Complex Numbers for IGCSE Edexcel Mathematics: Key Points | IGCSE Edexcel 数学：复数 考点精讲

📚 Complex Numbers for IGCSE Edexcel Mathematics | IGCSE Edexcel 数学：复数 考点精讲

Complex numbers appear in the IGCSE Edexcel Further Pure Mathematics course and are a vital tool for solving equations that have no real solutions. This guide covers all the key concepts—from the imaginary unit i to de Moivre’s theorem—to help you revise efficiently and tackle exam questions with confidence.

复数属于 IGCSE Edexcel 进阶纯数学（Further Pure Mathematics）的核心内容，是求解无实数解方程的关键工具。本文梳理虚数单位 i、复数运算、共轭复数、极坐标形式及棣莫弗定理等重要考点，助你高效复习、从容应考。

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1. The Imaginary Unit i | 虚数单位 i

The imaginary unit i is defined as the principal square root of -1, and its fundamental property is i² = -1. From this definition, i³ = i²·i = -i and i⁴ = (i²)² = 1.

虚数单位 i 定义为 -1 的主平方根，核心性质是 i² = -1。由此可推得 i³ = -i，i⁴ = 1。

i = √-1, i² = -1

Any real number multiple of i is called a pure imaginary number, e.g. 5i, -√2 i.

任何实数与 i 的乘积称为纯虚数，例如 5i、-√2 i。

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2. Complex Numbers: Standard Form a + bi | 复数的代数形式 a + bi

A complex number is written in the form z = a + bi, where a and b are real numbers. a is the real part Re(z), and b is the imaginary part Im(z). If b = 0 the number is real; if a = 0 it is purely imaginary.

复数写作 z = a + bi，其中 a, b 是实数。a 称为实部 Re(z)，b 称为虚部 Im(z)。若 b = 0，该数为实数；若 a = 0，则为纯虚数。

Two complex numbers are equal if and only if both their real parts and imaginary parts are equal: a + bi = c + di ⇒ a = c and b = d.

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 且 b = d。

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3. Addition and Subtraction | 加法和减法

To add or subtract complex numbers, simply combine the real parts and the imaginary parts separately: (a + bi) ± (c + di) = (a ± c) + (b ± d)i.

复数的加减，只需分别合并实部和虚部：(a + bi) ± (c + di) = (a ± c) + (b ± d)i。

For example, (3 + 2i) + (4 – 7i) = 7 – 5i, and (6 – i) – (2 + 3i) = 4 – 4i.

例如，(3 + 2i) + (4 – 7i) = 7 – 5i；(6 – i) – (2 + 3i) = 4 – 4i。

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4. Multiplication | 乘法

Multiply two complex numbers by expanding the brackets as in ordinary algebra and then simplify using i² = -1.

复数相乘时，像普通代数一样展开括号，再利用 i² = -1 进行化简。

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

For instance, (2 + 3i)(1 – 4i) = 2 – 8i + 3i – 12i² = 2 – 5i + 12 = 14 – 5i.

例如，(2 + 3i)(1 – 4i) = 2 – 8i + 3i – 12i² = 2 – 5i + 12 = 14 – 5i。

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5. Complex Conjugate | 共轭复数

The complex conjugate of z = a + bi is denoted by z̄ or z* and is defined as z̄ = a – bi. The product of a complex number and its conjugate is always a non‑negative real number: z·z̄ = a² + b².

复数 z = a + bi 的共轭记作 z̄ 或 z*，定义为 z̄ = a – bi。一个复数与其共轭的乘积总是非负实数：z·z̄ = a² + b²。

Conjugates are essential for division and for finding real solutions from complex expressions.

共轭复数在除法运算和从复式中提取实数结果时至关重要。

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6. Division | 除法

To divide one complex number by another, multiply both numerator and denominator by the conjugate of the denominator. This turns the denominator into a real number.

复数相除时，将分子分母同乘以分母的共轭复数，即可将分母转化为实数。

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

Example: (3 + i) / (2 – i) = (3 + i)(2 + i) / (2² + 1²) = (6 + 3i + 2i + i²) / 5 = (5 + 5i) / 5 = 1 + i.

例如：(3 + i) / (2 – i) = (3 + i)(2 + i) / (4 + 1) = (6 + 5i + i²) / 5 = (5 + 5i) / 5 = 1 + i。

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7. Solving Quadratic Equations with Real Coefficients | 实系数二次方程的复数解

For a quadratic equation ax² + bx + c = 0 with real coefficients and a negative discriminant (Δ = b² – 4ac < 0), the roots are a conjugate pair of complex numbers given by the quadratic formula:

实系数二次方程 ax² + bx + c = 0，当判别式 Δ = b² – 4ac < 0 时，其根为一对共轭复数，由求根公式给出：

x = [-b ± i√(4ac – b²)] / (2a)

For example, solve x² – 4x + 13 = 0: Δ = 16 – 52 = -36. The roots are x = (4 ± i√36) / 2 = 2 ± 3i.

例如，解方程 x² – 4x + 13 = 0：Δ = -36，根为 x = (4 ± 6i) / 2 = 2 ± 3i。

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8. Argand Diagram | Argand 图

An Argand diagram represents complex numbers as points or vectors on a plane, with the horizontal axis for the real part and the vertical axis for the imaginary part. The complex number z = a + bi is plotted at the point (a, b).

Argand 图将复数表示为平面上的点或向量，其中水平轴表示实部，垂直轴表示虚部。复数 z = a + bi 对应点 (a, b)。

This visual representation makes it easy to understand addition (parallelogram law) and the geometric meaning of modulus and argument.

这种直观表示有助于理解复数的加法（平行四边形法则）以及模和辐角的几何意义。

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9. Modulus and Argument | 模与辐角

The modulus of a complex number z = a + bi, written |z|, is the distance from the origin to the point (a, b) on the Argand diagram. It is calculated as |z| = √(a² + b²).

复数 z = a + bi 的模，记作 |z|，是 Argand 图中原点到点 (a, b) 的距离，计算公式为 |z| = √(a² + b²)。

The argument of z, denoted arg(z), is the angle θ that the vector makes with the positive real axis, usually measured in radians and given in the range (-π, π] or [0, 2π). tan θ = b/a, but careful adjustment of the quadrant is needed.

辐角 arg(z) 是该向量与正实轴之间的夹角 θ，通常以弧度为单位，主值范围为 (-π, π] 或 [0, 2π)。tan θ = b/a，但需根据象限进行调整。

For example, if z = 1 – i, then |z| = √(1+1) = √2, and arg(z) = -π/4 (or 7π/4).

例如，z = 1 – i，则 |z| = √2，arg(z) = -π/4（或 7π/4）。

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10. Modulus-Argument Form (Polar Form) | 极坐标形式

A complex number can be expressed in modulus-argument form using its modulus r and argument θ:

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), where r = |z|, θ = arg(z)

This form is extremely useful for multiplication, division and raising to powers. Always ensure the argument is correct for the quadrant.

这种形式对于乘法、除法及幂运算极为便利。务必根据象限确定正确的辐角。

Example: Express 1 – √3 i in polar form. r = √(1+3) = 2; tan θ = -√3, fourth quadrant ⇒ θ = -π/3, so z = 2(cos(-π/3) + i sin(-π/3)).

例如：将 1 – √3 i 表示为极坐标形式。r = 2，θ = -π/3，故 z = 2(cos(-π/3) + i sin(-π/3))。

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11. Multiplication and Division in Polar Form | 极坐标形式下的乘除运算

When multiplying two complex numbers in polar form, multiply their moduli and add their arguments. When dividing, divide the moduli and subtract the arguments.

极坐标形式下，两复数相乘时，模长相乘、辐角相加；相除时，模长相除、辐角相减。

z₁z₂ = r₁r₂ [cos(θ₁+θ₂) + i sin(θ₁+θ₂)]

z₁ / z₂ = r₁/r₂ [cos(θ₁-θ₂) + i sin(θ₁-θ₂)]

For instance, if z₁ = 3(cos π/3 + i sin π/3) and z₂ = 2(cos π/6 + i sin π/6), then z₁z₂ = 6(cos π/2 + i sin π/2) = 6i.

例如，z₁ = 3(cos π/3 + i sin π/3)，z₂ = 2(cos π/6 + i sin π/6)，则 z₁z₂ = 6(cos π/2 + i sin π/2) = 6i。

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12. de Moivre’s Theorem | 棣莫弗定理

For any integer n, de Moivre’s theorem states that if z = r(cos θ + i sin θ), then zⁿ = rⁿ (cos nθ + i sin nθ). This provides a quick way to compute powers of complex numbers.

对于任意整数 n，棣莫弗定理指出：若 z = r(cos θ + i sin θ)，则 zⁿ = rⁿ (cos nθ + i sin nθ)。这为计算复数的高次幂提供了捷径。

Example: Find (1 + i)⁴. First write 1 + i in polar form: r = √2, θ = π/4. Then (1 + i)⁴ = (√2)⁴ (cos π + i sin π) = 4(-1 + 0i) = -4.

例如：计算 (1 + i)⁴。先将 1 + i 化为极坐标形式：r = √2，θ = π/4。则 (1 + i)⁴ = (√2)⁴ (cos π + i sin π) = 4(-1) = -4。

The theorem is also the basis for finding roots of complex numbers when n is a rational exponent.

该定理也是求复数方根（当 n 为有理指数时）的基础。

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