Complex Numbers | AS进阶数学：复数

📚 Complex Numbers | AS进阶数学：复数

Complex numbers extend our number system beyond the real line, unlocking solutions to equations that were previously impossible. In the AQA International AS Further Mathematics specification, complex numbers form a cornerstone topic that connects algebra, geometry and trigonometry. This article provides a complete revision guide.

复数将数系从实数轴扩展到更广阔的平面，为原本无解的方程打开了新的可能。在 AQA 国际 AS 进阶数学考纲中，复数是连接代数、几何与三角学的基石性课题。本文为你提供一份完整的复习指南。

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1. What Are Complex Numbers? | 什么是复数？

The imaginary unit is defined as i = √(−1), so i² = −1. A complex number is written in the form z = a + bi, where a and b are real numbers. Here, a is called the real part, denoted Re(z), and b is the imaginary part, denoted Im(z).

虚数单位定义为 i = √(−1)，因此 i² = −1。复数写作 z = a + bi 的形式，其中 a 和 b 是实数。a 称为实部，记作 Re(z)；b 称为虚部，记作 Im(z)。

When b = 0, the complex number is purely real; when a = 0, it is purely imaginary. Every real number is therefore also a complex number, and the real numbers form a subset of the complex numbers. Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.

当 b = 0 时，复数为纯实数；当 a = 0 时，复数为纯虚数。因此每个实数都是复数，实数集是复数集的子集。两个复数相等，当且仅当它们的实部相等且虚部相等。

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2. Basic Arithmetic of Complex Numbers | 复数的基本运算

Two complex numbers can be added or subtracted by combining their real and imaginary parts separately. For example, (2 + 3i) + (4 − 5i) = 6 − 2i.

两个复数相加或相减时，将实部与虚部分别合并。例如，(2 + 3i) + (4 − 5i) = 6 − 2i。

Multiplication follows the distributive law, remembering that i² = −1. For instance, (2 + 3i)(4 − 5i) = 8 − 10i + 12i − 15i² = 8 + 2i + 15 = 23 + 2i.

乘法遵循分配律，并记住 i² = −1。例如，(2 + 3i)(4 − 5i) = 8 − 10i + 12i − 15i² = 8 + 2i + 15 = 23 + 2i。

(a + bi)(c + di) = (ac − bd) + (ad + bc)i

Note that the real part of the product is ac − bd because the two imaginary cross-terms are multiplied by i² = −1. Practise this pattern until it becomes automatic, as it appears in nearly every complex number question.

注意积的实部是 ac − bd，因为两个含 i 的交叉项需要乘以 i² = −1。反复练习这一模式直至熟练，因为它在几乎所有复数题中都会出现。

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3. Complex Conjugates | 共轭复数

For a complex number z = a + bi, its complex conjugate is z̄ = a − bi. The conjugate is obtained by changing the sign of the imaginary part only. Geometrically, z̄ is the reflection of z across the real axis.

对于复数 z = a + bi，其共轭复数为 z̄ = a − bi。共轭只需改变虚部的符号。从几何上看，z̄ 是 z 关于实轴的对称点。

The product of a complex number and its conjugate is always a non-negative real number:

复数与其共轭的乘积始终是一个非负实数：

(a + bi)(a − bi) = a² + b²

This property is the key to dividing complex numbers and to finding the modulus, as we will see below. It also explains why every quadratic with real coefficients and negative discriminant has a pair of conjugate complex roots.

这一性质是复数除法与求模的关键，下文会详细说明。它也解释了为什么判别式为负的实系数二次方程必有一对共轭复根。

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4. Division of Complex Numbers | 复数的除法

To divide complex numbers, multiply both numerator and denominator by the conjugate of the denominator. This rationalises the denominator, turning it into a real number.

两个复数相除时，将分子分母同时乘以分母的共轭复数。这样可以使分母’有理化’，化为实数。

z/w = z·w̄ / (w·w̄)

For example, divide (1 + 2i) by (3 − 4i). Multiply the numerator and denominator by 3 + 4i:

例如，计算 (1 + 2i) ÷ (3 − 4i)。分子分母同乘 3 + 4i：

(1 + 2i)(3 + 4i) / ((3 − 4i)(3 + 4i)) = (−5 + 10i) / 25 = −0.2 + 0.4i

Remember that the denominator becomes 3² + 4² = 25. Verify your final answer by multiplying it back by the original denominator; you should recover the original numerator.

记住分母变为 3² + 4² = 25。将结果乘回原分母即可验证答案是否正确，应当能还原原来的分子。

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5. Argand Diagrams | 阿尔冈图

An Argand diagram is a plane with a real axis (horizontal) and an imaginary axis (vertical). The complex number z = a + bi is represented as the point (a, b).

阿尔冈图是以实轴（水平）和虚轴（垂直）构成的平面。复数 z = a + bi 表示为点 (a, b)。

Adding two complex numbers corresponds to vector addition on the Argand diagram. For example, adding 3 + i to 1 + 2i gives the point 4 + 3i, which is the fourth vertex of the parallelogram formed by the origin and the two addends.

在阿尔冈图上，复数加法对应于向量加法。例如，将 3 + i 与 1 + 2i 相加得到点 4 + 3i，该点是原点与两个加数构成的平行四边形的第四个顶点。

Multiplication by i rotates a point 90° anticlockwise about the origin, since i has modulus 1 and argument 90°. This geometric interpretation is invaluable for understanding rotations and symmetry problems.

乘以 i 等价于将点绕原点逆时针旋转 90°，因为 i 的模为 1、辐角为 90°。这一几何解释对理解旋转与对称问题极为重要。

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6. Modulus and Argument | 模与辐角

The modulus of z = a + bi is the distance from the origin to the point (a, b):

复数 z = a + bi 的模是原点到点 (a, b) 的距离：

|z| = √(a² + b²)

The argument, arg(z) = θ, is the angle that the line from the origin to (a, b) makes with the positive real axis, measured anticlockwise in radians. It satisfies:

辐角 arg(z) = θ 是原点到 (a, b) 的连线与正实轴之间的夹角，按逆时针方向以弧度计量，满足：

tan θ = b/a, with −π < θ ≤ π

When finding the argument, always consider the quadrant in which the point lies. A calculator usually gives the principal value, but you must adjust the angle for points in the second and third quadrants by adding or subtracting π.

求辐角时，必须注意点所在的象限。计算器通常给出主值，但对于第二、

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