📚 Complex Numbers for IGCSE Mathematics | IGCSE数学复数考点精讲
Complex numbers are a fundamental extension of the real number system, introduced to solve equations that have no real solutions, such as x² + 1 = 0. In IGCSE Mathematics, you are expected to understand the basic algebra of complex numbers, represent them on an Argand diagram, and use them to find roots of quadratic equations. This revision guide covers all essential exam points, with clear explanations and examples.
复数是实数系的自然延伸，为解决像 x² + 1 = 0 这样无实数解的方程而引入。在 IGCSE 数学中，你需要掌握复数的基本代数运算、在阿尔冈图上的表示，以及用复数求二次方程的根。本考点精讲梳理了所有重要知识点，配合清晰的解释和例题，帮助你轻松应对考试。
1. Introduction to Complex Numbers | 复数简介
A complex number is of the form z = a + bi, where a and b are real numbers, and i is the imaginary unit satisfying i² = -1. The set of complex numbers extends the real numbers, allowing every polynomial equation to have a solution.
复数通常写作 z = a + bi 的形式，其中 a 和 b 是实数，i 是虚数单位，满足 i² = -1。复数集合是对实数的扩充，使得每一个多项式方程都有解。
z = a + bi, i² = -1
Here, a is called the real part and b is the imaginary part. Note that the imaginary part is the coefficient b, not bi. For example, in 3 – 4i, the real part is 3 and the imaginary part is –4.
这里 a 称为实部，b 称为虚部。注意虚部指的是系数 b，而不是 bi。例如，在 3 – 4i 中，实部是 3，虚部是 –4。
2. The Imaginary Unit i | 虚数单位 i
The symbol i represents √(-1). Its key property is i² = -1. From this, we can compute powers of i using a repeating cycle: i¹ = i, i² = -1, i³ = -i, i⁴ = 1, and then the pattern repeats every four powers.
符号 i 表示 √(-1)。其核心性质是 i² = -1。由此可以计算 i 的各次幂，它们以周期 4 循环：i¹ = i, i² = -1, i³ = -i, i⁴ = 1，之后每四次幂重复一次。
To evaluate iⁿ, divide n by 4 and use the remainder. For instance, i¹¹: 11 ÷ 4 gives remainder 3, so i¹¹ = i³ = -i.
要计算 iⁿ，用 n 除以 4，根据余数得出结果。例如 i¹¹：11 ÷ 4 余 3，所以 i¹¹ = i³ = -i。
i = √(-1); i² = -1; i³ = -i; i⁴ = 1
3. Real and Imaginary Parts | 实部与虚部
For a complex number z = a + bi, we write Re(z) = a and Im(z) = b. Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.
对复数 z = a + bi，记作 Re(z) = a，Im(z) = b。两个复数相等当且仅当它们的实部和虚部分别相等。
This concept is often used to solve for unknowns. Example: If x + yi = 5 – 2i, then x = 5 and y = –2.
这个概念常用于求解未知数。例如：若 x + yi = 5 – 2i，则 x = 5，y = –2。
A pure real number has Im(z) = 0; a pure imaginary number has Re(z) = 0.
纯实数的 Im(z) = 0；纯虚数的 Re(z) = 0。
4. Complex Conjugates | 共轭复数
The complex conjugate of z = a + bi is denoted by z* or z, and is defined as z* = a – bi. Geometrically, it is the reflection of z in the real axis on an Argand diagram.
复数 z = a + bi 的共轭复数记作 z* 或 z，定义为 z* = a – bi。在阿尔冈图上，共轭复数就是 z 关于实轴的反射。
Conjugates are useful in division and in finding the modulus. A key property: z × z* = a² + b², a real number.
共轭复数在除法运算和求模时非常有用。一个重要性质：z × z* = a² + b²，这是一个实数。
Also, the sum z + z* = 2a, which is twice the real part, and the difference z – z* = 2bi, a pure imaginary number.
此外，和 z + z* = 2a，即实部的两倍；差 z – z* = 2bi，为一纯虚数。
5. Modulus (Absolute Value) | 模（绝对值）
The modulus of a complex number z = a + bi is defined as |z| = √(a² + b²). It represents the distance from the origin to the point (a, b) on the Argand diagram.
复数 z = a + bi 的模定义为 |z| = √(a² + b²)。它表示阿尔冈图上从原点到点 (a, b) 的距离。
|z| = √(a² + b²)
Properties include: |z| ≥ 0; |z*| = |z|; |z₁ z₂| = |z₁| · |z₂|; and |z₁ / z₂| = |z₁| / |z₂| when z₂ ≠ 0.
性质包括：|z| ≥ 0；|z*| = |z|；|z₁ z₂| = |z₁| · |z₂|；以及当 z₂ ≠ 0 时，|z₁ / z₂| = |z₁| / |z₂|。
Exam tip: You may be asked to find the modulus of a complex expression. Always simplify first, then apply the formula.
考试提示：你可能需要求一个复数表达式的模。务必先化简，再套用公式。
6. Addition and Subtraction | 加减运算
To add or subtract complex numbers, combine the real parts and combine the imaginary parts separately. If z₁ = a + bi and z₂ = c + di, then z₁ + z₂ = (a + c) + (b + d)i, and z₁ – z₂ = (a – c) + (b – d)i.
复数的加减运算，分别合并实部和虚部。若 z₁ = a + bi，z₂ = c + di，则 z₁ + z₂ = (a + c) + (b + d)i，z₁ – z₂ = (a – c) + (b – d)i。
Example: (3 + 2i) + (4 – 5i) = 7 – 3i. Subtraction works similarly.
例如：(3 + 2i) + (4 – 5i) = 7 – 3i。减法类似。
This operation corresponds to vector addition on the Argand diagram, giving a parallelogram interpretation.
这种运算对应于阿尔冈图上的向量加法，可以用平行四边形法则理解。
7. Multiplication | 乘法运算
Multiply complex numbers as you would binomials, using the distributive property and the fact that i² = -1. For z₁ = a + bi and z₂ = c + di, we have z₁ z₂ = (ac – bd) + (ad + bc)i.
复数的乘法如同二项式展开，利用分配律以及 i² = -1。对于 z₁ = a + bi，z₂ = c + di，有 z₁ z₂ = (ac – bd) + (ad + bc)i。
(a + bi)(c + di) = ac + adi + bci + bdi² = (ac – bd) + (ad + bc)i
You can also multiply by a real number: k(a + bi) = ka + kbi. Multiplication by i rotates a complex number by 90° counter-clockwise on the Argand diagram.
也可以与实数相乘：k(a + bi) = ka + kbi。乘以 i 相当于在阿尔冈图上将复数逆时针旋转 90°。
8. Division and Rationalising the Denominator | 除法与分母有理化
To divide two complex numbers, multiply numerator and denominator by the conjugate of the denominator. This process turns the denominator into a real number.
两个复数相除，可将分子和分母同乘以分母的共轭复数，从而使分母变为实数。
(a + bi) / (c + di) = [(a + bi)(c – di)] / [(c + di)(c – di)] = [(ac + bd) + (bc – ad)i] / (c² + d²)
Example: (1 + 2i) / (3 – 4i) = [(1 + 2i)(3 + 4i)] / (9 + 16) = [(3 – 8) + (4 + 6)i] / 25 = (-5 + 10i)/25 = -1/5 + (2/5)i.
例如：(1 + 2i) / (3 – 4i) = [(1 + 2i)(3 + 4i)] / (9 + 16) = [(3 – 8) + (4 + 6)i] / 25 = (-5 + 10i)/25 = -1/5 + (2/5)i。
Always express the final answer in the standard form a + bi.
最终结果务必写成 a + bi 的标准形式。
9. Solving Quadratic Equations with Complex Roots | 解二次方程得出复数根
When a quadratic equation ax² + bx + c = 0 has a negative discriminant (b² – 4ac < 0), the roots are complex conjugates. They can be found using the quadratic formula, expressing the square root of the negative discriminant in terms of i.
当二次方程 ax² + bx + c = 0 的判别式 (b² – 4ac) 为负时，方程的根是一对共轭复数。可使用求根公式，将负判别式的平方根用 i 表示。
x = [ -b ± √(b² – 4ac) ] / 2a, 其中 √(负数) = i √(正数)
Example: Solve x² + 4x + 13 = 0. Discriminant = 16 – 52 = -36. So x = [ -4 ± √(-36) ] / 2 = [ -4 ± 6i ] / 2 = -2 ± 3i.
例如：解方程 x² + 4x + 13 = 0。判别式 = 16 – 52 = -36。所以 x = [ -4 ± √(-36) ] / 2 = [ -4 ± 6i ] / 2 = -2 ± 3i。
The two roots are -2 + 3i and -2 – 3i, a conjugate pair.
两个根为 -2 + 3i 和 -2 – 3i，是一对共轭复数。
10. The Argand Diagram | 阿尔冈图
The Argand diagram represents complex numbers as points or vectors on a plane. The horizontal axis is the real axis, and the vertical axis is the imaginary axis. The complex number z = a + bi is plotted at the point (a, b).
阿尔冈图用平面上的点或向量表示复数。横轴为实轴，纵轴为虚轴。复数 z = a + bi 对应点 (a, b)。
The modulus |z| is the distance from the origin to the point; the argument arg(z) is the angle with the positive real axis (usually not assessed in core IGCSE, but useful to know).
模 |z| 是原点到点的距离；辐角 arg(z) 是与正实轴的夹角（核心 IGCSE 通常不考查，但了解有助于理解）。
Addition and subtraction correspond to vector addition. Multiplication by i rotates 90° counter-clockwise.
加减法对应向量加法，乘以 i 相当于逆时针旋转 90°。
Exam questions often ask you to plot given complex numbers or their conjugates and to interpret geometric relationships.
考题常要求标出给定的复数或其共轭，并解释几何关系。
11. Equality of Complex Numbers | 复数相等
Two complex numbers a + bi and c + di are equal exactly when a = c and b = d. This simple property is powerful for solving equations involving complex numbers.
两个复数 a + bi 和 c + di 相等当且仅当 a = c 且 b = d。这个简单的性质是求解含复数方程的有力工具。
Example: Find real numbers x and y such that 2x + 3yi = 8 – 12i. Equating real parts: 2x = 8 → x = 4. Equating imaginary parts: 3y = -12 → y = -4.
例如：求实数 x 和 y 使 2x + 3yi = 8 – 12i。实部相等：2x = 8 → x = 4。虚部相等：3y = -12 → y = -4。
This technique also helps when dealing with complex conjugates and identities.
这一技巧在处理共轭复数以及恒等式时也非常有用。
12. Summary and Exam Tips | 总结与考试技巧
IGCSE complex numbers questions test your fluency with definitions, algebraic manipulation, and visual representation. Always write answers in the form a + bi, simplify square roots of negatives using i, and use conjugates for division. Sketch Argand diagrams accurately, labelling axes and points.
IGCSE 复数题目考查你对定义的熟练程度、代数运算能力以及图形表示。始终将答案写成 a + bi 的形式，用 i 化简负平方根，利用共轭进行除法。准确绘制阿尔冈图，标出坐标轴和关键点。
Common mistakes include forgetting i² = -1 during multiplication, misidentifying the imaginary part, and errors in rationalising denominators. Practise a variety of problems, including solving quadratics with complex roots and proving simple identities.
常见错误包括乘法时忘记 i² = -1，误判虚部，以及分母有理化出错。需做多样化的练习，包括解二次方程得复数根以及证明简单的恒等式。
Remember: The key is to treat i as an algebraic symbol obeying i² = -1. Once you master that, complex numbers become straightforward.
记住，关键在于把 i 看成一个满足 i² = -1 的代数符号。一旦掌握这点，复数就会变得简单直接。
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