📚 Mastering Complex Numbers & Functions for IB & WJEC | IB WJEC 数学：复变函数 考点精讲
Complex numbers form a cornerstone of advanced mathematics, bridging algebra, trigonometry, and calculus. In both the IB Analysis & Approaches Higher Level and WJEC Further Mathematics specifications, a deep understanding of complex numbers and their functions is essential for solving equations, interpreting loci, and applying transformations. This guide walks you through the key concepts, from algebraic manipulation to De Moivre’s theorem and complex mappings, equipping you with the tools needed for top marks.
复数是高等数学的基石，连接了代数、三角学和微积分。在 IB AA HL 和 WJEC 进阶数学中，深刻理解复数及其函数对于解方程、解释轨迹以及应用变换至关重要。本文将带你从代数运算到棣莫弗定理和复映射，逐一攻克关键概念，助你轻松夺取高分。
1. The Complex Number System | 复数系统
A complex number is any number of the form z = a + bi, where a, b ∈ ℝ and i² = −1. Here a is called the real part Re(z) and b the imaginary part Im(z). The set of all complex numbers is denoted by ℂ.
任何一个形如 z = a + bi 的数均为复数，其中 a, b 为实数，i² = −1。a 称为实部 Re(z)，b 称为虚部 Im(z)。全体复数的集合记作 ℂ。
Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. This simple rule underpins the comparison and equation-solving methods you will use throughout the topic.
两个复数相等当且仅当它们的实部相等且虚部相等。这一简单规则是整个主题中比较与解方程方法的基础。
The imaginary unit i follows the cyclical pattern i = i, i² = −1, i³ = −i, i⁴ = 1, and then repeats. Simplifying powers of i using this cycle is a skill tested frequently at the start of complex number problems.
虚数单位 i 的幂次遵循循环规律：i = i, i² = −1, i³ = −i, i⁴ = 1，之后重复。利用这一循环简化 i 的幂次是复数问题中常考的基本功。
2. Algebraic Operations with Complex Numbers | 复数的代数运算
Addition and subtraction are performed by combining the real parts and the imaginary parts separately: (a + bi) ± (c + di) = (a ± c) + (b ± d)i.
加减法分别合并实部与虚部：(a + bi) ± (c + di) = (a ± c) + (b ± d)i。
Multiplication uses the distributive law and the fact that i² = −1: (a + bi)(c + di) = ac + adi + bci + bdi² = (ac − bd) + (ad + bc)i. Be especially careful with the sign change in the real part.
乘法运用分配律并利用 i² = −1：(a + bi)(c + di) = ac + adi + bci + bdi² = (ac − bd) + (ad + bc)i。实部的符号变化是一处极易出错的地方，务必谨慎。
Division involves multiplying the numerator and denominator by the complex conjugate of the denominator. The conjugate of z = a + bi is z* = a − bi. The product z z* = a² + b² is always a real number, which is used to rationalise the denominator.
除法需要分子分母同乘以分母的共轭复数。z = a + bi 的共轭为 z* = a − bi。乘积 z z* = a² + b² 始终为实数，用于分母有理化。
For example, (2 + i)/(1 − i) = (2 + i)(1 + i)/(1² + 1²) = (2 + 2i + i − 1)/2 = (1 + 3i)/2 = 0.5 + 1.5i.
例如，(2 + i)/(1 − i) = (2 + i)(1 + i)/(1² + 1²) = (2 + 2i + i − 1)/2 = (1 + 3i)/2 = 0.5 + 1.5i。
3. The Argand Diagram and Modulus-Argument Form | 阿干特图与模-幅角形式
The complex number z = a + bi can be represented as the point (a, b) on an Argand diagram, where the x-axis is the real axis and the y-axis is the imaginary axis. This geometric interpretation turns complex arithmetic into vector operations.
复数 z = a + bi 可在阿干特图上表示为点 (a, b)，其中 x 轴为实轴，y 轴为虚轴。这一几何解释将复数运算转化为向量操作。
The modulus of z, denoted |z|, is the distance from the origin to the point: |z| = √(a² + b²). The argument of z, arg z, is the angle θ measured from the positive real axis to the line segment OZ, usually given in the range −π < θ ≤ π.
z 的模 |z| 是原点到该点的距离：|z| = √(a² + b²)。z 的幅角 arg z 是从正实轴到线段 OZ 的夹角 θ，通常取值范围为 −π < θ ≤ π。
In modulus-argument form, z = r(cos θ + i sin θ), where r = |z| and θ = arg z. This representation is pivotal for multiplication, division, and powers. The argument of a product satisfies arg(z₁z₂) = arg z₁ + arg z₂ (mod 2π).
模-幅角形式中，z = r(cos θ + i sin θ)，其中 r = |z|，θ = arg z。这一表示对于乘、除和幂运算至关重要。乘积的幅角满足 arg(z₁z₂) = arg z₁ + arg z₂ (mod 2π)。
To convert from Cartesian to polar form, use r = √(a² + b²) and θ = arctan(b/a) with quadrant adjustments. A sketch on the Argand diagram helps avoid sign errors.
从笛卡尔形式转换为极坐标形式，使用 r = √(a² + b²) 以及 θ = arctan(b/a) 并注意象限调整。在阿干特图上绘图有助于避免符号错误。
4. Euler’s Formula and Exponential Form | 欧拉公式与指数形式
Euler’s formula eiθ = cos θ + i sin θ is a fundamental bridge between trigonometric and exponential functions. It allows us to write a complex number in the compact exponential form z = r eiθ.
欧拉公式 eiθ = cos θ + i sin θ 是连接三角函数与指数函数的基本桥梁，让我们能够将复数写成紧凑的指数形式 z = r eiθ。
In this form, multiplication and division become remarkably simple: z₁z₂ = r₁r₂ ei(θ₁+θ₂) and z₁/z₂ = (r₁/r₂) ei(θ₁−θ₂). The modulus of the product is the product of moduli, and the argument of the product is the sum of arguments.
在这种形式下，乘法和除法变得极为简洁：z₁z₂ = r₁r₂ ei(θ₁+θ₂)，z₁/z₂ = (r₁/r₂) ei(θ₁−θ₂)。乘积的模等于模的乘积，乘积的幅角等于幅角之和。
The exponential form is particularly useful when expressing periodic phenomena and when integrating complex exponentials. It also clarifies the relationship between the roots of unity and regular polygons.
指数形式在表示周期现象以及积分复指数时特别有用。它也清晰揭示了单位根与正多边形之间的联系。
5. 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 theorem is essential for finding powers and roots of complex numbers and for deriving multiple-angle trigonometric identities.
棣莫弗定理指出，对于任何实数 n，(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)。该定理是求复数幂与根以及推导多倍角三角恒等式的关键工具。
Proof for integer n relies on induction and the angle addition formulas; the result can be extended to rational n using the fundamental definition of roots. The theorem also works in exponential form: (r eiθ)ⁿ = rⁿ einθ.
对于整数 n 的证明采用归纳法并依赖角度加法公式；利用根的基本定义可将结果扩展到有理数 n。该定理在指数形式下同样成立：(r eiθ)ⁿ = rⁿ einθ。
A common exam application is to express sin(nθ) or cos(nθ) in terms of powers of sin θ and cos θ by expanding (cos θ + i sin θ)ⁿ and equating real and imaginary parts.
常见的考试应用是通过展开 (cos θ + i sin θ)ⁿ 并匹配实部和虚部，将 sin(nθ) 或 cos(nθ) 用 sin θ 和 cos θ 的幂表示。
6. Roots of Complex Numbers | 复数的根
To find the n-th roots of a complex number z = r(cos θ + i sin θ), use the formula: the k-th root is r1/n [cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)] for k = 0, 1, 2, …, n−1. These n roots lie equally spaced on a circle of radius r1/n centred at the origin.
复数 z = r(cos θ + i sin θ) 的 n 次方根公式为：第 k 个根为 r1/n [cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)]，k = 0, 1, 2, …, n−1。这 n 个根均匀分布在以原点为中心、半径为 r1/n 的圆上。
The roots of unity are the solutions to zⁿ = 1. They are given by ωₙ = cos(2πk/n) + i sin(2πk/n) and form the vertices of a regular n-gon in the Argand diagram. The sum of all n-th roots of unity is zero.
单位根是方程 zⁿ = 1 的解。它们为 ωₙ = cos(2πk/n) + i sin(2πk/n)，在阿干特图中构成正 n 边形的顶点。所有 n 次单位根之和为零。
When solving equations like zⁿ = a + bi, first write the right-hand side in polar form, then apply the root formula. Always check that you list exactly n distinct roots within the principal argument range.
在解形如 zⁿ = a + bi 的方程时，先将右端写成极坐标形式，然后套用求根公式。务必确保列出恰好 n 个不同的根，并在主幅角范围内表示。
7. Loci in the Complex Plane | 复平面中的轨迹
A locus is a set of points that satisfy a given condition. In the complex plane, common loci include circles, perpendicular bisectors, and rays. Interpreting these geometrically saves time and reduces algebraic errors.
轨迹是满足给定条件的点集。在复平面中，常见的轨迹包括圆、垂直平分线和射线。从几何角度解释这些条件能节省时间并减少代数错误。
- Circle: |z − z₀| = r represents a circle with centre z₀ and radius r.
- 圆： |z − z₀| = r 表示以 z₀ 为圆心、r 为半径的圆。
- Perpendicular bisector: |z − z₁| = |z − z₂| is the perpendicular bisector of the line segment joining z₁ and z₂.
- 垂直平分线： |z − z₁| = |z − z₂| 是连接 z₁ 与 z₂ 的线段的垂直平分线。
- Ray: arg(z − z₀) = α gives a half-line starting at z₀ making angle α with the positive real axis.
- 射线： arg(z − z₀) = α 表示一条从 z₀ 出发、与正实轴夹角为 α 的射线。
For inequalities, shading regions becomes intuitive. For example, {z : |z| < 3} is the interior of a circle radius 3, while {z : 0 ≤ arg z ≤ π/2} is the first quadrant including the axes.
对于不等式，区域着色可直观理解。例如 {z : |z| < 3} 是半径为 3 的圆内部，而 {z : 0 ≤ arg z ≤ π/2} 是包含坐标轴的第一象限。
WJEC and IB exam questions frequently ask you to find the Cartesian equation of a locus, sketch it, and determine intersections. The key is to substitute z = x + yi and manipulate the modulus or argument conditions algebraically.
WJEC 和 IB 试题常要求找出轨迹的笛卡尔方程、作图并确定交点。关键步骤是代入 z = x + yi 并对模或幅角条件进行代数操作。
8. Complex Functions and Transformations | 复变函数与变换
A complex function f maps a complex number z to another complex number w = f(z). Simple transformations include translation w = z + c, dilation/rotation w = kz (k ∈ ℂ), and inversion w = 1/z.
复变函数将复数 z 映射为另一个复数 w = f(z)。简单的变换包括平移 w = z + c、缩放/旋转 w = kz (k ∈ ℂ) 以及反演 w = 1/z。
The mapping w = kz scales the modulus by |k| and rotates by arg k. Thus multiplication by i corresponds to a 90° anticlockwise rotation. The mapping w = z + c simply shifts every point by vector c.
映射 w = kz 将模缩放 |k| 倍，并将幅角旋转 arg k。因此乘以 i 等同于逆时针旋转 90°。映射 w = z + c 仅按向量 c 平移所有点。
The reciprocal mapping w = 1/z maps circles and lines to circles or lines (considering lines as circles through infinity). It involves an inversion in the unit circle followed by a reflection in the real axis.
倒数映射 w = 1/z 将圆和直线映射为圆或直线（将直线视为过无穷远点的圆）。它包含了关于单位圆的反演以及关于实轴的反射。
Questions may ask you to find the image of a given locus under a transformation. Substitute z in terms of w and rearrange the condition. For example, under w = 1/z, the circle |z| = 2 becomes the circle |w| = 1/2.
考题可能要求找出给定轨迹在某一变换下的像。可将 z 用 w 表示，并重新整理条件。例如，在 w = 1/z 的映射下，圆 |z| = 2 变为圆 |w| = 1/2。
9. Solving Polynomial Equations in ℂ | 复数域中多项式方程的求解
The Fundamental Theorem of Algebra guarantees that a polynomial of degree n has exactly n roots in ℂ, counted with multiplicity. If the coefficients are real, non-real roots always occur in conjugate pairs.
代数基本定理保证 n 次多项式在复数域中恰好有 n 个根（计重数）。若系数为实数，非实根必以共轭对形式出现。
For a quadratic az² + bz + c = 0 with real coefficients, if the discriminant is negative, the roots are z = [−b ± i√(4ac − b²)]/(2a). These are conjugate pairs.
对于实系数二次方程 az² + bz + c = 0，若判别式为负，则根为 z = [−b ± i√(4ac − b²)]/(2a)。它们是共轭对。
Cubic and quartic equations with real coefficients can be solved by first finding one real root (using factor theorem), then performing polynomial division to obtain a quadratic, and finally applying the quadratic formula.
实系数的三次和四次方程可先通过因式定理找到一实根，然后进行多项式除法得到二次式，最后运用求根公式。
If one complex root is known, its conjugate is also a root, which gives two factors. Use them to reduce the degree. IB and WJEC often set problems where you are given one complex root and must find the others and the original polynomial.
若已知一个复数根，则其共轭也必为根，从而得到两个因子。利用它们可降低次数。IB 和 WJEC 常出题给出一个复数根，要求找出其余根及原始多项式。
10. Trigonometric Identities via De Moivre | 利用棣莫弗定理推导三角恒等式
De Moivre’s theorem gives a powerful method for deriving expressions for cos(nθ) and sin(nθ) in terms of powers of cos θ and sin θ. Expand (cos θ + i sin θ)ⁿ using the binomial theorem, then equate real parts for cos(nθ) and imaginary parts for sin(nθ).
棣莫弗定理为用 cos θ 和 sin θ 的幂表示 cos(nθ) 和 sin(nθ) 提供了有力方法。用二项式定理展开 (cos θ + i sin θ)ⁿ，然后匹配实部得 cos(nθ)，虚部得 sin(nθ)。
Conversely, powers of sin θ and cos θ can be expressed as sums of sines and cosines of multiple angles, which is crucial for integration in further topics like WJEC FP2 or IB HL calculus options.
反之，sin θ 和 cos θ 的幂可表示为多倍角正弦余弦之和，这对于 WJEC FP2 或 IB HL 微积分选修中的积分问题至关重要。
For example, to express cos³θ in multiple angles, write cos³θ as ( (eiθ+e−iθ)/2 )³, expand and group terms to get (1/4)cos3θ + (3/4)cos θ.
例如，将 cos³θ 用多倍角表示，可写为 ( (eiθ+e−iθ)/2 )³，展开并合并项得到 (1/4)cos3θ + (3/4)cos θ。
Common identities like cos2θ = 2cos²θ − 1 or sin3θ = 3sinθ − 4sin³θ are easily derived and can be checked quickly. Practice both directions to master this exam favourite.
诸如 cos2θ = 2cos²θ − 1 或 sin3θ = 3sinθ − 4sin³θ 等常见恒等式容易推导并能快速验证。请双向练习以彻底掌握这一考试热门。
11. Exponential and Trigonometric Forms Connection | 指数形式与三角形式的联系
The identities cos θ = (eiθ + e−iθ)/2 and sin θ = (eiθ − e−iθ)/(2i) allow conversion between trigonometric expressions and complex exponentials. They are extremely useful in proving trigonometric identities and evaluating integrals.
恒等式 cos θ = (eiθ + e−iθ)/2 和 sin θ = (eiθ − e−iθ)/(2i) 使得三角表达式与复指数之间可以相互转换。它们在证明三角恒等式和计算积分时极为有用。
These formulas also define the hyperbolic functions when extended to complex arguments: cos(iz) = cosh z and sin(iz) = i sinh z, linking two seemingly separate worlds. While not always core in IB or WJEC, they appear in advanced complex analysis.
这些公式在推广到复变量时也定义了双曲函数：cos(iz) = cosh z 和 sin(iz) = i sinh z，将两个看似独立的领域联系起来。虽然并非 IB 或 WJEC 的核心，但它们出现在较深的复分析中。
Understanding this duality reinforces the algebraic structure and prepares you for university-level mathematics. Even at the pre-university stage, recognising eiπ + 1 = 0 as a special case of Euler’s identity is a satisfying conclusion to this topic.
理解这种对偶性可巩固代数结构，为大学数学做好准备。即使在预科阶段，认识到 eiπ + 1 = 0 是欧拉恒等式的特例，也是为本主题画上的圆满句号。
12. Exam Techniques and Common Pitfalls | 考试技巧与常见陷阱
Always check for the principal argument range: either (−π, π] or [0, 2π) depending on the specification. IB HL usually uses −π < θ ≤ π, while WJEC may accept either but expects consistency. State your angle convention clearly.
务必检查主幅角范围：视考纲不同，常用 (−π, π] 或 [0, 2π)。IB HL 通常采用 −π < θ ≤ π，WJEC 可能两种均可，但要求一致性。请清晰说明你的角度约定。
When dividing complex numbers, multiply numerator and denominator by the conjugate of the denominator. A common error is to forget that i² = −1, leading to sign mistakes in the final answer.
进行复数除法时，分子分母同乘以分母的共轭。常见错误是忘记 i² = −1，导致最终答案的符号有误。
For locus sketching, first translate the condition into geometrical language rather than expanding Cartesian equations immediately. This speeds up the process and reduces algebraic load.
绘制轨迹时，先将条件转化为几何语言，而不是立即展开为笛卡尔方程。这样能加快解题速度并减少代数负担。
In root-finding questions, check that your answers are indeed the n distinct roots. Using the periodicity 2π correctly is vital; missing roots will lose marks even if the method is sound.
在求根问题中，检查你的答案确为 n 个不同的根。正确利用 2π 的周期性至关重要；遗漏根即便方法正确也会丢分。
Finally, practice past paper questions that combine complex numbers with polynomials, trigonometry, and coordinate geometry. This holistic approach mirrors the integrated style of IB and WJEC assessments.
最后，多练综合复数、多项式、三角与坐标几何的历年真题。这种整体训练方式能完美匹配 IB 和 WJEC 考试的综合题型风格。
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