📚 IB & CIE Math: Complex Numbers – Key Points | IB CIE 数学：复变函数 考点精讲
Complex numbers form a vital topic in both IB Mathematics Analysis & Approaches HL and CIE A-Level Mathematics (Pure Mathematics 3 and Further Mathematics). They extend the real number system to include solutions of equations such as x² + 1 = 0 and provide powerful algebraic and geometric tools. This article distills the key examination points for complex numbers, guiding you through essential definitions, operations, theorems, and typical problem types.
复数是 IB 数学分析与方法 HL 和 CIE A-Level 数学（纯数学 3 及进阶数学）中的核心主题。它们将实数系扩展到包含 x² + 1 = 0 等方程的根，并提供了强有力的代数与几何工具。本文提炼了复数部分的关键考点，带领你梳理核心定义、运算、定理与常见题型。
1. Introduction to Complex Numbers | 复数介绍
A complex number is written as z = a + bi, where a, b ∈ ℝ and i is the imaginary unit satisfying i² = −1.
复数写作 z = a + bi，其中 a, b 为实数，i 为虚数单位且满足 i² = −1。
The real part is Re(z) = a and the imaginary part is Im(z) = b (not bi).
实部记为 Re(z) = a，虚部记为 Im(z) = b（注意不是 bi）。
Two complex numbers are equal if and only if their real and imaginary parts are respectively equal.
两个复数相等当且仅当它们的实部与虚部分别相等。
a + bi = c + di ⇔ a = c and b = d
2. Algebraic Operations | 代数运算
Addition and subtraction are performed component‑wise: (a + bi) ± (c + di) = (a ± c) + (b ± d)i.
加减法按实部、虚部分别进行：(a + bi) ± (c + di) = (a ± c) + (b ± d)i。
Multiplication expands using ordinary algebra and replaces i² with −1: (a + bi)(c + di) = (ac − bd) + (ad + bc)i.
乘法按常规代数展开并将 i² 换成 −1：(a + bi)(c + di) = (ac − bd) + (ad + bc)i。
Division uses the complex conjugate: multiply numerator and denominator by the conjugate of the denominator to obtain a real denominator.
除法利用共轭复数：分子分母同乘分母的共轭，使分母变为实数。
(a + bi) / (c + di) = [(a + bi)(c − di)] / (c² + d²)
3. Complex Conjugate | 共轭复数
The conjugate of z = a + bi is denoted by z* = a − bi (or z̄). It reflects the complex number across the real axis.
z = a + bi 的共轭复数记作 z* = a − bi（或 z̄），它可看作复数关于实轴的反射。
Key properties: z + z* = 2a (purely real), z − z* = 2bi (purely imaginary), and z z* = a² + b² = |z|² (always real and non‑negative).
重要性质：z + z* = 2a（纯实数），z − z* = 2bi（纯虚数），z z* = a² + b² = |z|²（始终为实数且非负）。
These are extremely useful for simplifying expressions and proving identities.
这些性质在化简表达式和证明恒等式时极为有用。
4. Modulus and Argument | 模与辐角
The modulus of z, |z| = √(a² + b²), gives the distance from the origin in the complex plane.
复数 z 的模 |z| = √(a² + b²)，表示复平面上到原点的距离。
The argument of z, arg(z) = θ, is the directed angle from the positive real axis to the line representing z, typically with tan θ = b/a. Care must be taken to select the correct quadrant.
辐角 arg(z) = θ 是从正实轴到表示 z 的射线的有向角，通常满足 tan θ = b/a，需注意选择正确的象限。
IB and CIE exams normally use the principal argument in the interval (−π, π] or [0, 2π) – check your syllabus convention.
IB 与 CIE 考试通常要求辐角主值在 (−π, π] 或 [0, 2π) 内，请根据考纲确认区间约定。
|z₁ z₂| = |z₁|·|z₂|, arg(z₁ z₂) = arg(z₁) + arg(z₂)
5. Modulus–Argument Form | 模‑辐角形式
Using modulus r and argument θ, any complex number can be expressed in polar form: z = r (cos θ + i sin θ). This representation makes multiplication and division geometric.
利用模 r 和辐角 θ，任何复数可写成极坐标形式：z = r (cos θ + i sin θ)。这种表示使乘除法具有几何意义。
Multiplication: multiply the moduli and add the arguments. Division: divide the moduli and subtract the arguments.
乘法：模相乘，辐角相加；除法：模相除，辐角相减。
z₁ z₂ = r₁ r₂ [cos(θ₁+θ₂) + i sin(θ₁+θ₂)]
This form is the foundation for de Moivre’s theorem and for solving equations like zⁿ = w.
该形式是德莫弗定理以及求解 zⁿ = w 等方程的基础。
6. Euler’s Formula and de Moivre’s Theorem | 欧拉公式与德莫弗定理
Euler’s formula links trigonometry and exponentials: eiθ = cos θ + i sin θ. Hence z = r eiθ.
欧拉公式将三角函数与指数函数联系起来：eiθ = cos θ + i sin θ，从而有 z = r eiθ。
de Moivre’s theorem follows directly: (cos θ + i sin θ)n = cos(nθ) + i sin(nθ) for any integer n.
德莫弗定理直接给出：对任意整数 n，(cos θ + i sin θ)n = cos(nθ) + i sin(nθ)。
This theorem is widely used to find powers of complex numbers, derive trigonometric identities, and compute nth roots.
该定理广泛用于求复数的乘方、推导三角恒等式以及计算 n 次方根。
7. Powers and Roots of Complex Numbers | 复数的乘方与开方
To raise a complex number to an integer power, put it in modulus‑argument form and apply de Moivre.
将复数写成模‑辐角形式并应用德莫弗定理即可求其整数次幂。
For the nth roots of z = r(cos θ + i sin θ), there are exactly n distinct roots given by:
对于 z = r(cos θ + i sin θ) 的 n 次方根，恰好有 n 个相异的根，公式为：
wk = r1/n [ cos((θ + 2πk)/n) + i sin((θ + 2πk)/n) ], k = 0, 1, …, n−1
The roots are equally spaced on a circle of radius r1/n in the complex plane. The special case z = 1 yields the nth roots of unity: e2πik/n.
这些根在复平面上均匀分布于半径为 r1/n 的圆上。特别地，z = 1 给出 n 次单位根：e2πik/n。
Knowing the sum of all nth roots of unity is zero can save time in exam questions.
所有 n 次单位根之和为零，记住这一性质可在考试中节省时间。
8. Polynomials with Real Coefficients | 实系数多项式方程
If a polynomial has real coefficients, any non‑real complex root appears with its complex conjugate. This means roots occur in conjugate pairs.
若多项式系数均为实数，则任一非实复数根必与其共轭成对出现。即复根以共轭对形式存在。
Thus, for a real cubic polynomial, either all three roots are real or there is one real root and one pair of conjugate complex roots.
因此，对于实系数三次多项式，要么三个根均为实数，要么有一个实根和一对共轭复根。
This fact helps factorise polynomials, construct equations from given roots, and find unknown coefficients.
这一事实有助于因式分解、由给定根构造方程以及求解未知系数。
9. Loci in the Complex Plane | 复平面上的轨迹
Loci problems appear frequently in IB and CIE exams. The equation |z − z₀| = r describes a circle with centre z₀ and radius r.
轨迹问题常见于 IB 与 CIE 考题。方程 |z − z₀| = r 表示以 z₀ 为圆心、r 为半径的圆。
|z − z₁| = |z − z₂| represents the perpendicular bisector of the segment joining z₁ and z₂.
|z − z₁| = |z − z₂| 表示连接 z₁ 与 z₂ 线段的垂直平分线。
The half‑line arg(z − z₀) = α (with a condition such as |z − z₀| > 0) is a ray starting at z₀.
arg(z − z₀) = α（并附条件如 |z − z₀| > 0）表示起点为 z₀ 的射线。
Inequalities involving modulus or argument describe regions: for example, |z − (2+3i)| ≤ 5 is a closed disc.
含有模或辐角的不等式描述区域：例如 |z − (2+3i)| ≤ 5 表示一个闭圆盘。
Sketching such loci accurately and carrying out algebraic derivations are essential exam skills.
准确绘制这些轨迹并进行代数推导是必备的应试技能。
10. Applications and Exam Tips | 应用与应试技巧
Use de Moivre to prove trigonometric identities such as expressing cos 3θ in terms of cos θ.
用德莫弗定理证明三角恒等式，例如将 cos 3θ 用 cos θ 表示。
When finding roots of zⁿ = w, write w in modulus‑argument form first, then add 2πk inside the argument to capture all roots.
求解 zⁿ = w 时，应先将 w 化为模‑辐角形式，并在辐角中加 2πk 以覆盖所有根。
Always state the principal argument clearly when required, and check whether your answer needs to be in the interval (−π, π] or [0, 2π).
需要时明确写出辐角主值，并检查你的答案是否在要求的区间 (−π, π] 或 [0, 2π) 内。
For polynomial problems, remember that conjugate pair roots guarantee a real quadratic factor with discriminant negative.
对于多项式问题，记住共轭复根保证可得到一个判别式为负的实系数二次因式。
Simplify calculations by using properties like z z* = |z|² and by working with modulus‑argument form for products and quotients.
利用 z z* = |z|² 等性质并在处理乘除时使用模‑辐角形式以简化计算。
Finally, practise sketching Argand diagrams quickly and accurately – a well‑drawn diagram can often reveal the solution before algebra does.
最后，多练习快速而准确地绘制阿尔冈图——一幅清晰的图示常常能让你在代数推导之前就看出答案。
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