📚 IGCSE Maths: Complex Numbers Key Points | IGCSE 数学：复数考点精讲
Complex numbers are a fundamental topic in IGCSE Additional Mathematics (0606). They extend the real number system to include solutions of equations such as x² = –1. Mastering imaginary numbers, operations, Argand diagrams, polar form, and De Moivre’s theorem is essential for success in the exam.
复数是 IGCSE 附加数学（0606）的基础课题。它们扩展了实数系统，使得像 x² = –1 这样的方程有解。掌握虚数、运算、阿尔冈图、极坐标形式以及棣莫弗定理对于考试成功至关重要。
1. Introduction to Imaginary Numbers | 虚数简介
The imaginary unit i is defined by i² = –1, or equivalently i = √(–1). This allows us to take square roots of negative numbers.
虚数单位 i 由 i² = –1 定义，或等价地 i = √(–1)。这使得我们可以对负数开平方根。
Powers of i follow a repeating cycle of period 4: i¹ = i, i² = –1, i³ = –i, i⁴ = 1, then i⁵ = i, and so on.
i 的幂以 4 为周期循环：i¹ = i, i² = –1, i³ = –i, i⁴ = 1，然后 i⁵ = i，依此类推。
Any negative real number –a (a > 0) has two square roots: √a i and –√a i. For example, √(–9) = ±3i.
任何负实数 –a（a > 0）都有两个平方根：√a i 和 –√a i。例如 √(–9) = ±3i。
2. Standard Form of Complex Numbers | 复数的标准形式
A complex number is written in standard form as z = a + bi, where a and b are real numbers. a is called the real part, Re(z), and b is called the imaginary part, Im(z).
复数写作标准形式 z = a + bi，其中 a 和 b 为实数。a 称为实部 Re(z)，b 称为虚部 Im(z)。
If b = 0, z is a real number. If a = 0 and b ≠ 0, z is a purely imaginary number, written as bi.
若 b = 0，则 z 为实数。若 a = 0 且 b ≠ 0，则 z 为纯虚数，写作 bi。
When writing complex numbers, the order ‘real part + imaginary part i’ must be kept; for instance, 3 – 2i is the standard form, not –2i + 3.
书写复数时，须保持“实部 + 虚部 i”的顺序；例如标准形式为 3 – 2i，而非 –2i + 3。
3. Equality and Basic Operations | 相等与基本运算
Two complex numbers a + bi and c + di are equal if and only if their real parts are equal (a = c) and their imaginary parts are equal (b = d).
两个复数 a + bi 和 c + di 相等当且仅当它们的实部相等（a = c）且虚部相等（b = d）。
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: Expand using FOIL and replace i² with –1: (a + bi)(c + di) = ac + adi + bci + bdi² = (ac – bd) + (ad + bc)i
乘法： 使用多项式展开并将 i² 替换为 –1：(a + bi)(c + di) = ac + adi + bci + bdi² = (ac – bd) + (ad + bc)i
4. Complex Conjugate | 共轭复数
The complex conjugate of z = a + bi is denoted by z* and defined as z* = a – bi. Geometrically, it is the reflection of z across the real axis.
复数 z = a + bi 的共轭复数记作 z*，定义为 z* = a – bi。几何上，它是 z 关于实轴的反射。
Key properties: z + z* = 2a (purely real), and z z* = a² + b² (also purely real). The product of a number and its conjugate gives the square of its modulus.
重要性质：z + z* = 2a（纯实数），z z* = a² + b²（亦为纯实数）。一个数与其共轭的乘积等于该数模的平方。
The conjugate of a sum, product, or quotient is the sum, product, or quotient of the conjugates: (z₁ ± z₂)* = z₁* ± z₂*, (z₁ z₂)* = z₁* z₂*, (z₁ / z₂)* = z₁* / z₂*.
和、积、商的共轭等于共轭的和、积、商：(z₁ ± z₂)* = z₁* ± z₂*，(z₁ z₂)* = z₁* z₂*，(z₁ / z₂)* = z₁* / z₂*。
5. Division of Complex Numbers | 复数除法
To divide one complex number by another, multiply both numerator and denominator by the conjugate of the denominator. This makes the denominator real.
复数相除时，将分子和分母同乘以分母的共轭复数，从而使分母化为实数。
Example: (3 + 2i) / (1 – i) = (3 + 2i)(1 + i) / ((1 – i)(1 + i)) = (3 + 3i + 2i + 2i²) / (1 – i²) = (3 + 5i – 2) / (1 + 1) = (1 + 5i) / 2 = 0.5 + 2.5i.
例子：(3 + 2i) / (1 – i) = (3 + 2i)(1 + i) / ((1 – i)(1 + i)) = (3 + 3i + 2i + 2i²) / (1 – i²) = (3 + 5i – 2) / (1 + 1) = (1 + 5i) / 2 = 0.5 + 2.5i.
Always write the final answer in standard form a + bi. Simplify fractions where possible.
最终答案务必写成标准形式 a + bi，并尽可能化简分数。
6. Argand Diagram | 阿尔冈图
The Argand diagram is a coordinate plane where the horizontal axis represents the real part and the vertical axis represents the imaginary part. A complex number a + bi is plotted as the point (a, b).
阿尔冈图是一个坐标平面，横轴表示实部，纵轴表示虚部。复数 a + bi 用点 (a, b) 表示。
Real numbers lie on the horizontal axis, purely imaginary numbers lie on the vertical axis. Conjugates appear as mirror images across the real axis.
实数落在横轴上，纯虚数落在纵轴上。共轭复数表现为关于实轴的镜像。
Addition and subtraction correspond to vector addition and subtraction in the diagram. Argand diagrams help visualise modulus and argument.
复数的加减法对应图上向量的加减法。阿尔冈图有助于直观理解模与辐角。
7. Modulus of a Complex Number | 复数的模
The modulus (or absolute value) of z = a + bi is denoted |z| and defined as the distance from the origin to the point (a, b) in the Argand diagram.
复数 z = a + bi 的模（或绝对值）记作 |z|，定义为阿尔冈图上原点到点 (a, b) 的距离。
Formula: |z| = √(a² + b²). Using the conjugate, |z|² = z z* = a² + b².
公式：|z| = √(a² + b²)。利用共轭，有 |z|² = z z* = a² + b²。
Properties: |z₁ z₂| = |z₁| |z₂|, |z₁ / z₂| = |z₁| / |z₂| (z₂ ≠ 0), |zⁿ| = |z|ⁿ for integer n.
性质：|z₁ z₂| = |z₁| |z₂|，|z₁ / z₂| = |z₁| / |z₂|（z₂ ≠ 0），|zⁿ| = |z|ⁿ（n 为整数）。
8. Argument of a Complex Number | 复数的辐角
The argument of a non-zero complex number z, denoted arg(z), is the angle θ that the line connecting z to the origin makes with the positive real axis, measured in radians.
非零复数 z 的辐角记为 arg(z)，是从原点到 z 的连线与正实轴之间的夹角，以弧度度量。
If z = a + bi, then tan θ = b / a, taking account of the quadrant in which the point lies. The principal argument is usually taken in the range –π < θ ≤ π (or sometimes 0 ≤ θ < 2π; check the exam specification).
若 z = a + bi，则 tan θ = b / a，需根据点所在象限确定角度。主值辐角通常取 –π < θ ≤ π（有时为 0 ≤ θ < 2π；请查阅考纲）。
Examples: arg(1 + i) = π/4, arg(–1) = π, arg(–i) = –π/2, arg(1 – √3 i) = –π/3. Always sketch an Argand diagram to avoid common quadrant mistakes.
例：arg(1 + i) = π/4，arg(–1) = π，arg(–i) = –π/2，arg(1 – √3 i) = –π/3。请始终画出阿尔冈图以避免象限错误。
9. Polar Form | 极坐标形式
A complex number can be expressed in polar form using its modulus r = |z| and argument θ = arg(z): z = r (cos θ + i sin θ). Sometimes this is abbreviated as z = r cis θ.
复数可用其模 r = |z| 和辐角 θ = arg(z) 表示为极坐标形式：z = r (cos θ + i sin θ)。有时简写为 z = r cis θ。
To convert from a + bi to polar form: compute r = √(a² + b²) and find θ such that cos θ = a/r and sin θ = b/r, adjusting for the correct quadrant.
从 a + bi 转换为极坐标形式：计算 r = √(a² + b²)，并求出满足 cos θ = a/r 且 sin θ = b/r 的 θ，注意象限。
To convert back: a = r cos θ, b = r sin θ. This representation is extremely useful for multiplication, division, and powers.
转换回代数形式：a = r cos θ, b = r sin θ。这种表示法对乘、除和乘方运算极为有用。
10. Multiplication & Division in Polar Form | 极坐标形式的乘除法
If z₁ = r₁ (cos θ₁ + i sin θ₁) and z₂ = r₂ (cos θ₂ + i sin θ₂), then their product is obtained by multiplying moduli and adding arguments:
若 z₁ = r₁ (cos θ₁ + i sin θ₁) 且 z₂ = r₂ (cos θ₂ + i sin θ₂)，则其乘积可通过模相乘、辐角相加得到：
z₁ z₂ = r₁ r₂ [cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)]
Division works similarly by dividing moduli and subtracting arguments:
除法类似，通过模相除、辐角相减得到：
z₁ / z₂ = (r₁ / r₂) [cos(θ₁ − θ₂) + i sin(θ₁ − θ₂)], r₂ ≠ 0
Always ensure the argument remains within the principal range, adding or subtracting 2π if needed.
注意保持辐角在所求的主值区间内，必要时加减 2π。
11. De Moivre’s Theorem | 棣莫弗定理
For any complex number in polar form and any integer n, De Moivre’s theorem states:
对于极坐标形式的任意复数以及任意整数 n，棣莫弗定理指出：
[r (cos θ + i sin θ)]ⁿ = rⁿ (cos nθ + i sin nθ)
This theorem is a powerful tool for raising complex numbers to integer powers and for finding trigonometric identities. For example, to find (1 + i)⁴: write 1 + i = √2 (cos π/4 + i sin π/4), then (1 + i)⁴ = (√2)⁴ (cos π + i sin π) = 4 (–1 + 0i) = –4.
该定理是求复数整数次幂和推导三角恒等式的有力工具。例如，计算 (1 + i)⁴：将 1 + i 写成 √2 (cos π/4 + i sin π/4)，则 (1 + i)⁴ = (√2)⁴ (cos π + i sin π) = 4 (–1 + 0i) = –4。
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