📚 Complex Numbers for A-Level CCEA Mathematics | A-Level CCEA 数学：复数 考点精讲
Complex numbers extend the real number system by introducing the imaginary unit i, defined such that i² = −1. This powerful concept allows us to solve equations that have no real solutions, such as x² + 1 = 0, and to model a wide range of physical and engineering phenomena. For CCEA A-Level Mathematics, mastering complex numbers means understanding their algebraic form, geometric representation on the Argand diagram, polar form, De Moivre’s theorem, and applications to polynomial equations and loci. This article provides a comprehensive, structured revision of all essential topics, with clear explanations and paired bilingual content to reinforce your learning.
复数通过引入虚数单位 i（满足 i² = −1）扩展了实数系统。这一强大的概念使我们能够求解没有实数解的方程，例如 x² + 1 = 0，并用于模拟众多物理和工程现象。对于 CCEA A-Level 数学，掌握复数意味着要理解其代数形式、在阿尔冈图上的几何表示、极坐标形式、棣莫弗定理，以及在多项式方程和轨迹中的应用。本文对所有核心考点进行了系统梳理，通过双语对照讲解帮助你巩固理解。
1. Introduction to Complex Numbers | 复数简介
A complex number is any number that can be expressed in the form z = a + bi, where a and b are real numbers, and i is the imaginary unit satisfying i² = −1.
复数是可以表示为 z = a + bi 形式的任何数，其中 a 和 b 是实数，i 是满足 i² = −1 的虚数单位。
The real part of z is denoted Re(z) = a, and the imaginary part is Im(z) = b (note that Im(z) is the real number b, not bi).
z 的实部记作 Re(z) = a，虚部记作 Im(z) = b（注意 Im(z) 是实数 b，而不是 bi）。
Complex numbers arise naturally when solving quadratic equations. For example, the equation x² + 1 = 0 gives x = ±√(−1) = ±i.
复数在求解二次方程时自然产生。例如，方程 x² + 1 = 0 的解为 x = ±√(−1) = ±i。
All real numbers are also complex numbers with an imaginary part of zero. Purely imaginary numbers have a real part of zero and take the form bi.
所有实数也是虚部为零的复数。纯虚数的实部为零，形式为 bi。
2. The Imaginary Unit and Powers of i | 虚数单位与 i 的幂
The definition i² = −1 leads to a cyclic pattern for higher powers of i. This cycle repeats every four powers.
由定义 i² = −1 可以推出 i 的高次幂存在周期性规律，每四次幂循环一次。
i¹ = i, i² = −1, i³ = i²·i = −i, i⁴ = (i²)² = 1, and then i⁵ = i, and so on.
i¹ = i，i² = −1，i³ = i²·i = −i，i⁴ = (i²)² = 1，然后 i⁵ = i，以此类推。
To simplify expressions like iⁿ, divide n by 4 and use the remainder to determine the equivalent power.
要简化形如 iⁿ 的表达式，可以将 n 除以 4，利用余数确定等价的幂。
For example, i¹⁰ has remainder 2 when 10 is divided by 4, so i¹⁰ = i² = −1.
例如，i¹⁰，10 除以 4 余 2，因此 i¹⁰ = i² = −1。
This property is fundamental when simplifying products, quotients, and powers of complex numbers in Cartesian form.
这一性质是简化复数代数形式下乘除和幂运算的基础。
3. Algebra of Complex Numbers in Cartesian Form | 代数形式的复数运算
Addition and subtraction are performed component-wise: (a + bi) ± (c + di) = (a ± c) + (b ± d)i.
加法和减法按分量进行：(a + bi) ± (c + di) = (a ± c) + (b ± d)i。
Multiplication uses the distributive law and 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 + adi + bci + bdi² = (ac − bd) + (ad + bc)i。
Division is achieved by multiplying numerator and denominator by the complex conjugate of the denominator, which makes the denominator a real number.
除法的实现方法是分子分母同乘以分母的共轭复数，使分母变为实数。
For (a + bi) ÷ (c + di), multiply by (c − di)/(c − di) to obtain [(a + bi)(c − di)] / (c² + d²).
对于 (a + bi) ÷ (c + di)，乘以 (c − di)/(c − di) 得到 [(a + bi)(c − di)] / (c² + d²)。
Equality of complex numbers means that two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.
复数相等意味着两个复数相等当且仅当它们的实部相等且虚部相等。
This principle is often used to solve equations involving complex numbers by equating real and imaginary parts.
这一原理常被用于通过比较实部和虚部来求解含有复数的方程。
4. Complex Conjugate and Modulus | 共轭复数与模
The complex conjugate of z = a + bi is denoted by z̄ or z* and is defined as z̄ = a − bi. Geometrically, it is a reflection of z in the real axis.
复数 z = a + bi 的共轭记作 z̄ 或 z*，定义为 z̄ = a − bi。几何上，它是 z 关于实轴的镜像。
Key properties: z + z̄ = 2a (purely real), z − z̄ = 2bi (purely imaginary), and z·z̄ = a² + b² = |z|².
关键性质：z + z̄ = 2a（纯实数），z − z̄ = 2bi（纯虚数），以及 z·z̄ = a² + b² = |z|²。
The modulus (or absolute value) of z, denoted |z|, is defined as |z| = √(a² + b²). It represents the distance from the origin to the point (a, b) on the complex plane.
z 的模（或绝对值）记作 |z|，定义为 |z| = √(a² + b²)。它表示复平面上从原点到点 (a, b) 的距离。
The conjugate distributes over sum, product, and quotient: (z₁ ± z₂)̄ = z̄₁ ± z̄₂, (z₁z₂)̄ = z̄₁z̄₂, (z₁/z₂)̄ = z̄₁/z̄₂ (z₂ ≠ 0).
共轭对和、积、商可分配：(z₁ ± z₂)̄ = z̄₁ ± z̄₂，(z₁z₂)̄ = z̄₁z̄₂，(z₁/z₂)̄ = z̄₁/z̄₂（z₂ ≠ 0）。
The modulus properties include |z₁z₂| = |z₁||z₂|, |z₁/z₂| = |z₁|/|z₂|, and the triangle inequality |z₁ + z₂| ≤ |z₁| + |z₂|.
模的性质包括 |z₁z₂| = |z₁||z₂|，|z₁/z₂| = |z₁|/|z₂|，以及三角不等式 |z₁ + z₂| ≤ |z₁| + |z₂|。
5. The Argand Diagram | 阿尔冈图
The Argand diagram is a plane where the horizontal axis represents the real part and the vertical axis represents the imaginary part of a complex number.
阿尔冈图是一个平面，其横轴表示复数的实部，纵轴表示复数的虚部。
Each complex number z = a + bi corresponds to a unique point (a, b) or a position vector from the origin to (a, b).
每个复数 z = a + bi 对应唯一一个点 (a, b) 或从原点到 (a, b) 的位置向量。
The distance from the origin to the point is the modulus |z|, and the angle measured from the positive real axis is the argument, denoted arg(z).
从原点到该点的距离是模 |z|，从正实轴测量的角度是辐角，记作 arg(z)。
The principal argument is usually taken in the interval (−π, π] or [0, 2π) depending on convention; CCEA typically uses (−π, π].
主辐角通常取在区间 (−π, π] 或 [0, 2π) 内，CCEA 习惯使用 (−π, π]。
The Argand diagram makes addition of complex numbers visually similar to vector addition, using the parallelogram law.
阿尔冈图使得复数的加法在视觉上类似于向量加法，运用平行四边形法则。
6. Polar Form and Argument | 极坐标形式与辐角
A complex number can be written in polar form as z = r(cos θ + i sin θ), where r = |z| and θ = arg(z).
复数可以写作极坐标形式 z = r(cos θ + i sin θ)，其中 r = |z|，θ = arg(z)。
To convert from Cartesian a + bi to polar form: r = √(a² + b²); θ is found using tan θ = b/a, adjusting the quadrant based on the signs of a and b.
从代数形式 a + bi 转换为极坐标形式：r = √(a² + b²)；θ 通过 tan θ = b/a 求出，并根据 a、b 的符号调整象限。
For example, z = 1 − i: r = √(1² + (−1)²) = √2; θ = arctan(−1/1) = −π/4 (since the point is in the fourth quadrant). So z = √2 (cos(−π/4) + i sin(−π/4)).
例如，z = 1 − i：r = √(1² + (−1)²) = √2；θ = arctan(−1/1) = −π/4（因为点在第四象限）。因此 z = √2 (cos(−π/4) + i sin(−π/4))。
Arguments differing by multiples of 2π represent the same direction, so the principal argument eliminates ambiguity.
相差 2π 整数倍的辐角表示同一方向，因此主辐角消除了歧义。
The form r(cos θ + i sin θ) is essential for multiplication, division, and exponentiation.
形式 r(cos θ + i sin θ) 对于乘法、除法和乘方至关重要。
7. Multiplication and Division in Polar Form | 极坐标形式的乘除法
If z₁ = r₁(cos θ₁ + i sin θ₁) and z₂ = r₂(cos θ₂ + i sin θ₂), then their product is z₁z₂ = r₁r₂ [cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)].
若 z₁ = r₁(cos θ₁ + i sin θ₁) 且 z₂ = r₂(cos θ₂ + i sin θ₂)，则它们的乘积为 z₁z₂ = r₁r₂ [cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)]。
Thus, multiplying complex numbers multiplies their moduli and adds their arguments.
因此，复数相乘，模相乘，辐角相加。
For division, z₁/z₂ = (r₁/r₂) [cos(θ₁ − θ₂) + i sin(θ₁ − θ₂)], provided z₂ ≠ 0. Moduli divide, arguments subtract.
对于除法，z₁/z₂ = (r₁/r₂) [cos(θ₁ − θ₂) + i sin(θ₁ − θ₂)]，其中 z₂ ≠ 0。模相除，辐角相减。
This geometric interpretation makes it easy to compute powers and roots later using De Moivre’s theorem.
这种几何解释使得之后利用棣莫弗定理计算乘方和开方变得简单。
It also explains why multiplying by i corresponds to a rotation by 90° anticlockwise on the Argand diagram.
这还解释了为何乘以 i 对应在阿尔冈图上逆时针旋转 90°。
8. De Moivre’s Theorem | 棣莫弗定理
De Moivre’s theorem states that for any integer n, (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ).
棣莫弗定理指出，对于任意整数 n，(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)。
This can be extended to any real n, but for A-Level, we primarily use integer powers and rational roots.
这可以拓展到任意实数 n，但在 A-Level 中，我们主要使用整数次幂和有理数次方根。
To raise a complex number to a power using De Moivre: write z = r(cos θ + i sin θ), then zⁿ = rⁿ (cos(nθ) + i sin(nθ)).
利用棣莫弗定理求复数的乘方：写出 z = r(cos θ + i sin θ)，则 zⁿ = rⁿ (cos(nθ) + i sin(nθ))。
The theorem is extremely useful for finding trigonometric identities, e.g., expressing cos 3θ in terms of cos θ by expanding (cos θ + i sin θ)³ and equating real parts.
该定理对于求三角恒等式非常有用，例如，通过展开 (cos θ + i sin θ)³ 并比较实部，可以用 cos θ 表示 cos 3θ。
Proof for positive integer n can be done by induction; the result also holds for negative integers by using the reciprocal and the conjugate.
对于正整数 n 的证明可用归纳法完成；通过倒数和共轭，该结果对于负整数同样成立。
9. Finding the nth Roots of a Complex Number | 求复数的 n 次方根
To solve zⁿ = w, where w is a given complex number, write w in polar form: w = r(cos θ + i sin θ).
要求解 zⁿ = w，其中 w 是一个给定的复数，先将 w 写成极坐标形式：w = r(cos θ + i sin θ)。
The n distinct roots are given by zₖ = ⁿ√r [cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)] for k = 0, 1, 2, …, n−1.
n 个不同的根由 zₖ = ⁿ√r [cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)] 给出，其中 k = 0, 1, 2, …, n−1。
Here, ⁿ√r denotes the real positive nth root of r. The principal argument of w is usually used for θ, but any argument differing by 2π yields the same set of roots.
这里 ⁿ√r 表示 r 的正实 n 次方根。w 的主辐角通常用作 θ，但任何相差 2π 的辐角都会产生相同的根集合。
These n roots are equally spaced around a circle of radius ⁿ√r in the complex plane, separated by an angle of 2π/n. They form the vertices of a regular n-gon.
这 n 个根均匀分布在复平面上半径为 ⁿ√r 的圆周上，彼此夹角为 2π/n。它们构成正 n 边形的顶点。
For example, the cube roots of unity (1) are the solutions to z³ = 1: 1, cos(2π/3) + i sin(2π/3), cos(4π/3) + i sin(4π/3), which are 1, −½ + i√3/2, −½ − i√3/2.
例如，单位元的立方根是方程 z³ = 1 的解：1，cos(2π/3) + i sin(2π/3)，cos(4π/3) + i sin(4π/3)，即 1, −½ + i√3/2, −½ − i√3/2。
10. Solving Polynomial Equations with Complex Roots | 解带复根的多项式方程
For polynomial equations with real coefficients, complex roots occur in conjugate pairs. If a + bi is a root, then a − bi is also a root.
对于实系数多项式方程，复根成共轭对出现。如果 a + bi 是一个根，那么 a − bi 也是一个根。
This fact allows us to deduce all roots when one complex root is known, and to factorise the polynomial into real linear and quadratic factors.
这一事实使得已知一个复根时能够推导出所有根，并将多项式分解为实线性因子和二次因子。
For example, if z = 2 + i is a root of a cubic with real coefficients, then 2 − i is also a root. The quadratic factor from these two roots is (z − (2 + i))(z − (2 − i)) = z² − 4z + 5.
例如，如果 z = 2 + i 是一个实系数三次方程的根，那么 2 − i 也是一个根。由这两个根构成的二次因子为 (z − (2 + i))(z − (2 − i)) = z² − 4z + 5。
The fundamental theorem of algebra states that every non-constant polynomial with complex coefficients has at least one complex root, and thus an nth-degree polynomial can be factored into n linear factors over the complex numbers.
代数基本定理指出，每个非常数的复系数多项式至少有一个复根，因此 n 次多项式可以在复数域上分解为 n 个线性因子。
In practical problems, we often use the relationships between roots and coefficients (sum of roots = −b/a, product of roots = ±constant term, etc.) to find unknowns.
在实际问题中，我们常常利用根与系数的关系（根之和 = −b/a，根之积 = ±常数项，等等）来求未知量。
11. Loci in the Complex Plane | 复平面上的轨迹
A locus is a set of points satisfying a given condition. In the complex plane, these conditions are often expressed using modulus and argument.
轨迹是满足给定条件的点的集合。在复平面上，这些条件常通过模和辐角表示。
The equation |z − a| = r represents a circle with centre at the complex number a and radius r.
方程 |z − a| = r 表示以复数 a 为圆心、半径为 r 的圆。
The inequality |z − a| < r describes the interior of that circle, while |z − a| > r describes the exterior.
不等式 |z − a| < r 描述该圆的内部，|z − a| > r 描述其外部。
The equation |z − a| = |z − b| represents the perpendicular bisector of the line segment joining a and b. It is the set of points equidistant from a and b.
方程 |z − a| = |z − b| 表示连接 a 和 b 的线段的垂直平分线。它是到 a 和 b 等距的点的集合。
The argument condition arg(z − a) = θ represents a half-line (ray) emanating from a, making an angle θ with the positive real direction. The point a itself is usually excluded.
辐角条件 arg(z − a) = θ 表示从 a 出发、与正实轴成 θ 角的半直线（射线）。点 a 本身通常被排除。
Combining modulus and argument conditions can describe more complex regions, such as segments, arcs, and annular regions.
结合模和辐角条件可以描述更复杂的区域，例如线段、圆弧和环形区域。
12. Applications and Exam Tips | 应用与应试技巧
Complex numbers are used in CCEA A-Level to solve polynomial equations, prove trigonometric identities, and describe transformations in the plane.
在 CCEA A-Level 中，复数用于求解多项式方程、证明三角恒等式以及描述平面上的变换。
When tackling exam questions, always consider whether polar or Cartesian form is more convenient. Use Cartesian for addition/subtraction and polar for multiplication/division/powers.
处理考题时，始终考虑使用极坐标形式还是代数形式更方便。加减法用代数形式，乘除和乘方用极坐标形式。
Pay careful attention to the argument quadrant. A sketch on the Argand diagram helps avoid sign errors.
要特别注意辐角的象限。在阿尔冈图上画草图有助于避免符号错误。
For roots of unity and similar problems, remember the symmetric geometry: the sum of all nth roots of unity is zero.
对于单位根及类似问题，记住对称几何性质：所有 n 次单位根的和为零。
Memorise the key identities: cos(−θ) = cos θ, sin(−θ) = −sin θ, and the relationship between conjugate and modulus.
记住关键恒等式：cos(−θ) = cos θ，sin(−θ) = −sin θ，以及共轭与模的关系。
Finally, practice interpreting locus descriptions: ‘circle’, ‘perpendicular bisector’, ‘half-line’ are the most common, and using algebraic manipulation to rewrite conditions in a familiar form.
最后，练习解读轨迹描述：“圆”、“垂直平分线”、“半直线”是最常见的，并练习使用代数变形将条件重写为熟悉的形式。
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