📚 Complex Numbers: OCR IGCSE Maths Exam Tips | IGCSE OCR 数学：复数 考点精讲
Complex numbers often appear in the OCR IGCSE Additional Mathematics syllabus (or Level 3 FSMQ), extending the number system beyond the reals. Although they are not part of the core IGCSE Mathematics curriculum, mastering complex numbers is essential for solving equations with negative discriminants and for building a foundation for A‑level topics. This guide walks you through every key skill you need, from basic operations to Argand diagrams, with paired bilingual explanations to strengthen your understanding.
复数通常出现在 OCR IGCSE 附加数学大纲（或 Level 3 FSMQ）中，它将数系扩展到实数之外。虽然复数不属于 IGCSE 核心数学的考查范围，但掌握复数对解决判别式为负的方程以及为 A‑level 内容打下基础至关重要。本指南将带你逐一攻克所有关键技能，从基本运算到阿尔冈图，并配有中英双语对照讲解，加深你的理解。
1. Introduction to Imaginary Numbers | 虚数入门
The number i is defined as the square root of –1, so i² = –1. This seemingly simple definition unlocks a whole new set of numbers. Any real multiple of i, such as 3i or –√2 i, is called a purely imaginary number. When solving quadratic equations, imaginary numbers appear the moment the discriminant becomes negative, allowing us to express consistent solutions that the real number system cannot provide.
数 i 被定义为 –1 的平方根，因此 i² = –1。这个看似简单的定义开启了一类全新的数。任何实数与 i 的乘积，例如 3i 或 –√2 i，都称为纯虚数。在解二次方程时，一旦判别式变为负数，虚数就会出现，使我们能够表达实数系无法提供的自洽解。
2. The Standard Form a+bi | 标准形式 a+bi
A complex number is written as z = a + bi, where a and b are real numbers. The part a is called the real part, Re(z), and b is called the imaginary part, Im(z). Do not confuse the imaginary part with the whole term b i ; Im(z) is just the coefficient b. Expressing numbers in the form a+bi is crucial because it makes addition, subtraction, and multiplication systematic and aligns with the structure required for exam answers.
一个复数写作 z = a + bi，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)。不要将虚部与整个 b i 项混淆，Im(z) 仅仅是系数 b。将数写成 a+bi 的形式至关重要，因为这能让我们系统地进行加减乘运算，且符合考试答案所要求的结构。
3. Adding and Subtracting Complex Numbers | 复数的加减法
To add complex numbers, simply combine their real parts and their imaginary parts separately: (a+bi) + (c+di) = (a+c) + (b+d)i. Subtraction works in the same way: (a+bi) – (c+di) = (a–c) + (b–d)i. Treat the i as a placeholder that follows algebraic rules. This is one of the easiest topics, but watch out for negative signs when subtracting brackets.
复数相加时，只需分别合并实部和虚部: (a+bi) + (c+di) = (a+c) + (b+d)i。减法同理: (a+bi) – (c+di) = (a–c) + (b–d)i。把 i 当作一个遵循代数规则的占位符即可。这是最简单的知识点之一，但在去括号做减法时要特别注意负号。
4. Multiplying Complex Numbers | 复数的乘法
Multiplication uses the distributive law and the fact that i² = –1. For (a+bi)(c+di), expand to get ac + ad i + bc i + bd i² = (ac – bd) + (ad + bc)i. The process is identical to multiplying two binomials. A common mistake is forgetting to replace i² with –1, so always simplify i² terms.
复数乘法需使用分配律和 i² = –1 这一事实。对于 (a+bi)(c+di)，展开得到 ac + ad i + bc i + bd i² = (ac – bd) + (ad + bc)i。整个过程与两个二项式相乘相同。常见的错误是忘记将 i² 替换为 –1，因此务必化简所有 i² 项。
(a+bi)(c+di) = (ac – bd) + (ad + bc)i
5. Complex Conjugates | 共轭复数
The complex conjugate of z = a+bi is z̄ = a – bi. Conjugates are useful because their product is always a real number: z z̄ = a² + b². You will use conjugates when dividing complex numbers and when finding real denominators. Also, if a polynomial has real coefficients, any complex roots occur in conjugate pairs, which is a popular exam deduction.
复数 z = a+bi 的共轭复数是 z̄ = a – bi。共轭复数非常有用，因为它们的乘积总是实数: z z̄ = a² + b²。当你进行复数除法或需要得到实数分母时，就会用到共轭。此外，如果一个多项式具有实系数，那么所有复数根会成对出现为共轭根，这是考试中常见的推论。
6. Dividing Complex Numbers | 复数的除法
To divide (a+bi) by (c+di), multiply the numerator and denominator by the conjugate of the denominator, (c–di). The denominator becomes c² + d², giving a real number. The result is [(ac+bd) + (bc–ad)i] / (c² + d²). Always write the final answer in the form a+bi, separating the real and imaginary parts. Practise this until you can avoid sign slip‑ups.
计算 (a+bi) ÷ (c+di) 时，需对分子和分母同乘分母的共轭 (c–di)。分母变为 c² + d²，成为一个实数。结果为 [(ac+bd) + (bc–ad)i] / (c² + d²)。最终答案务必写成 a+bi 的形式，将实部和虚部分开。要多加练习，直到练成不会出现符号失误。
(a+bi)/(c+di) = [(ac+bd) + (bc–ad)i] / (c²+d²)
| Operation | Formula |
|---|---|
| Addition | (a+bi)+(c+di) = (a+c)+(b+d)i |
| Multiplication | (a+bi)(c+di) = (ac–bd)+(ad+bc)i |
| Conjugate | If z = a+bi, then z̄ = a–bi |
| Division | (a+bi)/(c+di) = ( (ac+bd)+(bc–ad)i )/(c²+d²) |
7. Equality of Complex Numbers | 复数的相等
Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. That is, a+bi = c+di ⇔ a = c and b = d. This property is used to solve equations where the complex number is unknown or to compare coefficients. For instance, if (x+y) + (x–y)i = 5+3i, then we can set up two real equations to find x and y.
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 ⇔ a = c 且 b = d。这一性质常用来求解含未知复数的方程，或进行系数比较。例如，若 (x+y) + (x–y)i = 5+3i，则可以建立两个实数方程来求出 x 和 y。
8. Solving Equations with Complex Numbers | 复数方程的求解
Combine the equality rule with algebraic manipulation. Often you will be given an equation like (2+i)z = 5–i and asked to find z. Solve by dividing both sides by (2+i), which means writing z = (5–i)/(2+i) and then rationalising the denominator. Always express the final answer as a+bi. When two expressions are set equal, equate real and imaginary components to form simultaneous equations if needed.
将相等规则与代数运算结合起来。考题常给出类似 (2+i)z = 5–i 的式子，让你求 z。解法是两边同时除以 (2+i)，即写成 z = (5–i)/(2+i)，再对分母进行有理化。最终答案一定要写成 a+bi 的形式。当两个表达式相等时，如有必要，可以令实部和虚部分别相等，从而建立联立方程。
9. Square Roots and Negative Discriminants | 平方根与负判别式
The square root of a negative number is written using i. For example, √(–9) = 3i. This skill is essential when the quadratic formula gives a negative discriminant. Instead of stopping and writing “no real roots”, you can write the roots as complex numbers: x = [–b ± i√(4ac–b²)]/(2a). Remember to take the negative out of the square root as i.
负数的平方根借助 i 来表达。例如，√(–9) = 3i。当二次公式的判别式为负时，这一技能必备。此时不应停下写出“无实数根”，而可以把根表示为复数: x = [–b ± i√(4ac–b²)]/(2a)。记住将负号从根号内提出来变成 i。
If Δ = b²–4ac < 0, then x = [–b ± i√(–Δ)] / (2a)
10. Quadratic Equations with Complex Roots | 具有复数根的二次方程
When a quadratic equation has real coefficients, complex roots always appear in conjugate pairs. For example, if 2+3i is a root, then 2–3i is automatically the other root. You can reconstruct the quadratic using the sum and product of roots: x² – (sum) x + (product) = 0. This is a clever exam question type—given one complex root, you can find the whole equation or determine missing coefficients.
当二次方程具有实系数时，复数根总是成共轭对出现。例如，若 2+3i 是一个根，那么 2–3i 自动就是另一个根。你可以利用根的和与积重建该二次方程: x² – (两根之和) x + (两根之积) = 0。这是一种常见的巧妙考试题型——已知一个复数根，你可以求出整个方程或确定缺失的系数。
11. Modulus and Argument | 模与辐角
The modulus |z| of a complex number z = a+bi is its distance from the origin, |z| = √(a²+b²). The argument arg(z) is the angle θ (in radians or degrees) measured from the positive real axis to the line joining the origin to the point (a,b). Modulus and argument give a polar view of complex numbers. OCR questions often ask for the modulus of a rational expression, so remember |z₁/z₂| = |z₁|/|z₂|.
复数 z = a+bi 的模 |z| 是它到原点的距离，|z| = √(a²+b²)。辐角 arg(z) 是从正实轴到连接原点与点 (a,b) 的线段的有向角度 θ(弧度或度)。模与辐角给出了复数的极坐标视角。OCR 考题常要求计算某个有理表达式的模，因此要记住 |z₁/z₂| = |z₁|/|z₂|。
12. Argand Diagrams | 阿尔冈图（复平面）
An Argand diagram represents complex numbers as points or vectors on a plane, with the x‑axis as the real part and the y‑axis as the imaginary part. Plotting numbers, adding them geometrically (parallelogram rule), and visualising the modulus are all tested. When given loci such as |z – (a+bi)| = r, you are describing a circle; such geometric interpretation questions are highly examinable.
阿尔冈图将复数表示为平面上的点或向量，x 轴代表实部，y 轴代表虚部。绘制复数、利用平行四边形法则进行几何加法以及直观理解模等都可能出现在试题中。当题目给出诸如 |z – (a+bi)| = r 的轨迹方程时，描述的其实是一个圆；类似这样的几何解释问题极易被考到。
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