📚 A-Level WJEC Mathematics: Complex Numbers – Key Points Explained | 复数 考点精讲
Complex numbers are a fundamental extension of the real number system, essential for solving equations where square roots of negative numbers appear. In the WJEC A-Level Mathematics syllabus, you are required to understand the algebra of complex numbers, their geometric representation on the Argand diagram, and the powerful De Moivre's theorem for powers and roots. This article breaks down every key topic into clear, exam-focused explanations, with worked examples, standard results and the examiner traps that cost marks every summer.
复数是实数系的基本扩展，对于求解涉及负数平方根的方程至关重要。在 WJEC A-Level 数学教学大纲中，你需要理解复数的代数运算、它们在阿尔甘特图上的几何表示，以及用于幂和根的棣莫弗定理。本文将以清晰、针对考试的方式分解每个重要主题，配合例题、标准结论以及每年考试中最容易失分的陷阱。
1. What Is a Complex Number? | 什么是复数
enoted Re(z) = x, and the imaginary part is Im(z) = y. The set of all complex numbers is denoted ℂ, and the real numbers form a subset of ℂ those with y = 0). Equally, numbers of the form iy with y ≠ 0 are called purely imaginary.
虚数单位 i 由性质 i² = −1 定义。复数 z 可以写成 z = x + iy，其中 x 和 y 是实数。实部记为 Re(z) = x，虚部记为 Im(z) = y。所有复数构成的集合记作 ℂ，实数集是 ℂ 的子集（即 y = 0 的那些数）。同样，形如 iy 且 y ≠ 0 的数称为纯虚数。
no real solution, because no real number squares to a negative value. Extending the number system by adjoining i makes every quadratic solvable, and in fact every polynomial equation of degree n has exactly n roots in ℂ (counting repeats), a result known as the Fundamental Theorem of Algebra.
复数存在的理由很简单：像 x² + 1 = 0 这样的方程没有实数解，因为没有任何实数的平方为负。通过引入 i 扩展数系后，任何二次方程都可解；事实上，任何 n 次多项式方程在 ℂ 中恰好有 n 个根（计重数），这就是代数基本定理。
i² = −1, z = x + iy, Re(z) = x, Im(z) = y
write the answer in the standard form x ± iy. For example, z² + 2z + 5 = 0 gives z = (−2 ± √4 − 20) ÷ 2 = (−2 ± √−16) ÷ 2 = −1 ± 2i.
当二次方程的判别式为负时，把答案写标准形式 x ± iy。例如 z² + 2z + 5 = 0 给出 z = (−2 ± √4 − 20) ÷ 2 = (−2 ± √−16) ÷ 2 = −1 ± 2i。
2. Addition, Subtraction and Multiplication | 加法、减法与乘法
tion requires expanding brackets and then replacing every i² with −1: (a + bi)(c + di) = ac + adi + bci + bdi² = (ac − bd) + (ad + bc)i.
复数的运算把 i 视为满足 i² = −1 的符号。加减法按分量进行：(a + bi) ± (c + di) = (a ± c) + (b ± d)i。乘法需要展开括号，然后把每个 i² 替换为 −1：(a + bi)(c + di) = ac + adi + bci + bdi² = (ac − bd) + (ad + bc)i。
The powers of i follow a repeating cycle of length four, which is useful for simplifying large powers quickly:
i 的幂遵循长度为四的循环，这对快速化简高次幂非常有用：
- i¹ = i, i² = −1, i³ = −i, i⁴ = 1, then the pattern repeats. | i¹ = i，i² = −1，i³ = −i，i⁴ = 1，之后循环重复。
- To evaluate iⁿ, divide n by 4 and use the remainder. | 计算 iⁿ 时，用 n 除以 4，看余数。
Worked example. 0 = 22 − 7i. Notice that the answer is again in the form x + iy, which is expected because ℂ is closed under multiplication.
例题。求 (3 + 2i)(4 − 5i)。展开得：12 − 15i + 8i − 10i² = 12 − 7i + 10 = 22 − 7i。注意结果仍然形如 x + iy，这是因为 ℂ 对乘法封闭。
e (1 + i)² in the form x + iy. Here (1 + i)² = 1 + 2i + i² = 2i, which is purely imaginary — a common one-mark check that many candidates rush and get wrong.
在 WJEC 试卷中，常要求把 (1 + i)² 表示成 x + iy 的形式。这里 (1 + i)² = 1 + 2i + i² = 2i，是纯虚数——这是一分的送分题，但很多考生因为粗心而失分。
3. The Complex Conjugate and Division | 共轭复数与除法
The complex conjugate of z = x + iy is defined as z̄ = x − iy; you simply change the sign of the imaginary part. The key identity is z × z̄ ays a non-negative real number. This is exactly the identity that makes division possible.
复数 z = x + iy 的共轭定义为 z̄ = x − iy，只需改变虚部的符号。核心恒等式是 z × z̄ = (x + iy)(x − iy) = x² + y²，它总是非负实数。正是这个恒等式让除法成为可能。
To divide two complex numbers, multiply the numerator and the denominator by the conjugate of the denominator. This makes the denominator real, after which you can split the answer into real and imaginary parts.
要作复数除法，把分子和分母同时乘以分母的共轭。这样分母变成实数，之后再把结果拆成实部与虚部。
(a + bi) ÷ (c + di) = ((a + bi)(c − di)) ÷ (c² + d²)
Worked example. + 2i + 2i² = 1 + 5i, and the denominator becomes 1² + 1² = 2. So the answer is ½ + (5÷2)i.
例题。把 (3 + 2i) ÷ (1 − i) 表示成 x + iy。分子分母同乘 1 + i：分子为 (3 + 2i)(1 + i) = 3 + 3i + 2i + 2i² = 1 + 5i，分母为 1² + 1² = 2。所以答案是 ½ + (5÷2)i。
- Conjugate properties: (z̄)̄ = z, and (z₁ ± z₂)̄ = z̄₁ ± z̄₂, (z₁z₂)̄ = z̄₁ z̄₂. | 共轭性质：(z̄)̄ = z，且 (z₁ ± z₂)̄ = z̄₁ ± z̄₂，(z₁z₂)̄ = z̄₁ z̄₂。
- If z is real then z̄ = z; if z is purely imaginary then z̄ = −z. | 若 z 为实数则 z̄ = z；若 z 为纯虚数则 z̄ = −z。
- For a quadratic with real coefficients, complex roots always occur in conjugate pairs. | 对于实系数二次方程，复根总以共轭对出现。
4. Modulus and Argument | 模与辐角
The modulus of z = x + iy is its distance from the origin in the complex plane: |z| = √x² + y² ed anticlockwise. The principal argument satisfies −π < θ ≤ π and is written arg z = θ.
复数 z = x + iy 的模是它在复平面上到原点的距离：|z| = √x² + y²。辐角是原点指向 z 的连线与正实轴所成的角，按逆时针方向度量。主辐角满足 −π < θ ≤ π，记作 arg z = θ。
Finding the argument requires care with quadrants. The raw calculator value arctan(y ÷ x) only gives the correct angle when x > 0. Otherwise you must add or subtract π.
求辐角时必须注意象限。计算器给出的 arctan(y ÷ x) 只在 x > 0 时才直接是正确答案，否则需要加或减 π。
| Quadrant | 象限 | Signs | 符号 | Argument | 辐角 |
| First | 第一象限 | x > 0, y > 0 | θ = arctan(y ÷ x) |
| Second | 第二象限 | x < 0, y > 0 | θ = arctan(y ÷ x) + π |
| Third | 第三象限 | x < 0, y < 0 | θ = arctan(y ÷ x) − π |
| Fourth | 第四象限 | x > 0, y < 0 | θ = arctan(y ÷ x) |
Worked example. For z = −1 + √3 i, |z| = √1 + 3 = 2 and the point lies in the second quadrant. The reference angle is arctan(√3 ÷ 1) = π ÷ 3, so arg z = π − π ÷ 3 = 2π ÷ 3. Always sketch the point before computing — it takes five seconds and prevents nearly all argument errors.
例题。对 z = −1 + √3 i，|z| = √1 + 3 = 2，且该点位于第二象限。参考角为 arctan(√3 ÷ 1) = π ÷ 3，因此 arg z = π − π ÷ 3 = 2π ÷ 3。计算前先画草图——只需五秒，却能避免几乎所有辐角错误。
5. The Argand Diagram | 阿尔甘特图
rt. The number z = x + iy is plotted as the point (x, y). This geometric view turns algebraic statements into geometry, which is exactly what many WJEC questions exploit.
阿尔甘特图是一个二维坐标图，横轴表示实部，纵轴表示虚部。复数 z = x + iy 对应点 (x, y)。这种几何视角把代数陈述转化为几何问题，正是许多 WJEC 题目所利用的。
- Conjugation is a reflection in the real axis. | 取共轭相当于关于实轴作对称。
- |z| is the distance from the origin to the point. | |z| 是点到原点的距离。
- arg z is the angle from the positive real axis. | arg z 是从正实轴量起的角。
- Addition follows the parallelogram law: z₁ + z₂ is the fourth vertex of the parallelogram with vertices 0, z₁, z₁ + z₂, z₂. | 加法遵循平行四边形法则：z₁ + z₂ 是以 0、z₁、z₁ + z₂、z₂ 为顶点的平行四边形的第四个顶点。
- |z₁ − z₂| is the distance between the points representing z₁ and z₂. | |z₁ − z₂| 是 z₁ 与 z₂ 两点之间的距离。
The last point is the single most useful geometric fact at this level, because almost every locus question is a disguised statement about distance or angle.
最后一点是本阶段最有用的几何事实，因为几乎所有轨迹题本质上都是关于距离或角度的陈述。
6. Modulus-Argument (Polar) Form | 模辐角（极坐标）形式
ts modulus and argument gives the modulus-argument form: z = r(cos θ + i sin θ), where r = |z| e geometry, x = r cos θ and y = r sin θ, which links the Cartesian and polar descriptions.
用模与辐角表示 z 就得到模辐角形式：z = r(cos θ + i sin θ)，其中 r = |z|，θ = arg z。有些教材简写为 r cis θ，但在 WJEC 试卷上通常应写出完整的括号形式。由几何关系有 x = r cos θ，y = r sin θ，这联系了直角坐标与极坐标两种描述。
z = r(cos θ + i sin θ), r = √x² + y², tan θ = y ÷ x
The great advantage of polar form is that multiplication and division become simple rules:
极坐标形式的最大优势在于，乘法和除法变成简单规律：
- |z₁ z₂| = |z₁| |z₂| and arg(z₁ z₂) = arg z₁ + arg z₂. | |z₁ z₂| = |z₁| |z₂|，且 arg(z₁ z₂) = arg z₁ + arg z₂。
- |z₁ ÷ z₂| = |z₁| ÷ |z₂| and arg(z₁ ÷ z₂) = arg z₁ − arg z₂. | |z₁ ÷ z₂| = |z₁| ÷ |z₂|，且 arg(z₁ ÷ z₂) = arg z₁ − arg z₂。
- |z̄| = |z| and arg z̄ = −arg z. | |z̄| = |z|，且 arg z̄ = −arg z。
Worked example. 6 + π ÷ 3 = π ÷ 2, so z₁z₂ = 6(cos π ÷ 2 + i sin π ÷ 2) = 6i. No bracket expansion needed.
例题。相乘 z₁ = 2(cos π ÷ 6 + i sin π ÷ 6) 与 z₂ = 3(cos π ÷ 3 + i sin π ÷ 3)。积的模为 6，辐角为 π ÷ 6 + π ÷ 3 = π ÷ 2，所以 z₁z₂ = 6(cos π ÷ 2 + i sin π ÷ 2) = 6i。完全不需要展开括号。
7. Exponential Form and Euler's Relation | 指数形式与欧拉关系
Euler's relation states that eiθ = cos θ + i sin θ, so any complex number can be written compactly as z = reiθ n exponential and trigonometric functions, and it follows from comparing Maclaurin series.
欧拉关系指出 eiθ = cos θ + i sin θ，因此任何复数都可以紧凑地写成 z = reiθ。这不是巧合，而是指数函数与三角函数之间的深刻联系，可由麦克劳林级数比较得到。
ng a complex number by i rotates it by π ÷ 2 anticlockwise about the origin, because i = eiπ ÷ 2 and r eiθ × eiπ ÷ 2 = r e^(i(θ + π ÷ 2)). Similarly, multiplying by eiθ represents a rotation through θ.
一个特别优雅的推论：把复数乘以 i，相当于在复平面上绕原点逆时针旋转 π ÷ 2，因为 i = eiπ ÷ 2，而 r eiθ × eiπ ÷ 2 = r e^(i(θ + π ÷ 2))。同样地，乘以 eiθ 表示旋转角度 θ。
WJEC questions frequently ask you to describe a transformation geometrically. Typical answers include 'a rotation of π ÷ 2 anticlockwise about the origin', 'an enlargement of scale factor 2 about the origin combined with a rotation of π ÷ 3', and 'a reflection in the real axis'. Learn this vocabulary; vague wording loses method marks.
WJEC 题目经常要求你用几何语言描述变换。典型答案包括"绕原点逆时针旋转 π ÷ 2"、"以原点为中心、比例因子为 2 的放大并伴随 π ÷ 3 的旋转"以及"关于实轴的反射"。要掌握这类表述，含糊的措辞会丢掉方法分。
8. De Moivre's Theorem | 棣莫弗定理
De Moivre' in nθ). The theorem holds for negative integers as well, so it also covers reciprocals and negative powers.
棣莫弗定理指出，对任意整数 n，(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ。带上模之后，推广为 [r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ)。该定理对负整数同样成立，因此也覆盖倒数和负幂。
Worked example 1. Find (1 + i)⁸. Here r = √2 and θ = π ÷ 4, so (1 + i)⁸ = (√2 naive approach of expanding eight brackets is hopeless; De Moivre reduces it to one line.
例题 1。求 (1 + i)⁸。这里 r = √2，θ = π ÷ 4，所以 (1 + i)⁸ = (√2)⁸ (cos 8 × π ÷ 4 + i sin 8 × π ÷ 4) = 16(cos 2π + i sin 2π) = 16。直接展开八个括号几乎不可能，棣莫弗定理把它压缩成一行。
Worked example 2. θ, and using sin²θ = 1 − cos²θ, cos 3θ = 4cos³θ − 3cos θ. Comparing imaginary parts gives sin 3θ = 3sin θ − 4sin³θ.
例题 2。用 cos θ 表示 cos 3θ。由棣莫弗定理，(cos θ + i sin θ)³ = cos 3θ + i sin 3θ。展开后比较实部得 cos 3θ = cos³θ − 3cos θ sin²θ，再用 sin²θ = 1 − cos²θ，得 cos 3θ = 4cos³θ − 3cos θ。比较虚部则得 sin 3θ = 3sin θ − 4sin³θ。
- Always convert to polar form before applying the theorem. | 应用定理前务必先化为极坐标形式。
- Work in radians unless the question explicitly uses degrees. | 除非题目明确使用角度制，否则用弧度制。
- Simplify the final angle into the range −π < θ ≤ π. | 把最终角度化简到区间 −π < θ ≤ π 内。
9. Roots of a Complex Number | 复数的根
exactly n distinct nth roots. If z = r(cos θ + i sin θ), then the nth roots are given by zk = r1÷n [cos((θ + 2kπ) ÷ n) + i sin((θ + 2kπ) ÷ n)] for k = 0, 1, 2, …, n − 1. The '+ 2kπ' ut it you find only one root, and examiners deliberately award marks for the general form.
每个非零复数恰有 n 个不同的 n 次方根。若 z = r(cos θ + i sin θ)，则 n 次方根为 zk = r1÷n [cos((θ + 2kπ) ÷ n) + i sin((θ + 2kπ) ÷ n)]，其中 k = 0, 1, 2, …, n − 1。这里的" + 2kπ"至关重要：没有它就只找到一个根，而考官会专门为此给分。
Worked example. Solve z³ = 8i. Write 8i = 8(cos π ÷ 2 + i sin π ÷ 2), so r1÷3 ents are (π ÷ 2 + 2kπ) ÷ 3 for k = 0, 1, 2, giving π ÷ 6, 5π ÷ 6 and 3π ÷ 2. The roots are √3 + i, −√3 + i and −2i.
例题。解 z³ = 8i。把 8i 写成 8(cos π ÷ 2 + i sin π ÷ 2)，则 r1÷3 = 2，
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