Mastering Complex Functions for IGCSE CCEA Mathematics | IGCSE CCEA 数学：复变函数考点精讲

📚 Mastering Complex Functions for IGCSE CCEA Mathematics | IGCSE CCEA 数学：复变函数考点精讲

Complex functions, built on the foundations of complex numbers, are a challenging yet rewarding topic in CCEA IGCSE Mathematics. This guide breaks down every essential concept, from the imaginary unit to solving quadratic equations with complex roots, ensuring you are fully prepared for the exam.

复变函数以复数为基础，是 CCEA IGCSE 数学中颇具挑战但收获颇丰的课题。本指南将拆解每一个核心概念，从虚数单位到求解带复数根的二次方程，确保你为考试做好充分准备。

1. Introduction to Complex Numbers | 复数概述

Real numbers alone cannot provide solutions to equations such as x² = -1. To overcome this limitation, mathematicians introduced the set of complex numbers, which extends the real number system and guarantees that every polynomial equation has a root.

仅靠实数无法为 x² = -1 这样的方程提供解。为突破这一局限，数学家引入了复数集，拓展了实数系，并确保每一个多项式方程都有根。

In CCEA IGCSE further pure topics, complex numbers allow you to handle algebraic expressions that would otherwise have no meaning in the real world. They are the gateway to understanding advanced functions and transformations.

在 CCEA IGCSE 进阶纯数专题中，复数使你能够处理在实数范围内无意义的代数表达式。它们是理解高级函数与变换的门户。

2. The Imaginary Unit i | 虚数单位 i

The imaginary unit i is defined as the principal square root of -1. This single definition unlocks an entirely new number system.

虚数单位 i 定义为 -1 的主平方根。仅凭这一条定义，就开启了一个全新的数系。

i = √(-1)

A direct consequence is that i² = -1. You must remember this relation because it is used constantly when simplifying expressions involving i.

由此直接得出 i² = -1。你必须牢记这一关系，因为在化简含 i 的表达式时会频繁用到。

Unlike real numbers, i does not represent a quantity on the ordinary number line; it exists on a separate axis, giving rise to two-dimensional representations.

与实数不同，i 并不代表普通数轴上的量；它存在于独立的轴上，从而产生了二维表示法。

3. Complex Number Notation | 复数表示法

A complex number is expressed in standard rectangular form as z = a + bi, where a and b are real numbers. The value a is called the real part, denoted Re(z), and b is the imaginary part, denoted Im(z).

复数以标准直角形式表示为 z = a + bi，其中 a 和 b 为实数。a 称作实部，记作 Re(z)；b 称作虚部，记作 Im(z)。

For example, in z = 3 – 2i, Re(z) = 3 and Im(z) = -2. Notice that the imaginary part includes the coefficient only, without the i.

例如，对于 z = 3 – 2i，Re(z) = 3，Im(z) = -2。注意虚部仅指系数，不含 i。

A purely real number has b = 0, while a purely imaginary number has a = 0. Both are special cases of complex numbers.

纯实数满足 b = 0，纯虚数满足 a = 0。两者都是复数的特例。

4. Addition and Subtraction | 加法与减法

Adding and subtracting complex numbers is straightforward: simply combine the real parts and the imaginary parts separately.

复数的加减法十分简单：分别合并实部和虚部即可。

(a + bi) + (c + di) = (a + c) + (b + d)i

For subtraction, distribute the negative sign before combining: (a + bi) – (c + di) = (a – c) + (b – d)i.

做减法时，先分配负号再合并：(a + bi) – (c + di) = (a – c) + (b – d)i。

Example: (5 + 4i) + (2 – 7i) = 7 – 3i. Always present your final answer in the form a + bi.

例如：(5 + 4i) + (2 – 7i) = 7 – 3i。始终将最终答案写成 a + bi 的形式。

5. Multiplication | 乘法

Multiply complex numbers as you would multiply two binomials, then simplify by replacing every i² with -1.

像乘二项式那样乘复数，然后将每个 i² 替换为 -1 化简。

(a + bi)(c + di) = ac + adi + bci + bdi² = (ac – bd) + (ad + bc)i

For instance, calculate (2 + 3i)(1 – 2i): expand to get 2 – 4i + 3i – 6i². Since i² = -1, the term –6i² becomes +6. Combine: 2 + 6 = 8 and –4i + 3i = –i, giving 8 – i.

例如，计算 (2 + 3i)(1 – 2i)：展开得 2 – 4i + 3i – 6i²。由于 i² = -1，项 –6i² 变为 +6。合并：2 + 6 = 8，–4i + 3i = –i，结果为 8 – i。

Carefully handle the signs when simplifying; a common mistake is to forget that the product of the imaginary terms yields a negative real component.

化简时小心处理符号；常见错误是忘记虚数项的乘积会产生负的实部。

6. Complex Conjugate | 共轭复数

The complex conjugate of z = a + bi is denoted by z* or a bar over z, and is defined as z* = a – bi. It reflects the number across the real axis.

复数 z = a + bi 的共轭记作 z* 或 z 上加横线，定义为 z* = a – bi。它在实轴另一侧镜像反映。

z z* = (a + bi)(a – bi) = a² + b²

Note that the product always results in a real number. This property is essential for division and for finding the modulus.

注意此乘积总是实数。这一性质对除法和求模运算至关重要。

Conjugates are also used when solving polynomial equations: if a complex number is a root, its conjugate is also a root, provided the coefficients are real.

解多项式方程时也用到共轭：若系数为实数，复数根成对出现，其共轭也是根。

7. Division | 除法

To divide one complex number by another, multiply both the numerator and the denominator by the conjugate of the denominator. This eliminates the imaginary part from the denominator.

两个复数相除时，分子分母同乘分母的共轭。这样可消去分母中的虚数部分。

(a + bi) / (c + di) = [(a + bi)(c – di)] / (c² + d²)

Example: (3 + 2i) / (1 – i). Multiply top and bottom by (1 + i): (3+2i)(1+i) = 3 + 3i + 2i + 2i² = 3 + 5i – 2 = 1 + 5i. Denominator becomes 1² + 1² = 2. The quotient is (1/2) + (5/2)i.

举例：(3 + 2i) / (1 – i)。分子分母同乘 (1 + i)：(3+2i)(1+i) = 3+3i+2i+2i² = 3+5i–2 = 1+5i。分母变为 1²+1² = 2。商为 (1/2) + (5/2)i。

Always write the final answer as separate real and imaginary parts, using fractions if necessary.

最终答案务必写成独立的实部和虚部，必要时使用分数。

8. Modulus and Argument | 模与辐角

The modulus of z = a + bi is the distance from the origin in the complex plane. It is denoted |z| and calculated using Pythagoras’ theorem.

复数 z = a + bi 的模是从原点到该点的距离。记作 |z|，用勾股定理计算。

|z| = √(a² + b²)

The argument of z, denoted arg(z) or θ, is the angle made with the positive real axis. It is usually measured in radians, and you can find it using tan θ = b/a, taking care to select the correct quadrant.

辐角记作 arg(z) 或 θ，是与正实轴形成的夹角。通常以弧度为单位，可通过 tan θ = b/a 求得，并注意选取正确象限。

For example, z = 1 – i has modulus |z| = √(1² + (-1)²) = √2, and argument θ = –π/4 (or 7π/4) because the point lies in the fourth quadrant.

例如 z = 1 – i，模为 |z| = √(1² + (-1)²) = √2，辐角 θ = –π/4（或 7π/4），因为点位于第四象限。

9. Argand Diagram | 阿根图

The Argand diagram represents complex numbers as points on a plane, with the horizontal axis as the real part and the vertical axis as the imaginary part. This visualisation helps in understanding operations geometrically.

阿根图将复数表示为平面上的点，水平轴为实部，垂直轴为虚部。这种可视化有助于从几何角度理解运算。

Plotting z = a + bi yields the point (a, b). The modulus is the length of the vector from the origin, and the argument is its direction.

绘制 z = a + bi 得到点 (a, b)。模是原点到该点的向量长度，辐角是其方向。

Addition corresponds to vector addition, and multiplication by i rotates a point by 90° anticlockwise. These geometric interpretations often simplify problem‑solving in CCEA exam questions.

加法对应向量加法，乘以 i 使点逆时针旋转 90°。这些几何意义常能简化 CCEA 考题的求解。

10. Solving Quadratic Equations | 解二次方程

When the discriminant Δ = b² – 4ac is negative, the quadratic equation ax² + bx + c = 0 has two complex conjugate roots.

当判别式 Δ = b² – 4ac 为负时，二次方程 ax² + bx + c = 0 有两个共轭复根。

x = [ -b ± √(b² – 4ac) ] / 2a

For example, solve x² + 4x + 8 = 0. Here a = 1, b = 4, c = 8. The discriminant is 16 – 32 = -16. Using the quadratic formula: x = [ -4 ± √(-16) ] / 2 = [ -4 ± 4i ] / 2 = -2 ± 2i. The roots are complex conjugates.

例如解 x² + 4x + 8 = 0。此处 a = 1, b = 4, c = 8。判别式为 16 – 32 = -16。代入求根公式：x = [ -4 ± √(-16) ] / 2 = [ -4 ± 4i ] / 2 = -2 ± 2i。根为共轭复数。

Always express complex roots in the form a ± bi, and remember that the sum and product of the roots are real.

务必将复根写成 a ± bi 的形式，并记住两根之和与积为实数。

11. Complex Functions | 复变函数

A complex function takes a complex number as its input and produces a complex number as its output. You can think of it as feeding a + bi into a rule like f(z) = z² + 2z + 3.

复变函数以复数为输入，输出也为一复数。你可以将它理解为将 a + bi 代入形如 f(z) = z² + 2z + 3 的规则中。

To evaluate f(z) at a given value, substitute and simplify using i² = -1. Example: for f(z) = z² – 3z + 2, find f(2 + i). Replace z with 2 + i: (2 + i)² – 3(2 + i) + 2 = (4 + 4i + i²) – 6 – 3i + 2 = (4 + 4i – 1) – 6 – 3i + 2 = (3 + 4i) – 6 – 3i + 2 = (3 – 6 + 2) + (4i – 3i) = –1 + i.

求给定点处的函数值，代入并用 i² = -1 化简。举例：对于 f(z) = z² – 3z + 2，计算 f(2 + i)。将 z 替换为 2 + i：(2 + i)² – 3(2 + i) + 2 = (4 + 4i + i²) – 6 – 3i + 2 = (3 + 4i) – 6 – 3i + 2 = –1 + i。

These functions often appear in CCEA papers when exploring mappings or transformations. Always work step by step and keep real and imaginary parts organised.

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Mastering Complex Functions for IGCSE CCEA Mathematics | IGCSE CCEA 数学：复变函数 考点精讲