📚 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 numbers, linking trigonometry and algebra through the polar form. It allows us to compute powers and roots of complex numbers, derive triple-angle identities, and solve polynomial equations with complex roots. This article presents the theorem for AQA A-Level Mathematics, with proofs, examples, and common exam pitfalls.
棣莫弗定理是复数领域的基石，将三角学与代数通过极坐标形式紧密联系。它使我们能够计算复数的幂与根、推导三倍角公式，并求解含复根的方程。本文面向 AQA A-Level 数学，系统讲解定理的证明、应用及常见考试陷阱。
1. The Polar Form of a Complex Number | 复数的极坐标形式
omplex number z = x + iy can be written in polar form as z = r(cos θ + i sin θ), where r = |z| = √x² + y² and θ = arg z, defined modulo 2π. The shorthand cis θ = cos θ + i sin θ is often used.
在陈述棣莫弗定理之前，我们先回顾：任意非零复数 z = x + iy 可写成极坐标形式 z = r(cos θ + i sin θ)，其中 r = |z| = √x² + y²，θ = arg z（模 2π 定义）。常使用简写 cis θ = cos θ + i sin θ。
Example: z = 1 + i√3 has r = 2 and θ = π/3, so z = 2 cis(π/3).
例如：z = 1 + i√3 的模 r = 2，辐角 θ = π/3，因此 z = 2 cis(π/3)。
2. Statement of De Moivre’s Theorem | 棣莫弗定理的表述
For any real number θ and any integer n,
对任意实数 θ 和任意整数 n，有
(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)
ero. For n = 0, both sides equal 1 because (cos θ + i sin θ)⁰ = 1 and cos 0 + i sin 0 = 1.
等价地，极坐标形式为：[r cis θ]ⁿ = rⁿ cis(nθ)。该定理对所有整数 n 成立，包括负整数与零。当 n = 0 时，两边均为 1，因为 (cos θ + i sin θ)⁰ = 1 且 cos 0 + i sin 0 = 1。
3. Proof by Induction for Positive Integers | 对正整数 n 的归纳证明
induction. Base case n = 1: (cos θ + i sin θ)¹ = cos θ + i sin θ, which is trivially true.
我们通过归纳法证明 n ≥ 1 的情形。基础情形 n = 1：(cos θ + i sin θ)¹ = cos θ + i sin θ，显然成立。
, i.e. (cos θ + i sin θ)ᵏ = cos(kθ) + i sin(kθ). Multiply both sides by (cos θ + i sin θ):
假设定理对 n = k 成立，即 (cos θ + i sin θ)ᵏ = cos(kθ) + i sin(kθ)。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)
Expanding the product and using the addition formulae:
展开乘积并使用和角公式：
cos(kθ)cos θ − sin(kθ)sin θ + i[sin(kθ)cos θ + cos(kθ)sin θ] = cos[(k+1)θ] + i sin[(k+1)θ]
. By induction, it holds for all positive integers n. Negative integers follow by taking reciprocals; the details are left as an exercise.
因此结论对 n = k + 1 也成立。由归纳法，定理对所有正整数 n 成立。负整数情形可通过取倒数得到，具体步骤留作练习。
4. Applying De Moivre’s Theorem to Powers | 利用定理计算幂
To raise a complex number to a power, first convert to polar form, then raise the modulus to that power and multiply the argument by n (or k).
要计算复数的幂，首先化为极坐标形式，然后将模取 n 次幂、辐角乘以 n（或 k）。
Simplify (2 cis 15°)³. Using the theorem, (2 cis 15°)³ = 2³ cis(3 × 15°) = 8 cis 45° = 8(√2/2 + i√2/2) = 4√2 + 4√2 i.
例 1：化简 (2 cis 15°)³。使用定理：(2 cis 15°)³ = 2³ cis(3 × 15°) = 8 cis 45° = 8(√2/2 + i√2/2) = 4√2 + 4√2 i。
Example 2: Compute (1 + i)⁶. First write 1 + i = √2 cis(π/4). Then (1 + i)⁶ = (√2)⁶ cis(6 × π/4) = 8 cis(3π/2) = 8(0 − i) = −8i.
例 2：计算 (1 + i)⁶。先写出 1 + i = √2 cis(π/4)。则 (1 + i)⁶ = (√2)⁶ cis(6 × π/4) = 8 cis(3π/2) = 8(0 − i) = −8i。
For negative n, use [r cis θ]⁻ⁿ = r⁻ⁿ cis(−nθ). For instance, (1 + i)⁻² = (√2)⁻² cis(−2 × π/4) = ½ cis(−π/2) = −i/2.
对于负整数 n，使用 [r cis θ]⁻ⁿ = r⁻ⁿ cis(−nθ)。例如，(1 + i)⁻² = (√2)⁻² cis(−2 × π/4) = ½ cis(−π/2) = −i/2。
5. Finding Roots of Complex Numbers | 求复数的 n 次方根
-th roots of a complex number. If z = r cis θ, then the n distinct n-th roots are given by
棣莫弗定理用于求一个复数的所有 n 次方根。若 z = r cis θ，则 n 个不同的 n 次方根为
r1/n cis((θ + 2kπ)/n), k = 0, 1, 2, …, n − 1
Because the argument is periodic modulo 2π, the n roots are equally spaced around a circle of radius r1/n.
由于辐角以 2π 为周期，这 n 个根均匀分布在半径为 r1/n 的圆上。
Example: Find all cube roots of 8. Write 8 = 8 cis 0. Then the roots are 81/3 cis(2kπ/3) for k = 0, 1, 2, giving:
例：求 8 的所有三次方根。将 8 写作 8 cis 0。则根为 81/3 cis(2kπ/3)，k = 0, 1, 2，即：
2 cis 0 = 2, 2 cis(2π/3) = −1 + i√3, 2 cis(4π/3) = −1 − i√3
Geometrically, these are three points equally spaced on the circle of radius 2, which sum to zero.
几何上，这三点均匀分布在半径为 2 的圆上，且总和为零。
6. Roots of Unity and Solving Equations | 单位根与解方程
plex roots, called the n-th roots of unity. They are given by z = cis(2kπ/n) for k = 0, 1, … regular n-gon in the complex plane and have the property that their sum is zero for n > 1.
方程 zⁿ = 1 有 n 个复数根，称为 n 次单位根。它们由 z = cis(2kπ/n)（k = 0, 1, …, n − 1）给出。这些根在复平面上构成正 n 边形，且当 n > 1 时其和为零。
32. We get z = 2 cis(2kπ/5) for k = 0, 1, 2, 3, 4. Writing these in Cartesian form gives:
例：解 z⁵ = 32。得到 z = 2 cis(2kπ/5)，k = 0, 1, 2, 3, 4。化为笛卡尔形式：
| k | Root |
| 0 | 2 |
| 1 | 2(cos 72° + i sin 72°) ≈ 0.618 + 1.902i |
| 2 | 2(cos 144° + i sin 144°) ≈ −1.618 + 1.176i |
| 3 | 2(cos 216° + i sin 216°) ≈ −1.618 − 1.176i |
| 4 | 2(cos 288° + i sin 288°) ≈ 0.618 − 1.902i |
Note that the roots come in conjugate pairs because the polynomial has real coefficients.
注意根成共轭对出现，因为该多项式具有实系数。
7. Deriving Trigonometric Identities | 推导三角恒等式
De Moivre’s Theorem provides a systematic way to express cos(nθ) and sin(nθ) in terms of powers of cos θ and sin θ. Expand (cos θ + i sin θ)ⁿ by the binomial theorem, then equate real and imaginary parts.
棣莫弗定理提供了一种系统方法，将 cos(nθ) 和 sin(nθ) 表示为 cos θ 和 sin θ 的幂。利用二项式定理展开 (cos θ + i sin θ)ⁿ，再比较实部与虚部。
Example: For n = 3, expand:
例：对 n = 3 展开：
(cos θ + i sin θ)³ = cos³ θ + 3i cos² θ sin θ − 3 cos θ sin² θ − i sin³ θ
s θ. Equating imaginary parts gives sin 3θ = 3 cos² θ sin θ − sin³ θ = 3 sin θ − 4 sin³ θ.
取实部：cos 3θ = cos³ θ − 3 cos θ sin² θ。由 sin² θ = 1 − cos² θ，化简得 cos 3θ = 4 cos³ θ − 3 cos θ。比较虚部得 sin 3θ = 3 cos² θ sin θ − sin³ θ = 3 sin θ − 4 sin³ θ。
These identities are particularly useful when solving cubic equations, as shown in the next section.
这些恒等式在解三次方程时特别有用，如下一节所示。
8. Solving Trigonometric Equations with Multiple Angles | 解多倍角三角方程
, we can solve equations such as cos 3θ = 1/2 by setting z = cos θ + i sin θ and using z³.
棣莫弗定理也可以逆向使用：若已知 cos(nθ) 或 sin(nθ)，我们可以通过令 z = cos θ + i sin θ 并利用 z³ 来解方程，如 cos 3θ = 1/2。
π/9 + 2kπ/3 or θ = −π/9 + 2kπ/3. For k = 0, 1, 2, we obtain six solutions in the interval.
例如，解区间 [0, 2π) 内的 cos 3θ = ½。写出 3θ = ±π/3 + 2kπ，故 θ = π/9 + 2kπ/3 或 θ = −π/9 + 2kπ/3。取 k = 0, 1, 2，可得区间内六个解。
These equations appear frequently in mechanics (e.g. three forces in equilibrium) and in Fourier analysis.
这类方程在力学（如三力平衡）和傅里叶分析中经常出现。
9. Summing Series and Evaluating Expressions | 级数求和与表达式求值
Treat the series as the real and imaginary parts of a geometric series of complex numbers:
一个强大的应用是求形如 ∑ cos(kθ) 或 ∑ sin(kθ) 的三角级数。将该级数视为复数等比级数的实部与虚部：
S = 1 + eiθ + e2iθ + … + e(n−1)iθ = (1 − eniθ)/(1 − eiθ)
Then ∑ cos(kθ) = Re S and ∑ sin(kθ) = Im S. For example, summing cos(π/3) + cos(2π/3) + … + cos(nπ/3) can be done elegantly using this method.
此时 ∑ cos(kθ) = Re S，∑ sin(kθ) = Im S。例如，利用此方法可优雅地求 cos(π/3) + cos(2π/3) + … + cos(nπ/3) 之和。
This technique is a favourite in A-Level Further Mathematics and in some AQA papers, so practise converting between exponential and polar forms.
该技巧是 A-Level 进阶数学以及某些 AQA 试卷中的热门考点，务必练习指数形式与极坐标形式之间的转换。
10. Common Mistakes and Exam Tips | 常见错误与应试提示
Common mistakes include: using degrees and radians inconsistently; forgetting that the modulus must be raised to the n-th power; omitting the 2kπ when finding roots; and applying the theorem to non-integer n without adjusting the argument range.
常见错误包括：角度制与弧度制混用；忘记模必须取 n 次幂；求根时遗漏 2kπ；以及未调整辐角范围就对非整数 n 使用定理。
equired. Check that the sum of roots is zero for zⁿ = a when n > 1, as a quick self-check.
考试中务必写出极坐标形式步骤。求根时列出 k = 0 到 n − 1，并按题目要求给出极坐标与笛卡尔形式。对于 zⁿ = a（n > 1），可用所有根之和为零快速自我检验。
n (−π, π] for AQA, but sometimes 0 ≤ θ < 2π is specified. Adjust your answers accordingly.
还需注意主辐角：AQA 中主辐角在 (−π, π] 内，但有时题目要求 0 ≤ θ < 2π。请按题目要求调整答案。
11. Worked Exam-Style Problem | 典型考试例题精解
Problem (AQA-style): (a) Express z = 1 − i√3 in polar form. (b) Hence find z⁴. (c) Find the two square roots of z.
题目（AQA 风格）：(a) 将 z = 1 − i√3 化为极坐标形式。(b) 由此求 z⁴。(c) 求 z 的两个平方根。
Solution: (a) r = √(1² + (−√3)²) = 2. The argument satisfies cos θ = 1/2, sin θ = −√3/2, so θ = −π/3 (or 5π/3). Hence z = 2 cis(−π/3).
解：(a) r = √(1² + (−√3)²) = 2。辐角满足 cos θ = 1/2，sin θ = −√3/2，故 θ = −π/3（或 5π/3）。因此 z = 2 cis(−π/3)。
(b) z⁴ = 2⁴ cis(4 × (−π/3)) = 16 cis(−4π/3) = 16 cis(2π/3) = 16(−1/2 + i√3/2) = −8 + 8√3 i.
(b) z⁴ = 2⁴ cis(4 × (−π/3)) = 16 cis(−4π/3) = 16 cis(2π/3) = 16(−1/2 + i√3/2) = −8 + 8√3 i。
(c) The square roots are √2 cis((−π/3 + 2kπ)/2), k = 0, 1. For k = 0, √2 cis(−π/6) = √2(√3/2 − i/2) = (√6 − i√2)/2. For k = 1, √2 cis(5π/6) = √2(−√3/2 + i/2) = (−√6 + i√2)/2.
(c) 平方根为 √2 cis((−π/3 + 2kπ)/2)，k = 0, 1。当 k = 0 时，√2 cis(−π/6) = √2(√3/2 − i/2) = (√6 − i√2)/2。当 k = 1 时，√2 cis(5π/6) = √2(−√3/2 + i/2) = (−√6 + i√2)/2。
Checking: the two roots are negatives of each other, as expected for square roots of a complex number.
检验：两个根互为相反数，这符合复数平方根的预期性质。
12. Conclusion and Summary | 结论与总结
De Moivre’s Theorem is more than a formula: it unifies complex arithmetic, trigonometry and algebra. Mastery of its proof, the polar form, and its applications to powers, roots, identities and equations will earn you marks across multiple A-Level topic areas.
棣莫弗定理不仅仅是一个公式：它统一了复数运算、三角学与代数。掌握其证明、极坐标形式，以及其在幂、根、恒等式和方程中的应用，将帮助你在 A-Level 多个主题中获得分数。
nclude 2kπ for roots, and equate real/imaginary parts for identities. With regular practice, De Moivre’s Theorem becomes a reliable tool rather than a daunting one.
记住关键步骤：化为极坐标形式，应用 [r cis θ]ⁿ = rⁿ cis(nθ)，求根时添加 2kπ，并用实部虚部相等法推恒等式。通过经常练习，棣莫弗定理将从令人畏惧的难点变为得心应手的工具。
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