📚 Complex Numbers for CIE A-Level Maths: Key Points Revision | A-Level CIE 数学：复数 考点精讲
Complex numbers form a core topic in the CIE A-Level Mathematics (9709) syllabus, especially within the Pure Mathematics 3 (P3) paper. A thorough understanding of imaginary and complex numbers, their arithmetic, geometric representation on the Argand diagram, polar form, and De Moivre's theorem is essential for solving polynomial equations, interpreting loci, and tackling a wide range of examination problems. This revision guide highlights every key point you need to master for the CIE board, with paired English and Chinese explanations to support bilingual learners.
复数是 CIE A-Level 数学 (9709) 大纲中的核心主题，尤其出现在纯数学 3 (P3) 试卷中。透彻理解虚数与复数、它们的运算、阿尔冈图上的几何表示、极坐标形式以及棣莫弗定理，对于解多项式方程、解读轨迹以及应对各类考试题目至关重要。这份考点精讲专为 CIE 考试局梳理每一个你必须掌握的关键点，并配以英中双语解释，支持双语学习者。
1. What is a Complex Number? | 什么是复数？
e imaginary unit. The real part is denoted Re(z) = a, and the imaginary part is Im(z) = b.
复数通过引入虚部来扩展实数系统。它写成标准形式 z = a + bi，其中 a 和 b 是实数，i 是虚数单位。实部记为 Re(z) = a，虚部记为 Im(z) = b。
Complex numbers allow us to solve equations like x² + 1 = 0, which have no real solutions.
若 b = 0，该数是纯实数；若 a = 0 且 b ≠ 0，则是纯虚数。复数使我们能够求解 x² + 1 = 0 这样没有实数解的方程。
2. The Imaginary Unit i | 虚数单位 i
a pattern: i¹ = i, i² = −1, i³ = −i, i⁴ = 1, and then the cycle repeats every four powers.
虚数单位 i 由性质 i² = −1 定义。这一条定义支撑着所有复数运算。i 的高次幂以周期模式循环：i¹ = i, i² = −1, i³ = −i, i⁴ = 1，之后每四次循环一次。
because 10 ÷ 4 gives remainder 2. This trick saves time in expansions and simplifications.
要化简 iⁿ，用 n 除以 4 并利用余数。例如 i¹⁰ = i² = −1，因为 10 ÷ 4 余数为 2。这一技巧可在展开与化简中节省时间。
3. Basic Operations with Complex Numbers | 复数的基本运算
nd imaginary parts separately: (a + bi) ± (c + di) = (a ± c) + (b ± d)i. There is no cross‑term mixing, making these operations straightforward.
加法与减法通过分别合并实部和虚部来执行：(a + bi) ± (c + di) = (a ± c) + (b ± d)i。没有交叉项混合，因此这些运算十分直接。
= (ac − bd) + (ad + bc)i. Always replace i² with −1 and collect real and imaginary terms.
乘法利用分配律以及 i² = −1 这一事实：(a + bi)(c + di) = ac + adi + bci + bdi² = (ac − bd) + (ad + bc)i。务必用 −1 替换 i² 并合并实部与虚部项。
Division is achieved by multiplying numerator and denominator by the complex conjugate of the denominator, which turns the denominator into a real number. This process is detailed in the next section.
除法通过将分子和分母同时乘以分母的共轭复数来实现，这样分母会变成一个实数。此过程将在下一节详述。
4. Complex Conjugate | 共轭复数
The complex conjugate of z = a + bi is denoted as z̄ or z* and is defined as z̄ i. Geometrically, conjugation reflects a point across the real axis on the Argand diagram.
z = a + bi 的共轭复数记作 z̄ 或 z*，定义为 z̄ = a − bi。在几何上，取共轭相当于将点关于实轴反射到阿尔冈图上。
Conjugates possess useful properties: (z̄)̄ = z, z + z̄ = 2a (twice the real part), z − z̄ = 2bi (twice the imaginary part times i), and z · z̄ = a² + b² = |z|², which is always a non‑ two complex numbers, compute (a + bi)/(c + di) by multiplying top and bottom by (c − di).
共轭复数具有非常有用的性质：(z̄)̄ = z，z + z̄ = 2a（两倍实部），z − z̄ = 2bi（两倍虚部乘以 i），以及 z · z̄ = a² + b² = |z|²，这始终是一个非负实数。进行两复数相除时，计算 (a + bi)/(c + di) 只需将分子分母同时乘以 (c − di)。
5. The Argand Diagram | 阿尔冈图
i corresponds to a unique point (a, b) or a position vector from the origin to that point.
阿尔冈图是一个平面，其横轴表示实部，纵轴表示虚部。每一个复数 z = a + bi 都对应于一个唯一点 (a, b) 或从原点指向该点的位置向量。
This visual representation turns complex numbers into geometric objects, making addition equivalent to vector addition and helping to interpret modulus, argument, and loci. It also explains why the complex conjugate is a reflection in the real axis.
这种可视化表示将复数转化为几何对象，使得加法等价于向量加法，并有助于理解模、辐角及轨迹。这也解释了为什么共轭复数是关于实轴的反射。
6. Modulus and Argument | 模与辐角
The modulus of z = a + bi is the distance from the origin to the point (a, b), given by |z| = √a² + b². It is always non‑negative and represents the magnitude of the complex number. For a purely real number, |a| = √a².
复数 z = a + bi 的模是从原点到点 (a, b) 的距离，由 |z| = √a² + b² 给出。模始终非负，表示复数的大小。对于纯粹实数，|a| = √a²。
the range −π < θ ≤ π. Care must be taken to identify the correct quadrant using a sketch.
z 的辐角，记作 arg(z)，是位置向量与正实轴所成的角 θ，通常以弧度度量。习惯上，CIE 使用主辐角，范围在 −π < θ ≤ π。必须通过草图判断正确的象限。
s of a and b. For instance, when a < 0 and b > 0, the argument is π − α, where α = arctan(|b/a|).
如果 z = a + bi，那么 tan θ = b/a，但根据 a 和 b 的正负，需要对反正切的结果进行调整。例如，当 a < 0 且 b > 0 时，辐角为 π − α，其中 α = arctan(|b/a|)。
7. Polar Form of a Complex Number | 复数的极坐标形式
Using the modulus r = |z| and argument θ, a complex number can be expressed in polar form (also called modulus‑argument form):
z = r(cos θ + i sin θ)
利用模 r = |z| 和辐角 θ，复数可表示为极坐标形式（也称为模‑辐角形式）：
z = r(cos θ + i sin θ)
This representation is unique up to adding multiples of 2π to the argument, but the principal argument ensures a standard choice. The polar form is especially powerful for multiplication, division, and exponentiation, as it separates the magnitude from the direction.
这种表示在辐角加上 2π 的整数倍时并不唯一，但主辐角保证了标准选择。极坐标形式将大小与方向分离，特别适合乘法、除法和乘方运算。
To convert from rectangular to polar form, compute r = √a² + b² . To convert back, use a = r cos θ, b = r sin θ. These conversions are routine exam tasks.
从直角坐标转换到极坐标形式，需计算 r = √a² + b² 和 θ = arg(a + bi)。转回时使用 a = r cos θ, b = r sin θ。这些转换是常规的考试任务。
8. Multiplication and Division in Polar Form | 极坐标形式下的乘法与除法
nd add their arguments. If z₁ = r₁(cos θ₁ + i sin θ₁) and z₂ = r₂(cos θ₂ + i sin θ₂), then
z₁z₂ = r₁r₂[cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)]
当复数写成极坐标形式时，乘法变得简单：模相乘，辐角相加。若 z₁ = r₁(cos θ₁ + i sin θ₁) 且 z₂ = r₂(cos θ₂ + i sin θ₂)，则
z₁z₂ = r₁r₂[cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)]
Division follows a similar pattern: divide the moduli and subtract the arguments. For z₂ ≠ 0,
z₁ / z₂ = (r₁/r₂)[cos(θ₁ − θ₂) + i sin(θ₁ − θ₂)]
除法遵循类似模式：模相除，辐角相减。对于 z₂ ≠ 0，
z₁ / z₂ = (r₁/r₂)[cos(θ₁ − θ₂) + i sin(θ₁ − θ₂)]
These results can be proved using trigonometric addition formulas. They are extremely useful for simplifying expressions and for problems involving powers or roots of complex numbers. Always reduce the resulting argument to the principal range if required.
这些结果可用三角加法公式证明。它们对于简化表达式以及处理复数乘方或开方问题极为有用。需要时始终将结果辐角约化到主值范围。
9. De Moivre's Theorem | 棣莫弗定理
De Moivre's theorem is a direct extension of multiplication in polar form. For any integer n,
[r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ)
棣莫弗定理是极坐标乘法的一个直接推广。对于任意整数 n，
[r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ)
mials. First convert 1 + i to polar form: √2(cos π/4 + i sin π/4), then apply the theorem.
该定理在使用指数形式时对所有实数幂皆有效，但在 A-Level 阶段我们通常将 n 限制为整数。它使得快速计算乘方成为可能，例如无需二项式展开即可求出 (1 + i)⁶。首先将 1 + i 化为极坐标形式：√2(cos π/4 + i sin π/4)，然后应用定理。
De Moivre's theorem is also a tool to derive trigonometric identities, such as expressing cos 3θ and sin 3θ in terms of cos θ and sin θ, which is a common P3 question.
棣莫弗定理也是推导三角恒等式的工具，例如用 cos θ 和 sin θ 表示 cos 3θ 和 sin 3θ，这是 P3 中的常见题型。
10. Roots of Complex Numbers | 复数的根
To find the n nth roots of a complex number z = r(cos θ + i sin θ), we use the formula
wk = r1/n[cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)], k = 0, 1, … , n−1
要求复数 z = r(cos θ + i sin θ) 的 n 个 n 次方根，我们使用公式
wk = r1/n[cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)], k = 0, 1, … , n−1
Each distinct root is obtained by adding 2kπ/n to the argument before dividing by n. The roots are symmetrically arranged on the Argand diagram: they lie on a circle of radius r1/n and are equally spaced by an angle of 2π/n. This geometric property often simplifies sketching and problem solving.
通过将 (θ + 2kπ) 除以 n 前向辐角添加 2kπ/n，我们得到每个不同的根。这些根在阿尔冈图上对称排列：它们位于半径为 r1/n 的圆上，且彼此间隔 2π/n 的角。这一几何性质常能简化作图与解题。
For example, the cube roots of unity (solutions to z³ = 1) are 1, ei2π/3 (or cos(2π/3) + i sin(2π/3)), and ei4π/3. In rectangular form, they are 1, −½ + i(√3/2), and −½ − i(√3/2). Notice that two of them are complex conjugates.
例如，单位立方根（方程 z³ = 1 的解）是 1, ei2π/3 （即 cos(2π/3) + i sin(2π/3)）和 ei4π/3。在直角坐标形式下，它们是 1, −½ + i(√3/2) 和 −½ − i(√3/2)。注意其中两个互为共轭复数。
11. Solving Polynomial Equations | 解多项式方程
Complex numbers allow us to solve polynomial equations that have no real roots, and crucially, the Fundamental Theorem of Algebra guarantees that every polynomial equation of degree n has exactly n complex roots (counting multiplicity). For real‑coefficient polynomials, non‑real roots occur in conjugate pairs.
复数使我们能够求解没有实数根的多项式方程；重要的是，代数基本定理保证每一个 n 次多项式方程恰有 n 个复数根（计重数）。对于实系数多项式，非实数根以共轭对出现。
az + (a² + b²), which has real coefficients, and then perform polynomial division to locate any remaining roots.
若已知实系数多项式的一个复数根，例如 a + bi，则其共轭 a − bi 也是一个根。你可以求出二次因式 (z − (a+bi))(z − (a−bi)) = z² − 2az + (a² + b²)，该因式具有实系数，然后通过多项式除法找出其余根。
+ z + 1 = 0 are often solved by recognising them as geometric series or by using De Moivre' oots of z⁵ = 1, excluding z = 1. This topic bridges algebra and complex numbers elegantly.
类似 z⁴ + z³ + z² + z + 1 = 0 的方程，常通过将其识别为等比数列或使用棣莫弗定理求 z⁵ = 1 的根（排除 z = 1）来求解。这一主题优雅地连接了代数与复数。
12. Loci in the Complex Plane | 复平面上的轨迹
loci tested in CIE exams include circles, perpendicular bisectors, and rays. The equation |z − a| = r describes a circle with centre at the complex number a and radius r.
轨迹是阿尔冈图上满足给定条件的点集。CIE 考试中常考的轨迹包括圆、垂直平分线和射线。方程 |z − a| = r 表示以复数 a 为圆心、半径为 r 的圆。
The equation |z − a| = |z − b| ce it expresses points equidistant from a and b. The locus arg(z − a) = θ describes a half‑line (ray) from point a, making an angle θ with the positive real axis.
方程 |z − a| = |z − b| 表示连接点 a 与点 b 的线段的垂直平分线，因为它表达了到 a 和 b 距离相等的点。轨迹 arg(z − a) = θ 描述了一条从点 a 出发、与正实轴夹角为 θ 的射线。
More complicated loci may involve inequalities or combinations of conditions, e.g., |z − 3| e loci often requires shading regions or drawing dashed boundaries. Always treat the modulus as a distance and the argument as a direction.
更复杂的轨迹可能涉及不等式或多个条件的组合，例如 |z − 3| < 2 且 0 ≤ arg(z) ≤ π/2。绘制这些轨迹通常需要涂色区域或画出虚线边界。始终将模视为距离，将辐角视为方向。
Published by TutorHao | 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