📚 AQA Mathematics: Complex Numbers – Key Revision Points | AQA 数学：复变函数 考点精讲
This article covers the essential AQA specification content for complex numbers, including Cartesian and polar forms, modulus-argument calculations, de Moivre’s theorem, roots of complex numbers, polynomial equations, loci, and transformations of the complex plane. Every section pairs concise English explanations with their Chinese translations to help you master both terminology and problem-solving techniques.
本文涵盖 AQA 考纲中复变函数（复数）板块的核心考点：代数形式与极坐标形式、模-辐角运算、棣莫弗定理、复数根、多项式方程、Argand 图中的轨迹以及复平面变换。每个部分都采用中英对照讲解，帮助你精准掌握术语与解题思路。
1. Introduction to Complex Numbers | 复数的引入
A complex number is any number of the form z = x + i y, where x and y are real numbers and i = √−1 is the imaginary unit. The set of all complex numbers is denoted by ℂ. The real part of z is Re(z) = x and the imaginary part is Im(z) = y.
复数是指形如 z = x + i y 的数，其中 x 和 y 为实数，i = √−1 为虚数单位。所有复数构成的集合记作 ℂ。z 的实部为 Re(z) = x，虚部为 Im(z) = y。
nly if both their real parts and their imaginary parts are identical. The number 0 = 0 + 0i is the additive identity, and 1 = 1 + 0i is the multiplicative identity.
两个复数相等当且仅当它们的实部相等且虚部相等。0 = 0 + 0i 是加法单位元，1 = 1 + 0i 是乘法单位元。
2. Cartesian Form and Basic Operations | 代数形式与基本运算
Given z = a + i b and w = c + i d, addition and subtraction are performed component‑wise:
z + w = (a + c) + i(b + d) and z − w = (a − c) + i(b − d).
设 z = a + i b，w = c + i d，则加法与减法分别把实部和虚部相加/相减：z + w = (a + c) + i(b + d)，z − w = (a − c) + i(b − d)。
Multiplication treats i as an algebraic symbol but replaces i² with −1:
z × w = (a + i b)(c + i d) = ac + i ad + i bc + i²bd = (ac − bd) + i(ad + bc).
乘法时将 i 视作代数符号并利用 i² = −1：z × w = (ac − bd) + i(ad + bc)。
Be careful when simplifying powers of i: i² = −1, i³ = −i, i⁴ = 1, and the cycle repeats every four powers.
简化 i 的幂时要留意周期：i² = −1，i³ = −i，i⁴ = 1，此后每四次循环一次。
3. Complex Conjugate and Division | 共轭复数与除法
The complex conjugate of z = a + i b is z̅ = a − i b. Conjugation reflects a point across the real axis on the Argand diagram. Key properties: z + z̅ = 2 Re(z), z − z̅ = 2 i Im(z), and z × z̅ = a² + b², which is always a non‑negative real number.
复数 z = a + i b 的共轭为 z̅ = a − i b。共轭在 Argand 图中关于实轴作镜像对称。重要性质：z + z̅ = 2 Re(z)，z − z̅ = 2 i Im(z)，以及 z × z̅ = a² + b²，这是一个非负实数。
To divide two complex numbers, multiply the numerator and denominator by the conjugate of the denominator:
(a + i b) ÷ (c + i d) = [(a + i b)(c − i d)] / (c² + d²).
复数除法时，将分子分母同乘以分母的共轭复数：(a + i b) ÷ (c + i d) = [(a + i b)(c − i d)] / (c² + d²)。
4. The Argand Diagram | Argand 图
An Argand diagram represents complex numbers as points in a plane: the horizontal axis is the real axis (Re) and the vertical axis is the imaginary axis (Im). The point z = x + i y has coordinates (x, y). This visualisation turns algebraic problems into geometric ones, which is extremely useful for loci and transformations.
Argand 图将复数表示为平面上的点：横轴为实轴（Re），纵轴为虚轴（Im）。点 z = x + i y 对应坐标 (x, y)。这种可视化方法将代数问题转化为几何问题，在轨迹和变换题型中尤其有用。
The distance from the origin to the point is the modulus, and the angle measured from the positive real axis is the argument. These polar coordinates lay the foundation for the modulus‑argument form.
点到原点的距离是模，从正实轴出发的转角是辐角。这组极坐标构成了模‑辐角形式的基础。
5. Modulus and Argument | 模与辐角
For z = x + i y, the modulus is |z| = √x² + y². It gives the distance from the origin on the Argand diagram. The argument, denoted arg(z), is the angle θ measured anticlockwise from the positive real axis to the line joining the origin to z. The principal argument usually lies in (−π, π].
对于 z = x + i y，模为 |z| = √x² + y²，表示 Argand 图上点到原点的距离。辐角记作 arg(z)，是从正实轴按逆时针方向旋转到原点与该点连线的角度。主辐角通常取 (−π, π] 区间。
To find the argument, use tan θ = y / x but always consider the quadrant in which the point lies. A common mistake is to ignore the sign of x and y and just use arctan on a calculator.
求辐角时利用 tan θ = y / x，但必须根据点所在的象限进行修正。常见错误是不顾 x 与 y 的符号，直接用计算器的 arctan 结果。
6. Polar and Exponential Forms | 极坐标形式与指数形式
Using modulus r and argument θ, a complex number can be written in polar form:
z = r (cos θ + i sin θ).
利用模 r 和辐角 θ，复数可表示成极坐标形式：z = r (cos θ + i sin θ)。
By Euler’s formula, eiθ = cos θ + i sin θ, so we obtain the compact exponential form:
z = r eiθ.
根据欧拉公式 eiθ = cos θ + i sin θ，得到更加简洁的指数形式：z = r eiθ。
These forms make multiplication, division, and exponentiation much more straightforward than working in Cartesian coordinates. They are also essential for applying de Moivre’s theorem and finding roots.
这两种形式让乘除运算及乘方远优于直角坐标形式，也是应用棣莫弗定理和求根运算的核心工具。
7. Multiplication and Division in Polar Form | 极坐标形式下的乘除法
If z₁ = r₁ eiθ₁ and z₂ = r₂ eiθ₂, then:
z₁ × z₂ = r₁r₂ ei(θ₁ + θ₂) (moduli multiply, arguments add).
若 z₁ = r₁ eiθ₁, z₂ = r₂ eiθ₂，则 z₁ × z₂ = r₁r₂ ei(θ₁ + θ₂)（模相乘，辐角相加）。
For division:
z₁ ÷ z₂ = (r₁ / r₂) ei(θ₁ − θ₂) (moduli divide, arguments subtract).
除法规则：z₁ ÷ z₂ = (r₁ / r₂) ei(θ₁ − θ₂)（模相除，辐角相减）。
Geometrically, multiplying by a complex number w = r eiθ scales the original point by factor r and rotates it anticlockwise by angle θ. This geometric interpretation is frequently tested in transformation questions.
几何上，乘以一个复数 w = r eiθ 相当于将原向径缩放 r 倍并逆时针旋转 θ 角。这种几何解释在变换题中经常出现。
8. De Moivre’s Theorem | 棣莫弗定理
De Moivre’s theorem states that for any real number θ and integer n:
(cos θ + i sin θ)n = cos(nθ) + i sin(nθ).
棣莫弗定理指出，对任意实数 θ 和整数 n，有 (cos θ + i sin θ)n = cos(nθ) + i sin(nθ)。
This is a direct consequence of (eiθ)n = ei nθ. It is valid for negative integers as well, provided we interpret (cos θ + i sin θ)−n as the reciprocal raised to the positive power.
这一定理源于 (eiθ)n = ei nθ。它对负整数也成立，只需将 (cos θ + i sin θ)−n 理解为取倒数后再进行正整数次幂。
Typical exam uses include evaluating powers of complex numbers, expressing sin nθ or cos nθ in terms of powers of sin θ and cos θ, and simplifying expressions like (1 + i√3)⁶.
典型考题包括求复数的幂、将 sin nθ 或 cos nθ 用 sin θ、cos θ 的幂表示，以及化简如 (1 + i√3)⁶ 的算式。
When using de Moivre, always express the complex number in exact polar or exponential form first. Write the modulus as an integer power and then multiply the argument by n.
使用棣莫弗定理时，应先将复数写成精确的极坐标或指数形式，将模取 n 次方，再将辐角乘以 n。
9. Roots of Complex Numbers | 复数的 n 次根
To find the nth roots of a complex number z = r eiθ, set n√z = r1/n ei(θ + 2πk)/n for k = 0, 1, 2, …, n−1. Each value of k gives a distinct root. All n roots lie equally spaced on a circle of radius r1/n centred at the origin.
求复数 z = r eiθ 的 n 次根时，公式为 n√z = r1/n ei(θ + 2πk)/n，k = 0, 1, …, n−1。每个 k 值对应一个不同的根，所有根均匀分布在以原点为圆心、半径为 r1/n 的圆周上。
For example, the cube roots of 8 are 2 ei(0), 2 ei(2π/3), and 2 ei(4π/3), which correspond to 2, −1+i√3, and −1−i√3. The roots of unity (zn = 1) are especially common and produce regular polygons on the unit circle.
例如，8 的立方根为 2 ei(0)、2 ei(2π/3)、2 ei(4π/3)，即 2、−1+i√3 和 −1−i√3。单位根（zn = 1）尤为常见，它们在单位圆上构成正多边形。
In exams, always list all roots in the required form (Cartesian or exponential) and sketch their positions on an Argand diagram if asked. Remember that the principal root may not be the one with the smallest argument; check the specification’s definition.
考试中需列出所有根（按指定形式），若要求画图，需在 Argand 图上标出它们的位置。注意主根不一定对应最小辐角，应根据考纲定义来确定。
10. Solving Polynomial Equations | 求解多项式方程
AQA questions often ask you to solve cubic or quartic equations with real coefficients, where one complex root is given. The key fact is: if z = α is a root, then its conjugate z̅ = α̅ is also a root (non‑real roots occur in conjugate pairs). This allows you to factorise the polynomial into linear and quadratic factors.
AQA 常考实系数三次或四次方程，其中已知一个复数根。核心事实：若 z = α 是一个根，则其共轭 z̅ = α̅ 也是根（非实根成对共轭出现）。由此可将多项式分解为一次与二次因式。
For example, given that 2 + i is a root of z³ − 5z² + 9z − 5 = 0, we know 2 − i is also a root. Multiply the corresponding factors (z − (2+i))(z − (2−i)) = z² − 4z + 5, then divide the original cubic by this quadratic to find the remaining real root.
例如，已知 2 + i 是方程 z³ − 5z² + 9z − 5 = 0 的一个根，则 2 − i 也是根。将因式 (z − (2+i))(z − (2−i)) = z² − 4z + 5 相乘，再用原三次式除以该二次式即得剩余实根。
If the polynomial has complex coefficients, the conjugate root theorem does not apply. Instead, you may need to substitute z = x + i y and equate real and imaginary parts, or use algebraic manipulation.
若多项式系数本身为复数，共轭根定理不再适用。此时可设 z = x + i y 代入并令实部、虚部分别为零，或通过代数技巧求解。
11. Loci in the Argand Diagram | Argand 图中的轨迹
Loci are sets of points satisfying a given condition and are best handled by replacing z with x + i y or by interpreting modulus and argument geometrically. Common loci include:
轨迹是满足特定条件的点集，解题时可将 z 代换为 x + i y 或运用模与辐角的几何意义。常见轨迹包括：
- Circle: |z − a| = r (centre a, radius r).
圆：|z − a| = r（圆心 a，半径 r）。 - Perpendicular bisector: |z − a| = |z − b| (points equidistant from a and b).
垂直平分线：|z − a| = |z − b|（与 a、b 等距的点）。 - Half‑line: arg(z − a) = θ (ray starting at a, making angle θ with the positive real axis).
射线（半直线）：arg(z − a) = θ（以 a 为端点，与正实轴夹角为 θ 的射线）。 - Annulus / disc: r₁ ≤ |z − a| ≤ r₂.
圆环/圆盘：r₁ ≤ |z − a| ≤ r₂。
When sketching, always indicate whether the boundaries are included (solid line) or excluded (dashed line). Interpreting inequalities involving the argument requires particular care with the branch cut.
作图时需明确边界是否包含（实线）或不包含（虚线）。涉及辐角不等式时，要特别注意辐角分支切割线的
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