📚 A-Level Mathematics: Complex Numbers Key Points | A-Level 数学：复数考点精讲
Complex numbers extend the real numbers by introducing an imaginary unit i, where i² = – tic, polar representation, De Moivre’s theorem, and geometric interpretations is required.
复数通过引入虚数单位 i（满足 i² = –1）扩展了实数系，使我们能求解如 x² + 1 = 0 的方程。在 A-Level 进阶数学中，需要深入理解复数的四则运算、极坐标表示、棣莫弗定理以及几何意义。
1. Introduction to Complex Numbers | 复数介绍
can be written as z = a + bi, where a is the real part, b is the imaginary part, and a, b ∈ ℝ.
复数 z 可写作 z = a + bi，其中 a 为实部，b 为虚部，且 a, b 为实数。
Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.
两个复数相等，当且仅当它们的实部相等且虚部相等。
The set of all complex numbers is denoted by ℂ. Every real number is also a complex number with b = 0.
全体复数构成的集合记作 ℂ。每个实数（b = 0）也是复数。
2. Arithmetic with Complex Numbers | 复数运算
Adding and subtracting complex numbers: (a + bi) ± (c + di) = (a ± c) + (b ± d)i.
复数加减法：(a + bi) ± (c + di) = (a ± c) + (b ± d)i。
Multiplying uses the distributive law and the fact i² = –1: (a + bi)(c + di) = ac + adi + bci + bdi² = (ac – bd) + (ad + bc)i.
乘法利用分配律和 i² = –1：(a + bi)(c + di) = (ac – bd) + (ad + bc)i。
denominator by the complex conjugate of the denominator: (a + bi) / (c + di) = (a + bi)(c – di) / (c² + d²).
除法：分子分母同乘分母的共轭复数，化为实分母：(a + bi) / (c + di) = (a + bi)(c – di) / (c² + d²)。
The complex conjugate of z = a + bi is denoted by z̄ = a – bi. It reflects the point across the real axis.
z = a + bi 的共轭复数记作 z̄ = a – bi，它在复平面上关于实轴对称。
3. The Complex Plane (Argand Diagram) | 复平面（Argand 图）
Complex numbers can be represented as points or vectors on an Argand diagram, with the x-axis representing the real part and the y-axis representing the imaginary part.
复数可用 Argand 图上的点或向量表示，x 轴为实轴，y 轴为虚轴。
The distance from the origin to the point z = a + bi is the modulus |z| = √a² + b².
原点到点 z = a + bi 的距离为模 |z| = √a² + b²。
positive real axis is the argument, denoted arg(z), usually measured in radians such that –π < θ ≤ π.
与正实轴形成的夹角为辐角，记作 arg(z)，通常以弧度表示，取值范围 –π < θ ≤ π。
4. Modulus and Argument | 模与辐角
For z = a + bi, the modulus |z| = √a² + b² the size of the vector. The argument θ satisfies tanθ = b/a, with the quadrant considered.
对于 z = a + bi，模 |z| = √a² + b² 表示向量长度。辐角 θ 满足 tanθ = b/a，且需根据象限确定。
Properties of modulus: |z₁z₂| = |z₁||z₂|, |z₁/z₂| = |z₁|/|z₂|, |zⁿ| = |z|ⁿ.
模的性质：|z₁z₂| = |z₁||z₂|，|z₁/z₂| = |z₁|/|z₂|，|zⁿ| = |z|ⁿ。
g(z₁) + arg(z₂) (+ 2kπ for principal value adjustment). For division: arg(z₁/z₂) = arg(z₁) – arg(z₂).
辐角的性质：arg(z₁z₂) = arg(z₁) + arg(z₂)（注意调整主值）；除法：arg(z₁/z₂) = arg(z₁) – arg(z₂)。
5. Polar Form of Complex Numbers | 复数的极坐标形式
A complex number can be written in polar form: z = r(cosθ + i sinθ), where r = |z| and θ = arg(z).
复数可用极坐标形式表示：z = r(cosθ + i sinθ)，其中 r = |z|，θ = arg(z)。
Using Euler's formula eiθ = cosθ + i sinθ, the polar form is often written as z = reiθ. This compact form simplifies multiplication and exponentiation.
利用欧拉公式 eiθ = cosθ + i sinθ，极坐标形式常写作 z = reiθ，这种简洁形式便于乘法和幂运算。
Converting between Cartesian form a + bi and polar form r(cosθ + i sinθ) is a key skill in exam problems.
在直角形式 a + bi 和极坐标形式 r(cosθ + i sinθ) 之间进行转换是考试中的核心技能。
6. Multiplication and Division in Polar Form | 极坐标形式的乘除
r₂(cosθ₂ + i sinθ₂), then multiplication gives: z₁ z₂ = r₁ r₂ [cos(θ₁+θ₂) + i sin(θ₁+θ₂)].
若 z₁ = r₁(cosθ₁ + i sinθ₁)，z₂ = r₂(cosθ₂ + i sinθ₂)，则乘积为：z₁z₂ = r₁r₂ [cos(θ₁+θ₂) + i sin(θ₁+θ₂)]。
Division yields: z₁ / z₂ = (r₁ / r₂) [cos(θ₁ – θ₂) + i sin(θ₁ – θ₂)], provided r₂ ≠ 0.
除法得：z₁ / z₂ = (r₁ / r₂) [cos(θ₁ – θ₂) + i sin(θ₁ – θ₂)]，其中 r₂ ≠ 0。
Thus, multiplying complex numbers multiplies their moduli and adds their arguments; dividing divides the moduli and subtracts the arguments.
因此，复数相乘时模相乘、辐角相加；相除时模相除、辐角相减。
7. De Moivre's Theorem | 棣莫弗定理
For any integer n, (cosθ + i sinθ)n = cos(nθ) + i sin(nθ).
对任意整数 n，有 (cosθ + i sinθ)n = cos(nθ) + i sin(nθ)。
This theorem is proved by mathematical induction for integer n and can be extended to rational n when dealing with roots. It is a powerful tool for deriving trigonometric identities and solving complex equations.
该定理对整数 n 可通过数学归纳法证明，推广到有理数 n 时可用于求根。它是推导三角恒等式和求解复数方程的有力工具。
Using De Moivre’s theorem along with the binomial theorem allows expansion of cos(nθ) and sin(nθ) in powers of sinθ and cosθ.
联合棣莫弗定理与二项式定理，可将 cos(nθ) 和 sin(nθ) 展开为 sinθ 和 cosθ 的幂次形式。
8. Applications of De Moivre's Theorem | 棣莫弗定理的应用
s: For example, to express cos3θ in terms of cosθ, write cos3θ + i sin3θ = (cosθ + i sinθ)3, expand the right-hand side, and equate real parts.
恒等式推导：例如，用 cosθ 表示 cos3θ，可写 cos3θ + i sin3θ = (cosθ + i sinθ)3，展开右边并取实部。
Expansion: (c + i s)3 = c3 + 3c2(i s) + 3c(i s)2 + (i s)3 = (c3 – 3cs2) + i(3c2 s – s3), giving cos3θ = 4cos3θ – 3cosθ.
展开：(c + i s)3 = c3 + 3c2(i s) + 3c(i s)2 + (i s)3 = (c3 – 3cs2) + i(3c2 s – s3)，由此得 cos3θ = 4cos3θ – 3cosθ。
Similarly, sin3θ = 3sinθ – 4sin3θ. Such identities are commonly examined.
类似地，sin3θ = 3sinθ – 4sin3θ。这类恒等式在考试中经常出现。
9. Finding Roots of Unity | 单位根的求解
The n-th roots of unity are the complex numbers z such that zn De Moivre’s theorem, the solutions are z = cos(2kπ/n) + i sin(2kπ/n) for k = 0, 1, ..., n – 1.
n 次单位根是满足 zn = 1 的复数。利用棣莫弗定理，解为 z = cos(2kπ/n) + i sin(2kπ/n)，k = 0, 1, ..., n – 1。
These roots are equally spaced around the unit circle in the Argand diagram, forming the vertices of a regular n-gon.
这些根均匀分布在 Argand 图的单位圆上，构成正 n 边形的顶点。
For example, the cube roots of unity are 1, ω = –½ + i(√3/2), and ω² = –½ – i(√3/2), where 1 + ω + ω² = 0 and ω³ = 1.
例如，三次单位根为 1，ω = –½ + i(√3/2)，ω² = –½ – i(√3/2)，满足 1 + ω + ω² = 0 且 ω³ = 1。
Solving zn = a + bi: first express the right-hand side in polar form R(cosφ + i sinφ), then z = R1/n [cos((φ+2kπ)/n) + i sin((φ+2kπ)/n)].
求解 zn = a + bi：先将右边写成极坐标形式 R(cosφ + i sinφ)，则 z = R1/n [cos((φ+2kπ)/n) + i sin((φ+2kπ)/n)]。
10. Solving Equations with Complex Numbers | 解复数方程
When a real polynomial has a complex root, its conjugate is also a root. This applies to polynomials with real coefficients.
实系数多项式的复根必成共轭对出现。
To solve equations such as z² + z + 1 = 0, apply the quadratic formula: z = [–1 ± √1 – 4] / 2 = –½ ± i(√3/2). These are the primitive cube roots of unity.
解方程如 z² + z + 1 = 0，使用求根公式：z = [–1 ± √1 – 4] / 2 = –½ ± i(√3/2)，恰为三次单位根。
by equating real and imaginary parts. Let z = x + yi, separate real and imaginary pieces.
含复系数的三次或四次方程常通过代换或实部虚部分别相等来处理。设 z = x + yi，分离实虚部。
11. Loci in the Complex Plane | 复平面上的轨迹
A locus is a set of points satisfying a given condition. Common loci include:
轨迹是满足给定条件的点集。常见轨迹有：
- Circle: |z – a| = r, centred at complex number a with radius r.
- 圆：|z – a| = r，以复数 a 为圆心，半径 r。
- Perpendicular bisector: |z – a| = |z – b|, the set of points equidistant from a and b.
- 垂直平分线：|z – a| = |z – b|，到 a 与 b 距离相等的点集。
- Half-line: arg(z – a) = θ, a ray from a making an angle θ with the positive real direction.
- 射线：arg(z – a) = θ，从 a 出发、与正实轴成 θ 角的射线。
- Annulus: r₁ < |z – a| < r₂, region between two concentric circles.
- 圆环：r₁ < |z – a| < r₂，两同心圆之间的区域。
- Combining conditions: Shading regions that satisfy inequalities involving both modulus and argument.
- 组合条件：同时满足模和辐角不等式的区域，需按要求涂阴影。
Sketching loci and finding the maximum or minimum value of |z| or arg(z) within a region are typical exam tasks.
画出轨迹，并在区域内求 |z| 或 arg(z) 的最大/最小值是典型考题。
12. Exam Tips and Common Mistakes | 考试技巧与常见错误
Always check the principal argument range (–π, π]. When an argument falls outside, add or subtract 2π to bring it into the correct range.
始终检查辐角的主值范围 (–π, π]；若超出，加减 2π 使其落在主值区间。
In division, remember to multiply by the conjugate of the denominator. A sign error in the imaginary part of the conjugate is a frequent slip.
除法时记住乘分母的共轭；共轭虚部符号出错是常见疏漏。
When using De Moivre for roots, dive the argument by n and add 2kπ/n. Do not forget to cycle through k = 0, 1, …, n–1.
用棣莫弗定理求根时，要将辐角除以 n 并加上 2kπ/n；勿忘遍历 k = 0, 1, …, n–1。
For loci, clearly label key points and indicate whether boundaries are included (solid line) or excluded (dashed line). Show the required shading.
画轨迹时需清楚标出关键点，边界包含用实线、不包含用虚线，并正确涂阴影。
When proving trigonometric identities via De Moivre, equate real and imaginary parts correctly and apply binomial coefficients with care.
借棣莫弗定理证明三角恒等式时，精确匹配实部和虚部，妥善使用二项式系数。
Practise converting between Cartesian and polar forms quickly, as many problems rely on the speed and accuracy of these conversions.
多加练习直角形式与极坐标形式的互换，因为许多题目依赖于转换的速度与准确性。
Published by TutorHao | Further Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图