GCSE OCR Maths: Complex Numbers Key Points | GCSE OCR 数学：复数考点精讲

📚 GCSE OCR Maths: Complex Numbers Key Points | GCSE OCR 数学：复数考点精讲

Although complex numbers do not feature in the official OCR GCSE Mathematics specification, many ambitious students encounter them as a natural stepping stone towards A-Level and Further Maths. This revision guide breaks down the fundamentals of complex numbers in a clear, GCSE-friendly way – perfect for curious learners who want to stretch their understanding beyond the syllabus. We will explore what complex numbers are, how to perform basic operations, and how they connect to quadratic equations and geometry.

尽管复数并不在 OCR GCSE 数学的正式考试大纲中，但许多学有余力的学生已开始接触复数，以此作为衔接 A-Level 和进阶数学的桥梁。本考点精讲以 GCSE 水平能理解的方式，拆解复数的基本概念、运算法则，以及它们与二次方程和几何的联系，帮助渴望拓宽视野的学习者提前掌握这一重要内容。

1. What are Complex Numbers? | 复数是什么

A complex number is a number that can be written in the form a + bi, where a and b are real numbers, and i is the imaginary unit. The set of complex numbers extends the real numbers by including solutions to equations like x² = -1, which have no real solutions.

复数是可以写成 a + bi 形式的数，其中 a 和 b 是实数，而 i 是虚数单位。复数集是对实数的扩充，它使得像 x² = -1 这样在实数范围内无解的方程拥有了根。

z = a + bi (a, b ∈ ℝ)

2. The Imaginary Unit i | 虚数单位 i

The imaginary unit i is defined as the principal square root of -1. This single definition opens the door to all complex arithmetic. The powers of i follow a repeating pattern of period 4, which is very useful for simplifying expressions.

虚数单位 i 定义为 -1 的主平方根。仅仅这个定义就打开了所有复数运算的大门。i 的幂次遵循周期为 4 的循环规律，对于简化表达式十分有用。

i = √(-1) i² = -1 i³ = -i i⁴ = 1

i¹	i
i²	-1
i³	-i
i⁴	1

3. Standard Form and Notation | 标准形式与记法

We usually write a complex number as z = a + bi. The real number a is called the real part, denoted Re(z), and b is the imaginary part, Im(z). If b = 0, z is purely real; if a = 0, z is purely imaginary, like 5i or -√3 i.

我们通常将复数写作 z = a + bi。实数 a 称为实部，记作 Re(z)；b 则是虚部，记作 Im(z)。若 b = 0，z 为纯实数；若 a = 0，z 为纯虚数，例如 5i 或 -√3 i。

Example: For z = 3 + 4i, we have Re(z) = 3, Im(z) = 4.

示例： 对于 z = 3 + 4i，实部为 3，虚部为 4。

4. Adding and Subtracting Complex Numbers | 复数加减法

Addition and subtraction are performed separately on the real and imaginary parts. They mirror the way we add like terms in algebra. This makes these operations very intuitive for GCSE students comfortable with collecting like terms.

复数的加法和减法分别对实部和虚部进行，方法类似于代数中的合并同类项。这对于已经熟悉合并同类项的 GCSE 学生来说非常直观。

(a + bi) + (c + di) = (a + c) + (b + d)i

(a + bi) – (c + di) = (a – c) + (b – d)i

Example: (2 + 3i) + (5 – i) = (2+5) + (3-1)i = 7 + 2i

示例： (2 + 3i) + (5 – i) = 7 + 2i

5. Multiplying Complex Numbers | 复数乘法

To multiply two complex numbers, expand the brackets as you would with binomials (using FOIL) and remember that i² = -1. Always simplify the final result back to the standard a + bi form. This is often the first step where the interplay between real and imaginary parts becomes visible.

复数相乘时，像展开二项式一样使用括号展开（例如 FOIL 方法），并牢记 i² = -1。最后务必将结果化简回 a + bi 的标准形式。乘法运算通常是第一次清晰展现实部与虚部相互作用的地方。

Example: Multiply (3 + 2i)(1 – 4i)

示例： 计算 (3 + 2i)(1 – 4i)

First expand: 3×1 + 3×(-4i) + 2i×1 + 2i×(-4i) = 3 – 12i + 2i – 8i²

展开：3 – 12i + 2i – 8i²

Since i² = -1, -8i² = -8(-1) = 8. So we get 3 + 8 – 10i = 11 – 10i

由于 i² = -1，-8i² = 8，合并得 3 + 8 – 10i = 11 – 10i

6. The Complex Conjugate | 共轭复数

The complex conjugate of z = a + bi is written as z̄ (or sometimes z*) and is defined as a – bi. Conjugates have the special property that z × z̄ = a² + b², a real number. This is a crucial tool for division and for finding the modulus.

复数 z = a + bi 的共轭记作 z̄（有时也写为 z*），定义为 a – bi。共轭复数的特殊之处在于 z × z̄ = a² + b²，结果是一个实数。它是进行复数除法和求模的关键工具。

Example: Conjugate of 4 + 5i is 4 – 5i. Their product: (4+5i)(4-5i) = 16 – 25i² = 16 + 25 = 41.

示例： 4 + 5i 的共轭为 4 – 5i，乘积为 16 + 25 = 41。

7. Dividing Complex Numbers | 复数除法

To divide two complex numbers, multiply both the numerator and the denominator by the conjugate of the denominator. This technique, called realising the denominator, turns the denominator into a real number and allows the division to be performed just like rationalising surds.

进行复数除法时，将分子和分母同时乘以分母的共轭复数。这一技巧被称为分母实数化，它能使分母变为实数，从而完成除法，过程与有理化根式分母十分相似。

(a + bi) / (c + di) = (a+bi)(c-di) / (c²+d²)

Example: (1 + 2i) / (3 – i) = (1+2i)(3+i) / (9+1) = (3 + i + 6i + 2i²) / 10 = (3 + 7i – 2)/10 = (1 + 7i)/10 = 0.1 + 0.7i.

示例： (1+2i)/(3-i) 乘以共轭得 (1+2i)(3+i)/10 = (1 + 7i)/10 = 0.1 + 0.7i。

8. Modulus (Absolute Value) | 复数的模

The modulus of a complex number z = a + bi, written |z|, is its distance from the origin on the complex plane. It is calculated using Pythagoras’ theorem: |z| = √(a² + b²). The modulus is always a non-negative real number and gives a measure of size.

复数 z = a + bi 的模，记作 |z|，表示该点在复平面上到原点的距离。它可利用勾股定理计算：|z| = √(a² + b²)。模总是一个非负实数，用来衡量复数的大小。

|z| = √(a² + b²)

Example: For z = 3 + 4i, |z| = √(3² + 4²) = √25 = 5.

示例： z = 3 + 4i 的模为 √(9+16) = 5。

9. The Argand Diagram | 阿尔冈图

The Argand diagram is a coordinate system used to visualise complex numbers. The horizontal axis represents the real part, and the vertical axis represents the imaginary part. Every complex number corresponds to a unique point, making it much easier to understand addition, conjugation and the geometric meaning of the modulus.

阿尔冈图是一种用于可视化复数的坐标系。横轴表示实部，纵轴表示虚部。每个复数在该平面上对应唯一的一点，这有助于理解加减法的几何意义、共轭以及模的概念。

On an Argand diagram, the complex number a + bi is plotted at the point (a, b). The modulus is simply the length of the line from the origin to that point. Conjugates appear as reflections across the real axis.

在阿尔冈图中，复数 a + bi 标于 (a, b) 处。模就是从原点到该点的线段长度。共轭复数则表现为关于实轴的对称点。

10. Solving Quadratic Equations with Complex Roots | 解含复数根的二次方程

When the discriminant (b² – 4ac) of a quadratic equation ax² + bx + c = 0 is negative, the equation has no real roots – but it does have two complex conjugate roots. GCSE students may have been told ‘no real solution’; with complex numbers we can now find those solutions.

当二次方程 ax² + bx + c = 0 的判别式 b² – 4ac 为负时，该方程没有实数根，但却拥有两个共轭复数根。可能你在 GCSE 阶段曾被告知“无实数解”，现在借助复数就能找出这些解了。

Example: Solve x² + 4x + 13 = 0. Using the quadratic formula: x = [-4 ± √(16 – 52)] / 2 = [-4 ± √(-36)] / 2 = [-4 ± 6i] / 2 = -2 ± 3i.

示例： 解 x² + 4x + 13 = 0。求根公式给出 x = [-4 ± √(-36)]/2 = (-4 ± 6i)/2 = -2 ± 3i。

The roots are a pair of complex conjugates: -2 + 3i and -2 – 3i.

这两个根是一对共轭复数：-2 + 3i 和 -2 – 3i。

11. Key Points Summary | 考点总结

Complex numbers extend real numbers, allowing us to solve equations like x² = -1. The imaginary unit i satisfies i² = -1. Standard form is a + bi, with a real part and an imaginary part. Addition and subtraction are done by combining like terms; multiplication uses FOIL and the fact i² = -1; division is performed by multiplying by the conjugate of the denominator. The conjugate a – bi helps produce a real number when multiplied. The modulus |z| = √(a² + b²) gives the distance from the origin on the Argand diagram. Finally, negative discriminants lead to complex conjugate roots in quadratic equations.

复数是对实数的扩充，允许我们求解 x² = -1 这样的方程。虚数单位 i 满足 i² = -1。标准形式为 a + bi，包含实部与虚部。加减法通过合并同类项完成；乘法用展开律并利用 i² = -1；除法通过乘以分母的共轭实现。共轭复数 a – bi 相乘可得实数。模 |z| = √(a² + b²) 是阿尔冈图上原点到该点的距离。最后，判别式为负时，二次方程将得到一对共轭复数根。

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GCSE OCR Maths: Complex Numbers Key Points | GCSE OCR 数学：复数 考点精讲