Complex Numbers Revision for GCSE CCEA Mathematics | GCSE CCEA 数学：复数考点精讲

📚 Complex Numbers Revision for GCSE CCEA Mathematics | GCSE CCEA 数学：复数考点精讲

Welcome to this focused revision guide on complex numbers for CCEA GCSE Mathematics. You will learn what complex numbers are, how to perform arithmetic with them, how to find the conjugate and modulus, and how to visualise them on an Argand diagram. Each section pairs English explanations with Chinese translations to support bilingual learners and ensures you grasp the essential concepts tested in the CCEA specification.

欢迎学习CCEA GCSE数学复数专项复习指南。你将了解什么是复数，如何进行复数运算，如何求共轭与模，以及如何在阿根图上直观表示复数。每个小节都配有中英文对照讲解，帮助双语学习者牢固掌握CCEA考纲的核心概念。

1. The Imaginary Unit i | 虚数单位 i

The imaginary unit i is defined as the square root of –1. This means i² = –1. No real number squared gives a negative result, so i is the foundation that allows us to extend the number system beyond the real line.

虚数单位 i 定义为 –1 的平方根，即 i² = –1。任何实数的平方都不可能为负，因此 i 是将数系扩展到实数线之外的基础。

For example, √(–9) can be written as √9 × √(–1) = 3i. In CCEA questions, you will often need to simplify square roots of negative numbers using i before carrying out further operations.

例如，√(–9) 可以写成 √9 × √(–1) = 3i。在 CCEA 的考题中，你经常需要先将负数的平方根用 i 化简，再进行后续运算。

2. Definition of a Complex Number | 复数的定义

A complex number z can be written in the form a + bi, where a and b are real numbers. a is called the real part, and b is called the imaginary part. Both parts are ordinary real numbers; b multiplies the imaginary unit i.

复数 z 可以写成 a + bi 的形式，其中 a 和 b 均为实数。a 称为实部，b 称为虚部。两个部分都是普通的实数，虚部 b 是乘以虚数单位 i 的系数。

For instance, z = 4 – 5i has real part 4 and imaginary part –5. If b = 0, the complex number reduces to a real number; if a = 0, it becomes a purely imaginary number.

例如，z = 4 – 5i 的实部为 4，虚部为 –5。当 b = 0 时，复数退化为实数；当 a = 0 时，则为纯虚数。

3. Complex Conjugate | 共轭复数

The complex conjugate of z = a + bi is denoted by z* (or sometimes z) and is defined as z* = a – bi. It simply changes the sign of the imaginary part while keeping the real part unchanged.

复数 z = a + bi 的共轭复数记作 z*（有时也记作 z），定义为 z* = a – bi。它只改变虚部的符号，实部保持不变。

The conjugate is essential for division of complex numbers and for finding the modulus. Also, note that the product of a complex number and its conjugate is always a real number: z × z* = a² + b².

共轭复数在复数除法和求模时至关重要。另外，注意一个复数与其共轭的乘积恒为实数：z × z* = a² + b²。

4. Addition and Subtraction | 加法与减法

To add or subtract complex numbers, simply combine the real parts together and the imaginary parts together. Treat i as a variable but remember i² = –1 only when simplifying products.

复数的加减法只需将实部与实部相加减，虚部与虚部相加减。运算时可将 i 视为变量，但需记住只有当化简乘积时才会用到 i² = –1。

Example: (3 + 2i) + (1 – 5i) = (3 + 1) + (2 – 5)i = 4 – 3i. Subtraction is similar: (6 + 4i) – (2 – 3i) = (6 – 2) + (4 – (–3))i = 4 + 7i.

例如：(3 + 2i) + (1 – 5i) = (3 + 1) + (2 – 5)i = 4 – 3i。减法同理：(6 + 4i) – (2 – 3i) = (6 – 2) + (4 – (–3))i = 4 + 7i。

5. Multiplication of Complex Numbers | 复数的乘法

Multiply complex numbers exactly as you would expand brackets in algebra. Use the fact that i² = –1 to simplify the result. Every term in the first bracket must multiply every term in the second bracket.

复数乘法与代数中的括号展开完全相同。利用 i² = –1 化简最终结果。第一个括号中的每一项都要与第二个括号中的每一项相乘。

For example, (2 + 3i)(1 – 4i) = 2(1) + 2(–4i) + 3i(1) + 3i(–4i) = 2 – 8i + 3i – 12i². Since i² = –1, –12i² = –12(–1) = 12, giving 2 + 12 – 5i = 14 – 5i.

例如，(2 + 3i)(1 – 4i) = 2(1) + 2(–4i) + 3i(1) + 3i(–4i) = 2 – 8i + 3i – 12i²。由于 i² = –1，–12i² = –12(–1) = 12，得到 2 + 12 – 5i = 14 – 5i。

6. Division of Complex Numbers | 复数的除法

To divide one complex number by another, multiply both the numerator and the denominator by the conjugate of the denominator. This turns the denominator into a real number, making the division straightforward.

进行复数除法时，将分子和分母同乘以分母的共轭复数。这样可将分母化为实数，使除法变得简单明了。

Example: (3 + 2i) ÷ (1 – i). Multiply top and bottom by (1 + i): [(3 + 2i)(1 + i)] / [(1 – i)(1 + i)]. Denominator becomes 1² + 1² = 2. Numerator: 3 + 3i + 2i + 2i² = 3 + 5i – 2 = 1 + 5i. Result: ½ + (5/2)i.

例如：(3 + 2i) ÷ (1 – i)。上下同乘 (1 + i)：[(3 + 2i)(1 + i)] / [(1 – i)(1 + i)]。分母变为 1² + 1² = 2。分子：3 + 3i + 2i + 2i² = 3 + 5i – 2 = 1 + 5i。结果为 ½ + (5/2)i。

7. Modulus of a Complex Number | 复数的模

The modulus of z = a + bi, written as |z|, is the distance from the origin to the point (a, b) on the complex plane. It is calculated as |z| = √(a² + b²).

复数 z = a + bi 的模记作 |z|，表示复平面上点 (a, b) 到原点的距离。计算公式为 |z| = √(a² + b²)。

For z = 3 – 4i, the modulus is √(3² + (–4)²) = √(9 + 16) = √25 = 5. The modulus is always a non‑negative real number and equals √(z × z*).

对于 z = 3 – 4i，模为 √(3² + (–4)²) = √(9 + 16) = √25 = 5。模总是一个非负实数，且等于 √(z × z*)。

8. Argand Diagram | 阿根图

An Argand diagram is a coordinate plane used to represent complex numbers. The horizontal axis (x‑axis) represents the real part, and the vertical axis (y‑axis) represents the imaginary part.

阿根图是用来表示复数的坐标系。横轴（x 轴）代表实部，纵轴（y 轴）代表虚部。

The complex number a + bi is plotted as the point (a, b). For example, 4 + 3i is located at (4, 3). This visual approach helps you understand operations like addition (vector addition) and the geometric meaning of the modulus.

复数 a + bi 在图上对应点 (a, b)。例如，4 + 3i 位于 (4, 3)。这种可视化方法有助于理解复数的加法（向量加法）以及模的几何意义。

9. Quadratic Equations with Complex Roots | 具有复数根的二次方程

When solving quadratic equations ax² + bx + c = 0, if the discriminant Δ = b² – 4ac is negative, the roots are complex and always occur in conjugate pairs. The quadratic formula still works: x = [–b ± √(b² – 4ac)] / (2a).

解二次方程 ax² + bx + c = 0 时，如果判别式 Δ = b² – 4ac 为负，则根为复数且总是成对共轭出现。二次公式仍然有效：x = [–b ± √(b² – 4ac)] / (2a)。

Example: x² + 4x + 13 = 0. Here a = 1, b = 4, c = 13, Δ = 16 – 52 = –36. Then √(–36) = 6i, so roots are x = (–4 ± 6i)/2 = –2 ± 3i. Conjugate pair –2 + 3i and –2 – 3i.

例如：x² + 4x + 13 = 0。这里 a = 1, b = 4, c = 13，Δ = 16 – 52 = –36。于是 √(–36) = 6i，根为 x = (–4 ± 6i)/2 = –2 ± 3i，即共轭对 –2 + 3i 和 –2 – 3i。

10. Key Skills Summary | 核心技能总结

Make sure you can confidently simplify square roots of negative numbers, add, subtract, multiply and divide complex numbers, find the complex conjugate, calculate the modulus, and plot points on an Argand diagram. Recognising conjugate pairs when solving quadratics with negative discriminant is also crucial for CCEA GCSE questions.

确保你能够熟练化简负数的平方根，进行复数的加、减、乘、除，求共轭复数，计算模，并在阿根图上描点。当解二次方程遇到负判别式时，能识别共轭对同样是CCEA GCSE试题的关键。

Practice with past paper questions, and remember that all the arithmetic rules extend naturally from real numbers once you treat i² as –1. The conjugate trick for division and the Pythagoras‑style modulus formula are the two most frequently tested numerical skills.

多用真题练习，并记住：只要把 i² 视作 –1，所有运算规则都可以从实数自然推广。除法的共轭技巧和毕达哥拉斯式的模长公式是两项考查最频繁的计算技能。

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Complex Numbers Revision for GCSE CCEA Mathematics | GCSE CCEA 数学：复数 考点精讲