane | 复平面

The complex plane, also called an Argand diagram, is a way to represent complex numbers graphically. The horizontal axis is the real axis, and the vertical axis is the imaginary axis.

复平面，也叫阿尔冈图，是复数的一种图形表示方法。水平轴为实轴，垂直轴为虚轴。

+ 3i is located 2 units right along the real axis and 3 units up along the imaginary axis.

复数 z = a + bi 在平面中对应点 (a, b)。例如，z = 2 + 3i 位于实轴向右2个单位、虚轴向上3个单位的位置。

This geometric view is extremely useful for understanding the modulus and argument of a complex number, which we discuss later.

这种几何视角对于理解复数的模和辐角极其有用，我们将在后文中讨论。

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4. Equality of Complex Numbers | 复数的相等

their imaginary parts are equal. That is, a + bi = c + di if and only if a = c and b = d.

两个复数相等，当且仅当它们的实部相等且虚部相等。即 a + bi = c + di 当且仅当 a = c 且 b = d。

This property allows us to equate the real and imaginary parts separately when solving equations.

这一性质允许我们在解方程时分别令实部和虚部相等。

ple, if (x + 2) + (y - 1)i = 3 + 4i, then x + 2 = 3 and y - 1 = 4, giving x = 1 and y = 5.

例如，若 (x + 2) + (y - 1)i = 3 + 4i，则 x + 2 = 3 且 y - 1 = 4，解得 x = 1，y = 5。

This simple rule is a cornerstone for solving complex equations in A-Level algebra.

这个简单规则是A-Level代数中解复数方程的核心。

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5. Addition and Subtraction of Complex Numbers | 复数的加法与减法

To add or subtract complex numbers, we simply add or subtract the corresponding real parts and imaginary parts separately.

复数的加法或减法，只需分别对实部和虚部进行相加或相减。

If z₁ = a + bi and z₂ = c + di, then:

若 z₁ = a + bi，z₂ = c + di，则：

z₁ + z₂ = (a + c) + (b + d)i

z₁ - z₂ = (a - c) + (b - d)i

For example, (3 + 2i) + (1 - 5i) = 4 - 3i, and (3 + 2i) - (1 - 5i) = 2 + 7i.

例如，(3 + 2i) + (1 - 5i) = 4 - 3i，(3 + 2i) - (1 - 5i) = 2 + 7i。

Geometrically, addition corresponds to the parallelogram law for vectors in the complex plane.

在几何上，加法对应复平面中向量的平行四边形法则。

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6. Multiplication of Complex Numbers | 复数的乘法

and the product as we would with two linear expressions, and then use i² = -1 to simplify.

复数的乘法与展开两个一次式的方式相同，然后利用 i² = -1 进行化简。

If z₁ = a + bi and z₂ = c + di, then:

若 z₁ = a + bi，z₂ = c + di，则：

z₁z₂ = (ac - bd) + (ad + bc)i

This result follows directly:

该结果直接来源于：

(a + bi)(c + di) = ac + adi + bci + bdi² = (ac - bd) + (ad + bc)i

For example, (2 + 3i)(1 - i) = 2 - 2i + 3i - 3i² = 2 + i + 3 = 5 + i.

例如，(2 + 3i)(1 - i) = 2 - 2i + 3i - 3i² = 2 + i + 3 = 5 + i。

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7. Complex Conjugate | 共轭复数

The conjugate of a complex number z = a + bi is denoted by z̄ and is defined as z̄ = a - bi. In other words, we change the sign of the imaginary part.

复数 z = a + bi 的共轭复数记作 z̄，定义为 z̄ = a - bi。换句话说，我们改变虚部的符号。

For example, if z = 3 + 4i, then z̄ = 3 - 4i. Geometrically, the conjugate is a reflection of the point across the real axis.

例如，若 z = 3 + 4i，则 z̄ = 3 - 4i。在几何上，共轭复数是该点关于实轴的反射。

Key properties of conjugates are:

共轭复数的重要性质如下：

• z + z̄ = 2Re(z)
• z - z̄ = 2iIm(z)
• z × z̄ is always a real number
• If z = z̄, then z is real

• z + z̄ = 2Re(z)
• z - z̄ = 2iIm(z)
• z × z̄ 始终为实数
• 若 z = z̄，则 z 为实数

The product z × z̄ plays a central role in division and in finding the modulus.

乘积 z × z̄ 在除法以及求模中起着核心作用。

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8. Division of Complex Numbers | 复数的除法

To divide complex numbers, we multiply the numerator and denominator by the conjugate of the denominator. This transforms the denominator into a real number.

复数的除法需要将分子分母同时乘以分母的共轭复数，从而使分母变为实数。

Example: calculate (3 + 2i) ÷ (1 - i).

例如：计算 (3 + 2i) ÷ (1 - i)。

(3 + 2i) / (1 - i) = [(3 + 2i)(1 + i)] / [(1 - i)(1 + i)]

Simplify the numerator and denominator:

化简分子和分母：

= (3 + 3i + 2i + 2i²) / (1 - i²) = (3 + 5i - 2) / (1 + 1) = (1 + 5i) / 2 = 0.5 + 2.5i

So the result is a complex number in standard form.

因此结果是一个标准形式的复数。

The general formula, for z₂ ≠ 0, is:

一般公式为，当 z₂ ≠ 0 时：

(a + bi) / (c + di) = [(a + bi)(c - di)] / (c² + d²)

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9. Modulus and Argument | 模与辐角

e modulus of z = a + bi is the distance from the origin to the point (a, b). It is denoted |z| and is given by:

在复平面中，复数 z = a + bi 的模是从原点到点 (a, b) 的距离。它记作 |z|，公式为：

|z| = √a² + b²

Notice that |z|² = z × z̄, since (a + bi)(a - bi) = a² + b².

注意 |z|² = z × z̄，因为 (a + bi)(a - bi) = a² + b²。

The argument of z, denoted arg(z), is the angle θ between the positive real axis and the line joining the origin to the point (a, b), measured anticlockwise.

复数 z 的辐角，记作 arg(z)，是从正实轴逆时针旋转到原点和点 (a, b) 连线的夹角 θ。

A common way to compute the argument is:

计算辐角的常用方法是：

tan θ = b / a

However, the quadrant in which the number lies must be considered to determine the correct value of θ.

但是，必须考虑复数所在的象限，才能确定 θ 的正确取值。

For z = 1 + i, |z| = √2 and arg(z) = π/4 (or 45°).

对于 z = 1 + i，|z| = √2，arg(z) = π/4（即45°）。

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10. Solving Quadratic Equations with Complex Roots | 含复数根的二次方程

Quadratic equations with discriminant less than zero have no real solutions, but they always have two complex roots that are conjugates of each other.

判别式小于零的二次方程没有实数解，但恒有两个共轭复数根。

For ax² + bx + c = 0, the quadratic formula gives:

对于 ax² + bx + c = 0，二次求根公式给出：

x = (-b ± √b² - 4ac) / (2a)

minant as a positive number times -1, and use i² = -1. For example, solve x² + 2x + 5 = 0.

当 b² - 4ac < 0 时，我们将负判别式写成正数乘以 -1，并利用 i² = -1。例如，解 x² + 2x + 5 = 0。

x = (-2 ± √4 - 20) / 2 = (-2 ± √-16) / 2 = (-2 ± 4i) / 2 = -1 ± 2i

So the roots are -1 + 2i and -1 - 2i. Notice their sum is -2 and their product is 5, matching the coefficients b and c.

因此根为 -1 + 2i 和 -1 - 2i。注意它们的和为 -2，积为 5，分别与系数 b 和 c 匹配。

For real quadratic equations, non-real roots always occur in conjugate pairs. This fact is extremely useful for reconstructing a quadratic equation from its roots.

对于实系数二次方程，非实数根总是成对共轭出现。这一事实对于从根反推二次方程极其有用。

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