📚 Complex Numbers: Key Concepts | 复变函数考点精讲
A complex number extends the real number system by including the square root of negative one. In IGCSE AQA Further Mathematics, understanding complex numbers opens the door to solving equations that have no real solutions and provides a powerful geometric perspective. This revision guide covers all the essential concepts you need: the imaginary unit, standard form, arithmetic operations, conjugates, the Argand diagram, modulus and argument, and solving quadratic equations with complex roots.
复数通过引入负一的平方根扩展了实数体系。在 IGCSE AQA 进阶数学中，理解复数能帮助你解决没有实数解的方程，并提供了一种强大的几何视角。本复习指南涵盖所有核心考点：虚数单位、标准形式、四则运算、共轭复数、阿尔冈图、模与辐角，以及带复数根的二次方程求解。
1. Introduction to Complex Numbers | 复数简介
A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i is the imaginary unit. Complex numbers allow us to solve equations like x² + 1 = 0, which has no real solution. They are fundamental in advanced algebra, physics and engineering.
复数是可以表示为 a + bi 形式的数，其中 a 和 b 是实数，i 是虚数单位。复数使我们能够求解像 x² + 1 = 0 这样没有实数解的方程。它们在高等代数、物理和工程学中至关重要。
2. The Imaginary Unit i | 虚数单位 i
The symbol i is defined by the property that i² = –1. It is not a real number, but it behaves consistently within the rules of algebra. From this definition, we can find higher powers of i following a cyclic pattern: i¹ = i, i² = –1, i³ = –i, i⁴ = 1, and then the cycle repeats every four powers.
符号 i 的定义是 i² = –1。它不是一个实数，但在代数运算中表现一致。根据这个定义，我们可以得出 i 的高次幂遵循循环规律：i¹ = i，i² = –1，i³ = –i，i⁴ = 1，然后每四个幂次循环一次。
i² = –1
For example, i⁵ = i⁴ · i = 1 · i = i, and i⁶ = i⁴ · i² = 1 · (–1) = –1.
例如，i⁵ = i⁴ · i = 1 · i = i，而 i⁶ = i⁴ · i² = 1 · (–1) = –1。
3. Standard Form and Real/Imaginary Parts | 标准形式及实部与虚部
A complex number is written as z = a + bi, where a is called the real part, denoted Re(z), and b is the imaginary part, denoted Im(z). Both a and b are real numbers. If b = 0, the number is purely real; if a = 0, it is purely imaginary.
复数写作 z = a + bi，其中 a 称为实部，记作 Re(z)；b 称为虚部，记作 Im(z)。a 和 b 都是实数。若 b = 0，该数为纯实数；若 a = 0，则为纯虚数。
For instance, in 3 + 4i, Re = 3 and Im = 4. In the exam, always present your answer in the form a + bi unless directed otherwise.
例如，在 3 + 4i 中，实部为 3，虚部为 4。考试时若有要求，请始终将答案写成 a + bi 形式。
4. Addition and Subtraction | 加法与减法
To add or subtract complex numbers, simply add or subtract the real parts and the imaginary parts separately. For z₁ = a + bi and z₂ = c + di:
复数的加减法：分别对实部和虚部进行加减。设 z₁ = a + bi，z₂ = c + di：
z₁ + z₂ = (a + c) + (b + d)i
z₁ – z₂ = (a – c) + (b – d)i
Example: (2 + 5i) + (3 – i) = (2+3) + (5–1)i = 5 + 4i. Similarly, (2+5i) – (3–i) = (2–3) + (5–(–1))i = –1 + 6i.
示例：(2 + 5i) + (3 – i) = (2+3) + (5–1)i = 5 + 4i。同样，(2+5i) – (3–i) = (2–3) + (5–(–1))i = –1 + 6i。
5. Multiplication | 乘法
Multiplication of complex numbers uses the distributive law and the fact that i² = –1. Multiply as you would binomials: (a + bi)(c + di) = ac + adi + bci + bdi². Then replace i² with –1 and combine real and imaginary terms.
复数乘法使用分配律以及 i² = –1。像二项式相乘那样展开：(a + bi)(c + di) = ac + adi + bci + bdi²，再将 i² 换成 –1，合并实部和虚部。
(a + bi)(c + di) = (ac – bd) + (ad + bc)i
Example: (2 + 3i)(4 – 5i) = 2·4 + 2·(–5i) + 3i·4 + 3i·(–5i) = 8 – 10i + 12i – 15i² = 8 + 2i – 15(–1) = 8 + 2i + 15 = 23 + 2i.
示例：(2 + 3i)(4 – 5i) = 8 – 10i + 12i – 15i² = 8 + 2i + 15 = 23 + 2i。
6. Complex Conjugate | 复共轭
The complex conjugate of z = a + bi is denoted z̅ or z* and is defined as a – bi. Geometrically, it reflects the point across the real axis. The product of a complex number and its conjugate is always a real number: z · z̅ = a² + b². This property is essential for division.
复数 z = a + bi 的共轭复数表示为 z̅ 或 z*，定义为 a – bi。几何上，它是点关于实轴的反射。一个复数与其共轭的乘积恒为实数：z · z̅ = a² + b²。此性质对除法至关重要。
Example: If z = 3 + 4i, then z̅ = 3 – 4i, and z·z̅ = (3)² + (4)² = 9 + 16 = 25.
示例：若 z = 3 + 4i，则 z̅ = 3 – 4i，且 z·z̅ = 3² + 4² = 25。
7. Division | 除法
To divide two complex numbers, multiply numerator and denominator by the conjugate of the denominator. This realises the denominator, allowing the result to be expressed in standard form a + bi.
进行复数除法时，将分子和分母同时乘以分母的共轭。这样可使分母实数化，从而将结果写为标准形式 a + bi。
(a + bi) / (c + di) = [(a + bi)(c – di)] / (c² + d²)
Example: Simplify (1 + 2i) / (3 – i). Multiply top and bottom by 3 + i:
示例：化简 (1 + 2i) / (3 – i)。分子分母同乘 3 + i：
(1+2i)(3+i) = 1·3 + 1·i + 2i·3 + 2i·i = 3 + i + 6i + 2i² = 3 + 7i – 2 = 1 + 7i. Denominator: (3)² + (1)² = 10. So the result is (1/10) + (7/10)i.
分子：(1+2i)(3+i) = 3 + i + 6i + 2i² = 3 + 7i – 2 = 1 + 7i。分母：(3)² + 1² = 10。结果为 (1/10) + (7/10)i。
8. Argand Diagram | 阿尔冈图
The Argand diagram represents complex numbers as points or vectors in a plane, with the x-axis as the real axis and the y-axis as the imaginary axis. A complex number z = a + bi is plotted at the point (a, b). This visualisation helps understand operations like addition (vector parallelogram) and modulus.
阿尔冈图将复数表示为平面上的点或向量，x 轴为实轴，y 轴为虚轴。复数 z = a + bi 绘制在点 (a, b) 处。这种可视化有助于理解加法（向量平行四边形法则）以及模等概念。
| Point | Real Part | Imaginary Part |
| 3 + 4i | 3 | 4 |
| –2 + i | –2 | 1 |
| –1 – 3i | –1 | –3 |
Plotting complex numbers on an Argand diagram is a common exam question. Ensure you label axes clearly.
在阿尔冈图上描点是常见考题。请务必清晰标注坐标轴。
9. Modulus and Argument | 模与辐角
The modulus of z = a + bi, written |z|, is the distance from the origin to the point (a, b). It is calculated as |z| = √(a² + b²). The argument, arg(z), is the angle θ made with the positive real axis, measured in degrees or radians. Usually –π < θ ≤ π (or 0 ≤ θ < 360°).
复数 z = a + bi 的模，记作 |z|，是从原点到点 (a, b) 的距离，计算公式为 |z| = √(a² + b²)。辐角 arg(z) 是复数与正实轴之间的夹角 θ，以度或弧度表示，通常取 –π < θ ≤ π（或 0 ≤ θ < 360°）。
|z| = √(a² + b²), tan θ = b/a
Example: For z = 1 + i, |z| = √(1²+1²) = √2. Argument: tan θ = 1/1 = 1 → θ = 45° (π/4 rad). With negative a or b, care must be taken to determine the correct quadrant.
示例：z = 1 + i 的模为 √2，辐角 tan θ = 1 → θ = 45°。若 a 或 b 为负，需根据象限确定正确的角度。
10. Solving Quadratic Equations | 解二次方程
When a quadratic equation ax² + bx + c = 0 has discriminant b² – 4ac < 0, the roots are complex. They can be written as a conjugate pair using the imaginary unit. The quadratic formula still holds:
当二次方程 ax² + bx + c = 0 的判别式 b² – 4ac < 0 时，根为复数。可使用虚数单位写成共轭对。求根公式仍然成立：
x = [ –b ± √(b² – 4ac) ] / 2a
Replace √(negative number) with i√(positive number). Example: Solve x² + 4x + 13 = 0. Discriminant: 16 – 52 = –36. So x = [–4 ± √(–36)] / 2 = [–4 ± 6i] / 2 = –2 ± 3i. The solutions are –2 + 3i and –2 – 3i.
将 √(负数) 替换为 i√(正数)。示例：求解 x² + 4x + 13 = 0。判别式 = 16 – 52 = –36，故 x = [–4 ± 6i] / 2 = –2 ± 3i。
This topic often appears in papers, and examiners expect answers in the form p ± qi.
此题在试卷中出现频繁，阅卷人期望给出 p ± qi 形式的答案。
11. Key Points Summary | 考点总结
- Complex numbers are expressed as a + bi with i² = –1. / 复数的形式为 a + bi，其中 i² = –1。
- Add/subtract by combining real and imaginary parts separately. / 加减时分别合并实部和虚部。
- Multiply using FOIL and simplify i² = –1. / 乘法利用二项式展开，并化简 i² = –1。
- The conjugate of a + bi is a – bi; product is a² + b². / a + bi 的共轭为 a – bi，乘积为 a² + b²。
- Division uses the conjugate to realise the denominator. / 除法通过共轭使分母有理化。
- Argand diagram plots Re on x-axis, Im on y-axis. / 阿尔冈图以 x 轴为实轴，y 轴为虚轴。
- Modulus = distance from origin; argument = angle with positive real axis. / 模 = 到原点的距离；辐角 = 与正实轴的夹角。
- For negative discriminant, quadratic roots are complex conjugates. / 判别式为负时，二次方程的根为共轭复数。
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