Complex Numbers: A Complete Guide for IB Mathematics | IB数学复数完全指南

📚 Complex Numbers: A Complete Guide for IB Mathematics | IB数学复数完全指南

is and Approaches HL course, complex numbers are essential for exploring polynomial equations, geometric transformations, and advanced proof techniques. This article provides a thorough yet accessible walkthrough of the core concepts, from basic arithmetic to De Moivre’s theorem and Euler’s formula, complete with worked examples and exam‑style reasoning.

复数通过引入虚数单位，将实数的概念进行了扩展，使得诸如 x² + 1 = 0 这样没有实数根的方程也能获得解。在IB数学分析与方法（HL）课程中，复数是探索多项式方程、几何变换和高等证明技巧的核心内容。本文将对核心概念进行一次透彻而又通俗易懂的梳理，从基本运算到棣莫弗定理与欧拉公式，并配有详尽的例题和符合考试要求的推理过程。

1. What is a Complex Number? | 什么是复数？

are real numbers and i is the imaginary unit. The set of all complex numbers is denoted by ℂ hen b = 0 the complex number reduces to a real number, so the real numbers are a subset of ℂ.

复数是形如 a + bi 的表达式，其中 a 和 b 为实数，i 为虚数单位。所有复数构成的集合记作 ℂ。实部 Re(z) 为 a，虚部 Im(z) 为 b。当 b = 0 时复数退化为实数，因此实数是复数集的子集。

e attempt to solve quadratic equations that have a negative discriminant, for example x² = –1. The invention of i with the fundamental property i² = – the square root of negative numbers and opens up a rich algebraic and geometric landscape.

当我们试图求解判别式为负的二次方程，例如 x² = –1 时，就需要引入复数。创造出满足基本性质 i² = –1 的虚数单位 i，赋予负数开平方运算实际意义，并开启了一个丰富的代数与几何世界。

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2. The Imaginary Unit i | 虚数单位 i

The imaginary unit i is defined by the equation i² = –1. From this definition it follows that i³ = i² · i = –i, i⁴ = (i²)² = 1, and the powers of i repeat in a cycle of four: i, –1, – l when simplifying higher powers of i, for example i²⁵ = i²⁴ · i = (i⁴)⁶ · i = 1⁶ · i = i.

虚数单位 i 由方程 i² = –1 定义。据此可得 i³ = i² · i = –i，i⁴ = (i²)² = 1，i 的幂次以 4 为周期循环：i、–1、–i、1。这种周期性在化简 i 的高次幂时非常有用，例如 i²⁵ = i²⁴ · i = (i⁴)⁶ · i = 1⁶ · i = i。

ebraic symbol that obeys regular arithmetic rules, with the additional simplification i² = –1. There is no need to memorise a separate set of rules – just remember to replace every occurrence of i² with –1 and simplify the expression.

把 i 当作一个遵循通常运算规则的代数符号来处理即可，只需多加一条化简规则 i² = –1。无需死记硬背另一套法则——只要记得每出现 i² 就用 –1 替换，然后化简表达式。

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3. Algebraic Form and Notation | 代数形式与符号表示

The standard algebraic form of a complex number is z = a + bi, where a, b ∈ ℝ. For instance, z₁ = 3 + 4i, z₂ = –2 – 5i, and z₃ = √2 i are all valid complex numbers; in the last case a = 0 and b = √2, making it a purely imaginary number.

复数的标准代数形式为 z = a + bi，其中 a, b ∈ ℝ。例如 z₁ = 3 + 4i、z₂ = –2 – 5i 以及 z₃ = √2 i 都是合法的复数；在最后一个例子中 a = 0 而 b = √2，因此它是一个纯虚数。

f their real parts are equal and their imaginary parts are equal. That is, a + bi = c + di ⇔ of equating real and imaginary parts, a powerful technique for solving complex equations.

两个复数相等，当且仅当它们的实部相等且虚部相等。即 a + bi = c + di ⇔ a = c 且 b = d。这一简单事实是“方程两边实虚部分别相等”方法的基础，是为解决复数方程的有力技巧。

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4. Addition and Subtraction | 加法与减法

reating the real and imaginary parts separately. For z₁ = a + bi and z₂ = c + di, we have:

z₁ + z₂ = (a + c) + (b + d)i

z₁ – z₂ = (a – c) + (b – d)i

复数加法与减法通过对实部和虚部分别运算来实现。设 z₁ = a + bi，z₂ = c + di，则有：

z₁ + z₂ = (a + c) + (b + d)i

z₁ – z₂ = (a – c) + (b – d)i

For example, (5 + 2i) + (3 – 7i) = (5+3) + (2–7)i = 8 – 5i. Similarly, (4 – i) – (1 + 6i) = (4–1) + (–1–6)i = 3 – irror vector addition in the plane, a connection that becomes clear on the Argand diagram.

例如，(5 + 2i) + (3 – 7i) = (5+3) + (2–7)i = 8 – 5i。类似地，(4 – i) – (1 + 6i) = (4–1) + (–1–6)i = 3 – 7i。这些运算与平面中的向量加法完全对应，这种联系在阿甘图上将变得一目了然。

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5. Multiplication | 乘法

plication of complex numbers uses the distributive law (FOIL) along with the identity i² = –1. For z₁ = a + bi and z₂ = c + di:

z₁ · z₂ = (a + bi)(c + di) = ac + ad i + bc i + bd i² = (ac – bd) + (ad + bc)i

复数乘法利用分配律（FOIL法则）以及恒等式 i² = –1 进行。设 z₁ = a + bi，z₂ = c + di：

z₁ · z₂ = (a + bi)(c + di) = ac + ad i + bc i + bd i² = (ac – bd) + (ad + bc)i

Let’s multiply (2 + 3i) and (1 – 4i): 2·1 = 2, 2·(–4i) = –8i, 3i·1 = 3i, 3i·(–4i) = –12 i² = +12. Summing gives (2+12) + (–8+3)i = 14 – 5i. Always replace i² with –1 to complete the simplification.

我们来计算 (2 + 3i) 与 (1 – 4i) 的乘积：2·1 = 2，2·(–4i) = –8i，3i·1 = 3i，3i·(–4i) = –12 i² = +12。将各项相加得到 (2+12) + (–8+3)i = 14 – 5i。务必用 –1 替换 i² 来完成化简。

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6. Complex Conjugate | 共轭复数

The complex conjugate of z = a + bi is denoted by z* or z̅ and is defined as z* = a – nt across the real axis. The product of a complex number and its conjugate is always a non‑negative real number: z z* = (a + bi)(a – bi) = a² + b².

复数 z = a + bi 的共轭记作 z* 或 z̅，定义为 z* = a – bi。从几何上看，取共轭相当于将点关于实轴反射。一个复数与其共轭的乘积总是一个非负实数：z z* = (a + bi)(a – bi) = a² + b²。

modulus. For example, the sum z + z* = 2a gives twice the real part, and the difference z – = z, and the conjugate of a sum or product obeys (z₁+z₂)* = z₁*+z₂* and (z₁z₂)* = z₁* z₂*.

共轭复数在除法和证明模的性质时极为有用。例如，和 z + z* = 2a 给出实部的两倍，差 z – z* = 2bi 给出虚部的两倍乘以 i。此外，(z*)* = z，并且和与积的共轭满足 (z₁+z₂)* = z₁*+z₂* 以及 (z₁z₂)* = z₁* z₂*。

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7. Division | 除法

his converts the denominator into a real number. For z₁ = a + bi and z₂ = c + di (z₂ ≠ 0):

z₁ / z₂ = (a + bi)/(c + di) = ((a + bi)(c – di)) / ((c + di)(c – di)) = ((ac + bd) + (bc – ad)i) / (c² + d²)

要对两个复数做除法，将分子与分母同时乘以分母的共轭。这样可将分母转化为实数。设 z₁ = a + bi，z₂ = c + di (z₂ ≠ 0)，则：

z₁ / z₂ = (a + bi)/(c + di) = ((a + bi)(c – di)) / ((c + di)(c – di)) = ((ac + bd) + (bc – ad)i) / (c² + d²)

As an illustration, divide 3 + 2i by 1 – or and denominator by 1 + i: ((3+2i)(1+i)) / (1² + 1²) = (3 + 3i + 2i + 2i²) / 2 = (3 + 5i – 2) / 2 = (1 + 5i) / 2 = 0.5 + 2.5i.

举一个例子：计算 (3 + 2i) ÷ (1 – i)。将分子分母同乘以 1 + i：((3+2i)(1+i)) / (1² + 1²) = (3 + 3i + 2i + 2i²) / 2 = (3 + 5i – 2) / 2 = (1 + 5i) / 2 = 0.5 + 2.5i。

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8. The Argand Diagram | 阿甘图（复数平面）

bi is represented by the point (a, b) or by the position vector from the origin to (a, b).

阿甘图将复数表示为平面中的点或向量。水平轴为实轴（Re），竖直轴为虚轴（Im）。复数 z = a + bi 用点 (a, b) 或从原点指向 (a, b) 的位置向量来表示。

tion, following the parallelogram law. Multiplication by i rotates a vector by 90° counter‑clockwise, because i(a + bi) = –b + ai, which is a quarter‑ metric interpretation is a cornerstone of many IB exam problems involving transformations.

复数加法对应向量加法，遵循平行四边形法则。乘以 i 会将向量逆时针旋转 90°，因为 i(a + bi) = –b + ai，相当于原向量旋转四分之一圈。这种几何解释是许多IB考试中涉及变换问题的基石。

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9. Modulus and Argument | 模长与辐角

The modulus of z = a + bi, denoted |z|, is the distance from the origin to the point (a, b):

|z| = √a² + b²

复数 z = a + bi 的模，记作 |z|，是原点到点 (a, b) 的距离：

|z| = √a² + b²

e θ that the position vector makes with the positive real axis, usually measured in radians and taken in the principal range (– an(b/a); when a < 0 careful quadrant adjustment is needed. The argument of 0 is undefined.

复数 z 的辐角 arg(z) 是其位置向量与正实轴之间的夹角 θ，通常以弧度为单位，并取主值区间 (–π, π]。对于 a > 0 的情况，θ = arctan(b/a)；当 a < 0 时需要进行象限修正。复数 0 的辐角无定义。

Key properties of modulus include |z₁z₂| = |z₁||z₂| and |z₁/z₂| = |z₁|/|z₂|. For arguments, arg(z₁z₂) = arg(z₁)+arg(z₂) (mod 2π) and arg(z₁/z₂) = arg(z₁)–arg(z₂). These multiplicative properties make polar form extremely powerful.

模的重要性质包括 |z₁z₂| = |z₁||z₂| 以及 |z₁/z₂| = |z₁|/|z₂|。对于辐角，arg(z₁z₂) = arg(z₁)+arg(z₂)（模 2π），arg(z₁/z₂) = arg(z₁)–arg(z₂)。这些乘法性质使得极坐标形式威力无穷。

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10. Polar Form and De Moivre's Theorem | 极坐标形式与棣莫弗定理

A complex number can be expressed in polar form using its modulus r = |z| and argument θ:

z = r (cos θ + i sin θ)

复数可以用它的模 r = |z| 和辐角 θ 表示成极坐标形式：

z = r (cos θ + i sin θ)

De Moivre’s theorem states that for any integer n:

(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ

棣莫弗定理指出，对任意整数 n 有：

(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ

This theorem is proved by induction for positive integers and extends to negative integers using the reciprocal. It allows the rapid computation of powers and roots of complex numbers. For instance, to compute (1 + i)⁶, first write 1 + i in polar form: √2 (cos π/4 + i sin π/4). Then (1 + i)⁶ = (√2)⁶ (cos 6π/4 + i sin 6π/4) = 8 (cos 3π/2 + i sin 3π/2) = 8 (0 – i) = –8i.

该定理对正整数可通过归纳法证明，并借助倒数推广至负整数。它使得复数的乘方与开方运算变得极为快捷。例如，要计算 (1 + i)⁶，先将 1 + i 写成极坐标形式：√2 (cos π/4 + i sin π/4)。于是 (1 + i)⁶ = (√2)⁶ (cos 6π/4 + i sin 6π/4) = 8 (cos 3π/2 + i sin 3π/2) = 8 (0 – i) = –8i。

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11. Euler's Formula | 欧拉公式

Euler’s formula establishes a profound link between trigonometric and exponential functions:

e^iθ = cos θ + i sin θ

欧拉公式在三角函数与指数函数之间搭建了一座深刻的桥梁：

e^iθ = cos θ + i sin θ

Using Euler’s formula, the polar form simplifies to z = r e^iθ ication and division straightforward: z₁z₂ = r₁r₂ e^(i(θ₁+θ₂)) and z₁/z₂ = (r₁/r₂) e^(i(θ₁–θ₂)). It also unifies De Moivre’s theorem, since (e^iθ)ⁿ = e^{i nθ} = cos nθ + i sin nθ.

利用欧拉公式，极坐标形式可简化为 z = r e^iθ。这种指数记法使乘除运算一目了然：z₁z₂ = r₁r₂ e^(i(θ₁+θ₂))，z₁/z₂ = (r₁/r₂) e^(i(θ₁–θ₂))。它还将棣莫弗定理统一起来，因为 (e^iθ)ⁿ = e^{i nθ} = cos nθ + i sin nθ。

Euler’s identity e^iπ most beautiful equations in mathematics, connecting five fundamental constants. In IB problems, Euler’s formula is frequently used to switch between polar and exponential forms, especially when dealing with complex roots of unity.

欧拉恒等式 e^iπ + 1 = 0 常被誉为数学中最优美的方程之一，它将五个基本常数联系在一起。在IB考题中，欧拉公式经常用于在极坐标与指数形式之间进行转换，尤其是在处理单位复数根时。

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12. Roots of Complex Numbers | 复数求根

To find the n‑th roots of a complex number w = r (cos φ + i sin φ) = r e^iφ, we solve zⁿ = w. The n distinct roots are given by:

z_k = r^1/n [ cos((φ + 2kπ)/n) + i sin((φ + 2kπ)/n) ], k = 0, 1, …, n–1

求复数 w = r (cos φ + i sin φ) = r e^iφ 的 n 次方根，即解 zⁿ = w。n 个互不相同的根由下式给出：

z_k = r^1/n [ cos((φ + 2kπ)/n) + i sin((φ + 2kπ)/n) ], k = 0, 1, …, n–1

These roots are equally spaced around a circle of radius r^1/n r the cube roots of 1. With r = 1, φ = 0, we obtain z₀ = 1, z₁ = cos(2π/3) + i sin(2π/3) = –½ + i(√3/2), and z₂ = cos(4π/3) + i sin(4π/3) = –½ – i(√3/2). Collectively they are denoted 1, ω, ω², where ω = e^2πi/3.

这些根均匀分布在半径为 r^1/n 的圆上，彼此间隔的角度为 2π/n。一道经典的IB考题是求 1 的立方根。取 r = 1, φ = 0，得到 z₀ = 1，z₁ = cos(2π/3) + i sin(2π/3) = –½ + i(√3/2)，以及 z₂ = cos(4π/3) + i sin(4π/3) = –½ – i(√3/2)。它们常被记作 1, ω, ω²，其中 ω = e^2πi/3。

Understanding roots of complex numbers is essential for solving polynomial equations and for geometric problems involving regular n‑gons inscribed in a circle. The symmetry of the roots often simplifies calculations of sums and products, a recurring theme in higher‑level IB questions.

理解复数根对于求解多项式方程以及涉及圆内接正 n 边形的几何问题至关重要。根的对称性常常能简化求和与求积的计算，这是IB高等级考题中反复出现的主题。

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