IB CCEA Mathematics: Complex Numbers & Functions – Core Revision Guide | IB CCEA 数学：复变函数考点精讲

📚 IB CCEA Mathematics: Complex Numbers & Functions – Core Revision Guide | IB CCEA 数学：复变函数考点精讲

Welcome to this comprehensive revision guide on complex numbers and functions, tailored specifically for the IB and CCEA examination specifications. In this article, you will find clear explanations of every key topic, from algebraic operations to De Moivre’s theorem and loci, together with carefully chosen examples and common pitfalls. Mastering these ideas will build a solid foundation for tackling both routine and challenging paper questions.

欢迎阅读这份针对 IB 与 CCEA 考试大纲的复变函数考点精讲。本文涵盖了从代数运算到棣莫弗定理和轨迹在内的每一个核心主题，配有精选例题和常见错误分析。掌握这些内容将为顺利解答试卷中的基础题和难题打下坚实基础。

1. Introduction to Complex Numbers | 复数简介

The need for complex numbers arises when we try to solve equations such as x² + 1 = 0, which have no solution in the set of real numbers. We define the imaginary unit i by i² = -1. A complex number z can then be written as z = a + bi, where a, b ∈ ℝ. The set of all complex numbers is denoted by ℂ.

当我们试图求解在实数范围内无解的方程（例如 x² + 1 = 0）时，复数便应运而生。我们定义虚数单位 i，满足 i² = -1。复数 z 可以写作 z = a + bi，其中 a 和 b 均为实数。所有复数构成的集合记作 ℂ。

2. Algebraic Form and Operations | 代数形式与四则运算

Given two complex numbers z₁ = a + bi and z₂ = c + di, we define addition, subtraction and multiplication as follows:

给定两个复数 z₁ = a + bi 和 z₂ = c + di，其加法、减法和乘法定义如下：

Addition: (a + bi) + (c + di) = (a + c) + (b + d)i
加法：(a + bi) + (c + di) = (a + c) + (b + d)i
Subtraction: (a + bi) – (c + di) = (a – c) + (b – d)i
减法：(a + bi) – (c + di) = (a – c) + (b – d)i
Multiplication: (a + bi)(c + di) = (ac – bd) + (ad + bc)i, using i² = -1.
乘法：(a + bi)(c + di) = (ac – bd) + (ad + bc)i，利用了 i² = -1。

For division, we multiply numerator and denominator by the complex conjugate of the denominator to obtain a real denominator.

在除法中，我们通过将分子和分母同时乘以分母的共轭复数，使分母变为实数。

3. Complex Conjugate and Modulus | 共轭复数与模

If z = a + bi, its complex conjugate is denoted by z̄ = a – bi. The conjugate is obtained by changing the sign of the imaginary part. Geometrically, it is a reflection of z across the real axis.

如果 z = a + bi，它的共轭复数记作 z̄ = a – bi。共轭复数通过改变虚部符号得到。从几何角度看，它是 z 关于实轴的镜像。

The modulus of z, written |z|, is the distance from the origin to the point (a, b) in the complex plane: |z| = √(a² + b²). Key properties include |z̅| = |z|, z·z̅ = |z|², and |z₁z₂| = |z₁|·|z₂|.

复数 z 的模，记为 |z|，是复平面上原点到点 (a, b) 的距离：|z| = √(a² + b²)。重要性质包括 |z̅| = |z|、z·z̅ = |z|² 以及 |z₁z₂| = |z₁|·|z₂|。

4. Argument and Principal Argument | 辐角与主辐角

The argument of a non-zero complex number z, written arg(z), is the angle θ formed by the positive real axis and the line segment from the origin to z. The angle is measured anticlockwise. The principal argument, denoted Arg(z), usually lies in the interval (-π, π] or [0, 2π) depending on the convention used in your syllabus – be sure to check the specification.

非零复数 z 的辐角，记作 arg(z)，是正实轴与原点到 z 的连线所成的角 θ，按逆时针方向度量。主辐角 Arg(z) 通常取在区间 (-π, π] 或 [0, 2π)，具体取决于教学大纲的规定，请务必查阅考试说明。

For z = a + bi, tan θ = b/a, but you must determine the correct quadrant. For example, if a < 0 and b > 0, θ is in the second quadrant, so θ = π + arctan(b/a) (or 180° + arctan(b/a)).

对于 z = a + bi，tan θ = b/a，但需要正确判断象限。例如，当 a < 0 且 b > 0 时，θ 位于第二象限，因此 θ = π + arctan(b/a)（或 180° + arctan(b/a)）。

5. Polar Form and Euler’s Formula | 极坐标形式与欧拉公式

A complex number can be expressed in polar form using its modulus r and argument θ: z = r(cos θ + i sin θ). This is extremely useful for multiplication, division, and exponentiation. Euler’s formula states that e^iθ = cos θ + i sin θ, giving the compact exponential form z = re^iθ.

复数可以用模 r 和辐角 θ 表示为极坐标形式：z = r(cos θ + i sin θ)。这对于乘法、除法和乘方极其有用。欧拉公式表明 e^iθ = cos θ + i sin θ，从而得到了简洁的指数形式 z = re^iθ。

Multiplication in polar form: if z₁ = r₁e^iθ₁ and z₂ = r₂e^iθ₂, then z₁z₂ = r₁r₂e^{i(θ₁+θ₂)}. Division gives (r₁/r₂)e^{i(θ₁-θ₂)}. These rules greatly simplify problems involving powers and roots.

极坐标形式下的乘法：若 z₁ = r₁e^iθ₁ 且 z₂ = r₂e^iθ₂，则 z₁z₂ = r₁r₂e^{i(θ₁+θ₂)}。除法则得到 (r₁/r₂)e^{i(θ₁-θ₂)}。这些法则极大地简化了涉及乘方和开方的问题。

6. De Moivre’s Theorem | 棣莫弗定理

De Moivre’s theorem states that for any real number n, (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ). This can be extended to rational n when finding roots, but careful treatment of multiple values is required. The theorem can be proved by mathematical induction for integer n or derived from Euler’s formula.

棣莫弗定理指出，对于任意实数 n，有 (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)。在求方根时，该定理可推广到有理数 n，但需要小心处理多值的情况。该定理对整数 n 可用数学归纳法证明，或从欧拉公式推导得到。

A typical exam question asks you to express sin 3θ in terms of sin θ using De Moivre’s theorem: expand (cos θ + i sin θ)³, then equate imaginary parts of the expanded binomial with sin 3θ.

典型的考题会要求你利用棣莫弗定理将 sin 3θ 用 sin θ 表示：展开 (cos θ + i sin θ)³，然后将二项展开式的虚部与 sin 3θ 对应相等。

7. 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:

求复数 z = r(cos θ + i sin θ) 的 n 次方根时，使用以下公式：

z_k = r^1/n [cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)] for k = 0, 1, …, n-1

z_k = r^1/n [cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)]，其中 k = 0, 1, …, n-1

Geometrically, these n roots lie equally spaced on a circle of radius r^1/n centred at the origin. The difference between successive arguments is 2π/n. This symmetry is often exploited in sketching or solving polynomial equations.

从几何上看，这 n 个方根均匀分布在以原点为圆心、半径为 r^1/n 的圆上。相邻辐角之差为 2π/n。这一对称性常用于作图和求解多项式方程。

8. Loci in the Complex Plane | 复平面上的轨迹

Loci problems test your ability to translate geometric conditions into complex equations. The most common forms are:

轨迹问题考察你将几何条件转化为复数方程的能力。最常见的形式有：

|z – a| = r: a circle with centre a and radius r.
|z – a| = r: 以 a 为圆心、半径为 r 的圆。
|z – a| = |z – b|: the perpendicular bisector of the segment joining a and b.
|z – a| = |z – b|: 连接 a 与 b 的线段的垂直平分线。
arg(z – a) = θ: a half-line from a making angle θ with the positive real axis.
arg(z – a) = θ: 从 a 出发、与正实轴成角 θ 的射线。

Combining these conditions with inequalities leads to regions, such as shaded segments, discs or sectors. Always draw a diagram and interpret ‘less than’ or ‘greater than’ carefully.

将这些条件与不等式结合便得到区域，例如阴影扇形、圆盘或扇区。务必先画草图，并仔细理解“小于”和“大于”的含义。

9. Solving Polynomial Equations | 解多项式方程

Complex numbers allow us to solve any polynomial equation. The Fundamental Theorem of Algebra guarantees that every non-constant polynomial with complex coefficients has at least one complex root. Real polynomials have complex roots in conjugate pairs: if a + bi is a root, so is a – bi.

复数使我们能够求解任意多项式方程。代数基本定理保证了每个非常数的复系数多项式至少有一个复数根。实系数多项式的复数根成共轭对出现：如果 a + bi 是根，那么 a – bi 也是根。

When solving cubics or quartics, given one complex root, you can find its conjugate and then factorise the polynomial using (z – (a+bi))(z – (a-bi)) = z² – 2a z + (a²+b²), which is a real quadratic factor.

在求解三次或四次方程时，若已知一个复数根，可先找出其共轭根，然后利用 (z – (a+bi))(z – (a-bi)) = z² – 2a z + (a²+b²) 这一实二次因式进行因式分解。

10. Exponential and Trigonometric Functions of Complex Numbers | 复数的指数函数与三角函数

Using Euler’s formula, we define the complex exponential function e^z for z = x + iy as e^x(cos y + i sin y). This satisfies the usual index laws and is periodic with period 2πi. The complex trigonometric functions are then defined by:

借助欧拉公式，可将复数指数函数 e^z（其中 z = x + iy）定义为 e^x(cos y + i sin y)。它满足通常的指数律，且以 2πi 为周期。在此基础上定义复数三角函数：

cos z = (e^iz + e^-iz) / 2, sin z = (e^iz – e^-iz) / (2i)

cos z = (e^iz + e^-iz) / 2，sin z = (e^iz – e^-iz) / (2i)

These definitions lead to identities such as cos(iy) = cosh y and sin(iy) = i sinh y, linking trigonometric and hyperbolic functions. While this material may appear in some advanced IB/CCEA modules, check your syllabus for the required depth.

这些定义衍生出如 cos(iy) = cosh y 和 sin(iy) = i sinh y 的恒等式，将三角函数与双曲函数联系起来。虽然这部分内容可能出现在某些 IB / CCEA 的高级模块中，请依据教学大纲确认所需掌握的深度。

11. Applications and Worked Examples | 应用与典型例题

Example 1: Given z = 1 + √3 i, express z in polar form. Solution: |z| = 2, arg(z) = π/3, so z = 2(cos(π/3) + i sin(π/3)) = 2e^iπ/3.

例题 1：已知 z = 1 + √3 i，将 z 表示为极坐标形式。解：|z| = 2，arg(z) = π/3，因此 z = 2(cos(π/3) + i sin(π/3)) = 2e^iπ/3。

Example 2: Find the four fourth roots of -16. Write -16 = 16(cos π + i sin π). Roots: 2[cos((π+2πk)/4) + i sin((π+2πk)/4)] for k=0,1,2,3, giving 2e^iπ/4, 2e^i3π/4, 2e^i5π/4, 2e^i7π/4.

例题 2：求 -16 的四个四次方根。将 -16 写成 16(cos π + i sin π)。方根：2[cos((π+2πk)/4) + i sin((π+2πk)/4)]，k=0,1,2,3，得到 2e^iπ/4、2e^i3π/4、2e^i5π/4、2e^i7π/4。

Example 3: Sketch the locus |z – 2i| = 3. This is a circle centred at 2i with radius 3.

例题 3：画出轨迹 |z – 2i| = 3。这是一个以 2i 为圆心、半径为 3 的圆。

12. Common Mistakes and Exam Tips | 常见错误与考试技巧

Mistake 1: Forgetting to adjust the argument for the correct quadrant. Always sketch the complex number on an Argand diagram to determine the principal argument.

错误 1：忘记根据象限调整辐角。务必在阿尔冈图上画出复数，以确定主辐角。

Mistake 2: Confusing z̅ with -z. The conjugate reflects across the real axis, while -z is a rotation of π (half-turn) about the origin.

错误 2：混淆 z̅ 和 -z。共轭是关于实轴的反射，而 -z 是绕原点旋转 π（半圈）。

Mistake 3: When finding roots, only giving one root. Remember that an nth root has n distinct values unless stated otherwise.

错误 3：求方根时只写出一个根。记住，n 次方根有 n 个不同的值，除非题目另有说明。

Exam Tip: Present your working step by step. In loci questions, clearly state the geometric interpretation and draw a diagram even if the question doesn’t explicitly ask for it – it can earn method marks and prevent errors.

考试技巧：逐步展示解题过程。在轨迹问题中，即使题目未明确要求，也要清晰写出几何意义并画出草图，这既能赢得过程分，又能避免错误。

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IB CCEA Mathematics: Complex Numbers & Functions – Core Revision Guide | IB CCEA 数学：复变函数 考点精讲