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De Moivre’s Theorem and Its Applications | 棣莫弗定理及其应用

2026-08-02 22:50 · 本文部分内容由AI辅助创作

📚 De Moivre’s Theorem and Its Applications | 棣莫弗定理及其应用

De Moivre’s Theorem is a cornerstone of complex number algebra in the IB Mathematics: Analysis and Approaches HL course. It elegantly links trigonometry with the arithmetic of complex numbers, enabling us to raise complex numbers in polar form to integer powers, extract roots, and tackle a variety of trigonometric identities and equations. This article unpacks the theorem, its proof for positive integers, its natural extension to negative and rational exponents, and a wealth of practical applications – from computing high powers and nth roots to proving multiple-angle formulas and summing trigonometric series. Each section is built around key IB-style skills, offering fully worked reasoning in both English and Chinese, so you can follow the logic comfortably while mastering the bilingual terminology.

棣莫弗定理是 IB 数学分析与方法（AA）HL 课程中复数代数的基石。它巧妙地将三角学与复数运算联系起来，使我们能够将极坐标形式的复数进行整数次乘方、开方，并解决各种三角恒等式与方程。本文详细解析该定理、正整数情形的证明、向负整数和有理指数情形的自然推广，以及从计算高次幂和 n 次方根，到证明倍角公式与求和三角级数等丰富应用。每个小节围绕 IB 题型核心技能构建，提供中、英文配对讲解，让你在掌握双语术语的同时，顺畅地理解整个逻辑。

1. Complex Numbers in Polar Form | 复数的极坐标形式

Every complex number z = x + yi can be located on the Argand plane by its modulus r = |z| = √x² + y² and its argument θ = arg(z), measured from the positive real axis. The polar (or modulus– The compact notation r cis θ (where cis θ = cos θ + i sin θ) is often used to save space.

每个复数 z = x + yi 都可在阿根图上用模 r = |z| = √x² + y² 和辐角 θ = arg(z)（从正实轴量起）定位。极坐标（或模–辐角）形式为 z = r (cos θ + i sin θ)。这种表示把乘法转化为模相乘、辐角相加：若 z₁ = r₁(cos θ₁ + i sin θ₁)、z₂ = r₂(cos θ₂ + i sin θ₂)，则 z₁z₂ = r₁r₂ [cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)]。为简洁起见，常用缩写 r cis θ（其中 cis θ = cos θ + i sin θ）。

The polar form is indispensable for applying De Moivre’s Theorem because it isolates the modulus and the trigonometric part. Always ensure the argument is in radians unless the question specifies degrees, and remember that adding multiples of 2π does not change the complex number.

极坐标形式是应用棣莫弗定理不可或缺的基础，因为它分离了模与三角部分。务必确保辐角以弧度为单位（除非题目指定角度制），并记住加上 2π 的整数倍不改变复数本身。

2. Statement of De Moivre’s Theorem | 棣莫弗定理的陈述

ⁿ cis (nθ) when the modulus r is included. In IB exams, the theorem is most commonly used for integer exponents, but it can be extended to rational exponents when careful branch choices are made.

对任意整数 n，棣莫弗定理断言：(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)。等价地，包含模时为 (r cis θ)ⁿ = rⁿ cis (nθ)。在 IB 考试中，该定理最常用于整数指数，但在谨慎选取分支的情况下可推广至有理指数。

The formula reveals a simple pattern: raising a unit-modulus complex number to a power n multiplies its argument by n, while the modulus raised to the power scales the length. This makes computing high powers of complex numbers extremely efficient compared to expanding binomials.

该公式展示了一条简单规律：将模为 1 的复数进行 n 次乘方，其辐角变为原来的 n 倍，而模的 n 次幂则缩放长度。这使得计算复数的高次幂远比展开二项式高效。

3. Proof by Induction for Positive Integers | 正整数情形的数学归纳法证明

)ᵏ = cos(kθ) + i sin(kθ). Then for n = k + 1, we multiply both sides by (cos θ + i sin θ):

当 n = 1 时，左边为 (cos θ + i sin θ)¹ = cos θ + i sin θ，右边为 cos(1·θ) + i sin(1·θ) = cos θ + i sin θ，基础情形成立。假设 n = k 时命题成立： (cos θ + i sin θ)ᵏ = cos(kθ) + i sin(kθ)。则 n = k + 1 时，两边同乘 (cos θ + i sin θ)：

(cos θ + i sin θ)ᵏ⁺¹ = [cos(kθ) + i sin(kθ)] (cos θ + i sin θ)

m holds for n = k + 1. By mathematical induction, it is valid for all positive integers n.

展开乘积并利用和角公式 cos A cos B − sin A sin B = cos(A + B) 及 sin A cos B + cos A sin B = sin(A + B)，得到 cos(kθ + θ) + i sin(kθ + θ) = cos[(k+1)θ] + i sin[(k+1)θ]。于是定理对 n = k + 1 成立。由数学归纳法，该定理对所有正整数 n 均成立。

This inductive proof relies on the additive property of arguments and the trigonometric identities. IB often expects students to reproduce this proof for integer n, so practice writing it out clearly.

这一归纳证明依赖于辐角的可加性以及三角恒等式。IB 常要求学生能写出整数 n 情形的证明过程，因此请熟练清晰地写出。

4. Extension to Negative and Rational Exponents | 扩展到负整数和有理数指数

), this becomes cos(−mθ) + i sin(−mθ). Hence (cos θ + i sin θ)⁻ᵐ = cos(−mθ) + i sin(−mθ), confirming the theorem for negative integers.

对于负整数，令 n = −m 其中 m > 0。我们有 (cos θ + i sin θ)⁻ᵐ = 1 / (cos θ + i sin θ)ᵐ = 1 / [cos(mθ) + i sin(mθ)]。分子分母同乘 cos(mθ) − i sin(mθ)（共轭复数），得到 cos(mθ) − i sin(mθ)。利用奇偶性，cos(−mθ) = cos(mθ)、sin(−mθ) = −sin(mθ)，上式变为 cos(−mθ) + i sin(−mθ)。因此 (cos θ + i sin θ)⁻ᵐ = cos(−mθ) + i sin(−mθ)，验证了负整数情形。

For rational exponents of the form 1/q (q a positive integer), the theorem can be used to find qth roots. If we set (cos θ + i sin θ)1/q ). This implies qφ = θ + 2kπ, so φ = (θ + 2kπ)/q, producing q distinct roots for k = 0, 1, … nvolves multiple values and must be handled with the appropriate principal argument range.

对于形如 1/q（q 为正整数）的有理指数，定理可用于求 q 次方根。若设 (cos θ + i sin θ)1/q = cos φ + i sin φ，两边同时取 q 次幂得到 cos θ + i sin θ = cos(qφ) + i sin(qφ)。这意味着 qφ = θ + 2kπ，故 φ = (θ + 2kπ)/q，由此产生 k = 0, 1, …, q−1 时的 q 个不同方根。有理指数的完整陈述涉及多值性，必须在适当的主辐角范围内处理。

5. Computing Powers of Complex Numbers | 复数乘方的计算

To evaluate (√3 + i)⁵, first convert to polar form. Modulus: r = √((√3)² + 1²) = √3 + 1 = 2. Argument: tan θ = 1/√3 ⇒ θ = π/6 (since both real and imaginary parts are positive). So √3 (π/6). By De Moivre, (2 cis (π/6))⁵ = 2⁵ cis (5π/6) = 32 [cos(5π/6) + i sin(5π/6)] = 32 (−√3/2 + i·1/2) = −16√3 + 16i.

要计算 (√3 + i)⁵，先化为极坐标形式。模：r = √((√3)² + 1²) = √3 + 1 = 2。辐角：tan θ = 1/√3 ⇒ θ = π/6（实部和虚部均为正）。所以 √3 + i = 2 cis (π/6)。由棣莫弗定理，(2 cis (π/6))⁵ = 2⁵ cis (5π/6) = 32 [cos(5π/6) + i sin(5π/6)] = 32 (−√3/2 + i·1/2) = −16√3 + 16i。

(−π < θ ≤ π) unless otherwise specified, adding or subtracting multiples of 2π as needed.

这种方法比展开二项式再合并同类项要快得多。除非另有要求，一般应将辐角化至主值范围 (−π < θ ≤ π)，必要时加减 2π 的整数倍。

6. Finding nth Roots of Complex Numbers | 求复数的 n 次方根

The nth roots of a complex number z = r cis θ are given by z1/n = r1/n cis ((θ + 2kπ)/n) for k = 0, 1, 2, …, n−1. Geometrically, these roots lie equally spaced on a circle of radius r1/n centred at the origin, with an angular separation of 2π/n. They form the vertices of a regular n-gon.

复数 z = r cis θ 的 n 次方根由公式 z1/n = r1/n cis ((θ + 2kπ)/n), k = 0, 1, 2, …, n−1 给出。在几何上，这些方根等间距地分布在以原点为中心、半径为 r1/n 的圆周上，其辐角间隔为 2π/n，构成正 n 边形的顶点。

Example: e : Find the cube roots of −8. Write −8 = 8 cis π (since modulus 8, argument π). Then the cube roots are 81/3 cis ((π + 2kπ)/3) = 2 cis (π/3 + 2kπ/3). For k = 0: 2 cis (π/3) = 1 + i√3; k = 1: 2 cis π = −2; k = 2: 2 cis (5π/3) = 1 − i√3. These three roots form an equilateral triangle on the circle of radius 2.

例：求 −8 的立方根。将 −8 写作 8 cis π（模为 8，辐角为 π）。则立方根为 81/3 cis ((π + 2kπ)/3) = 2 cis (π/3 + 2kπ/3)。k = 0 时：2 cis (π/3) = 1 + i√3；k = 1 时：2 cis π = −2；k = 2 时：2 cis (5π/3) = 1 − i√3。这三个根在半径为 2 的圆上构成等边三角形。

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7. Linking Roots to Unit Circle Geometry | 方根与单位圆几何的联系

1, the nth roots of unity are especially neat: zⁿ = 1 gives roots cis(2kπ/n) for k = 0,1,… en, and always including z = 1. Their sum is zero, and they are key to factorising zⁿ − 1.

当模为 1 时，n 次单位根尤为简洁：zⁿ = 1 给出的根为 cis(2kπ/n)，k = 0,1,…,n−1。这些点位于单位圆上，当 n 为偶数时关于实轴对称，且总包含 z = 1。它们的和为零，并且在因式分解 zⁿ − 1 中起关键作用。

he right-hand side to polar form. The regular polygon property makes sketching straightforward and helps verify solutions.

IB 常要求学生画出这些方根，或利用它们求解诸如 z⁵ = 1 − i 之类的方程——首先将右端化为极坐标形式。正多边形特性使草图绘制直观，并有助于验证解答。

8. Proving Trigonometric Identities | 证明三角恒等式

Equating real and imaginary parts yields: cos 2θ = cos²θ − sin²θ, sin 2θ = 2 sin θ cos θ.

棣莫弗定理为倍角公式提供了一条代数路径。例如，(cos θ + i sin θ)² = cos 2θ + i sin 2θ。然而直接展开得：(cos θ + i sin θ)² = cos²θ − sin²θ + i(2 sin θ cos θ)。令实部和虚部分别相等，即得：cos 2θ = cos²θ − sin²θ，sin 2θ = 2 sin θ cos θ。

= 3 cos²θ sin θ − sin³θ = 3 sin θ − 4 sin³θ. These are the standard triple-angle formulas.

同理，对于 (cos θ + i sin θ)³ = cos 3θ + i sin 3θ，用二项式定理展开得 cos³θ + 3i cos²θ sin θ − 3 cos θ sin²θ − i sin³θ。比较实部：cos 3θ = cos³θ − 3 cos θ sin²θ = 4 cos³θ − 3 cos θ（利用 cos²θ + sin²θ = 1）。比较虚部：sin 3θ = 3 cos²θ sin θ − sin³θ = 3 sin θ − 4 sin³θ。这就是标准的三倍角公式。

as polynomials in cos θ and sin θ. IB questions often ask for the derivation up to n = 5.

这一方法可推广到一般情形：利用二项式定理展开 (cos θ + i sin θ)ⁿ，分离实部与虚部即可将 cos nθ 和 sin nθ 表示成关于 cos θ 和 sin θ 的多项式。IB 考题常要求推导至 n = 5。

9. Expressing sinⁿθ and cosⁿθ as Multiple Angles | 将 sinⁿθ、cosⁿθ 表示为倍角之和

= (z³ + 1/z³) + 3(z + 1/z) = 2 cos 3θ + 6 cos θ. Hence cos³θ = (1/4) cos 3θ + (3/4) cos θ.

反向操作，我们可以将正弦和余弦的幂表示为倍角余弦与正弦之和。令 z = cos θ + i sin θ，则 z + 1/z = 2 cos θ，z − 1/z = 2i sin θ。例如， os θ)³ = (z + 1/z)³ = z³ + 3z + 3/z + 1/z³ = (z³ + 1/z³) + 3(z + 1/z) = 2 cos 3θ + 6 cos θ。因此 cos³θ = (1/4) cos 3θ + (3/4) cos θ。

θ + 3/8. These expansions are invaluable for integrating powers of trigonometric functions – an application that IB often links to.

类似地，对于 sin⁴θ：(2i sin θ)⁴ = (z − 1/z)⁴。展开得 z⁴ − 4z² + 6 − 4/z² + 1/z⁴ = (z⁴ + 1/z⁴) − 4(z² + 1/z²) + 6 = 2 cos 4θ − 8 cos 2θ + 6。由于 i⁴ = 1，我们得到 16 sin⁴θ = 2 cos 4θ − 8 cos 2θ + 6，故 sin⁴θ = (1/8) cos 4θ − (1/2) cos 2θ + 3/8。这些展开式在积分三角函数的幂时极为宝贵——IB 常将二者关联起来考。

10. Solving Equations with Complex Roots | 求解带复数根的方程

(excluding z = 1). The non-real roots are cis(2π/5), cis(4π/5), cis(6π/5), and cis(8π/5).

像 zⁿ = a + bi 这样的方程可直接将右端化为极坐标形式并应用 n 次方根公式求解。更复杂的方程，如 z⁴ + z³ + z² + z + 1 = 0，可通过两边同乘 (z − 1) 得到 z⁵ − 1 = 0，再求 5 次单位根（排除 z = 1）。非实根为 cis(2π/5)、cis(4π/5)、cis(6π/5)、cis(8π/5)。

When asked to "find all complex solutions", ensure you give the complete set of n roots, expressed in exact polar or Cartesian form as required. The symmetry of roots helps deduce conjugate pairs: for a polynomial with real coefficients, non-real complex roots occur in conjugate pairs.

遇到“求所有复数解”的指令时，务必给出完整的 n 个根集，按题目要求用精确的极坐标或直角坐标形式表示。根的对称性有助于推断共轭对：实系数多项式的非实复根总是成对共轭出现。

11. Applications in Summation of Series | 在级数求和中的应用

the formula for the sum of a geometric series, can sum trigonometric series such as C = ∑k=0n cos kθ and S = ∑k=0n sin kθ. Consider the complex sum Σ cis(kθ) = Σ (cis θ)k )θ)) / (1 − cis θ), provided cis θ ≠ 1. By simplifying the quotient using half-angle identities, real and imaginary parts give closed forms for C and S.

结合几何级数求和公式，棣莫弗定理可对诸如下列三角级数求和：C = ∑k=0n cos kθ 和 S = ∑k=0n sin kθ。考虑复数和 Σ cis(kθ) = Σ (cis θ)k，这是一个公比为 cis θ 的等比数列。于是 C + iS = (1 − cis((n+1)θ)) / (1 − cis θ)，前提是 cis θ ≠ 1。利用半角恒等式化简该商，分离实部和虚部即得 C 与 S 的封闭形式。

Another classic example is summing binomial coefficients times cosines: ∑k=0n aking the real part yields 2ⁿ cosⁿ(θ/2) cos(nθ/2). Such manipulations are common in IB HL Paper 2 questions linking complex numbers with series.

另一个经典例题是求二项式系数与余弦乘积的和：∑k=0n C(n,k) cos kθ。将此视为 (1 + cis θ)ⁿ 的实部，而 (1 + cis θ)ⁿ = [2 cos(θ/2) cis(θ/2)]ⁿ = 2ⁿ cosⁿ(θ/2) cis(nθ/2)。取实部即得 2ⁿ cosⁿ(θ/2) cos(nθ/2)。这类操作常见于 IB HL 试卷二中将复数与级数结合的题目。

12. Summary and Exam Tips | 总结与考试技巧

cos nθ + i sin nθ for integer n is a must-know tool in IB AA HL. Always start by writing complex numbers in polar form, checking the quadrant to ensure the argument is correct. When extracting roots, remember to add 2kπ before dividing by n, and list all n distinct values. Use the theorem to prove identities, express powers of trig functions as multiple angles, and evaluate integrals. In summations, consider constructing a complex geometric progression.

棣莫弗定理 (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ（n 为整数）是 IB AA HL 必须熟练掌握的工具。解题时先将复数化为极坐标形式，检查象限确保辐角正确。开方时，务必将 2kπ 加上后再除以 n，并列出所有 n 个不同的值。借助该定理可证明恒等式、将三角函数的幂表示为倍角、计算积分。在级数求和中，可考虑构造复数等比数列。

tesian coordinates to arguments (e.g., for a point in the second quadrant, θ = π − arctan(|y/x| ins consistent with the principal value. Practice sketching the roots on an Argand diagram – it reinforces the geometry and helps catch missing solutions.

常见陷阱包括：将直角坐标转换为辐角时忘记符号（如第二象限的点，θ = π − arctan(|y/x|)）。使用负指数时，注意保持辐角与主值一致。多练习在阿根图上绘制方根——这能巩固几何直观并帮助发现漏解。

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