📚 De Moivre’s Theorem: Powers of Complex Numbers | 棣莫弗定理：复数的幂运算
De Moivre’s theorem is one of the most elegant and powerful results in complex number theory. It provides a direct method for raising complex numbers to integer, negative, and fractional powers, converting what would otherwise be messy algebraic expansions into simple trigonometric manipulations. This topic is a core component of the AQA A-Level Mathematics specification and frequently appears in both pure mathematics and further mathematics examinations.
棣莫弗定理是复数理论中最优雅且最强大的结果之一。它为计算复数的整数次、负整数次和分数次幂提供了一种直接的方法，将原本繁琐的代数展开转化为简单的三角运算。这一主题是 AQA A-Level 数学大纲的核心内容，在纯数学和进阶数学考试中都频繁出现。
1. The Polar Form of a Complex Number | 复数的极坐标形式
Before we can apply de Moivre’s theorem, we must express complex numbers in polar form. A complex number z = a + bi can be represented in polar form as z = r(cos θ + i sin θ), where r = √(a² + b²) is the modulus and θ = arctan(b/a) is the argument. The argument is measured in radians and is usually taken in the principal range -π < θ ≤ π.
在应用棣莫弗定理之前，我们必须先将复数表示为极坐标形式。复数 z = a + bi 可以表示为极坐标形式 z = r(cos θ + i sin θ)，其中 r = √(a² + b²) 是模长，θ = arctan(b/a) 是辐角。辐角以弧度为单位，通常取主值范围 -π < θ ≤ π。
For example, the complex number z = 1 + i√3 has modulus r = √(1² + (√3)²) = 2 and argument θ = arctan(√3) = π/3. Thus z = 2(cos(π/3) + i sin(π/3)). This form is essential because multiplication of complex numbers becomes simple: moduli multiply and arguments add.
例如，复数 z = 1 + i√3 的模长为 r = √(1² + (√3)²) = 2，辐角 θ = arctan(√3) = π/3。因此 z = 2(cos(π/3) + i sin(π/3))。这种形式至关重要，因为复数的乘法变得简单：模长相乘，辐角相加。
z = r(cos θ + i sin θ), where r = |z| and θ = arg(z)
2. Statement of de Moivre’s Theorem | 棣莫弗定理的表述
De Moivre’s theorem states that for any real number θ and any integer n, the following relationship holds: (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ). More generally, for a complex number in polar form z = r(cos θ + i sin θ), we have zⁿ = rⁿ(cos(nθ) + i sin(nθ)).
棣莫弗定理指出，对于任意实数 θ 和任意整数 n，以下关系成立：(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)。更一般地，对于极坐标形式的复数 z = r(cos θ + i sin θ)，我们有 zⁿ = rⁿ(cos(nθ) + i sin(nθ))。
The theorem also holds when n is a rational number, though in that case the result is multi-valued due to the periodic nature of the trigonometric functions. When n is an integer, the result is unique. The power of this theorem lies in its ability to transform exponentiation — a multiplicative operation — into rotation and scaling — geometric operations.
当 n 为有理数时，该定理同样成立，但由于三角函数的周期性，此时结果是多值的。当 n 为整数时，结果是唯一的。该定理的强大之处在于它将幂运算——一种乘法操作——转化为旋转和缩放——两种几何操作。
(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ) for n ∈ ℤ
3. Proof by Induction | 数学归纳法证明
The most rigorous proof of de Moivre’s theorem for positive integers uses mathematical induction. The base case n = 1 is trivially true: (cos θ + i sin θ)¹ = cos θ + i sin θ. We then assume the statement holds for n = k and prove it for n = k + 1.
棣莫弗定理对正整数的最严谨证明使用数学归纳法。基础情形 n = 1 显然成立：(cos θ + i sin θ)¹ = cos θ + i sin θ。然后我们假设命题对 n = k 成立，并证明它对 n = k + 1 也成立。
Assume (cos θ + i sin θ)ᵏ = cos(kθ) + i sin(kθ). Multiplying both sides by (cos θ + i sin θ), we obtain (cos θ + i sin θ)ᵏ⁺¹ = [cos(kθ) + i sin(kθ)][cos θ + i sin θ]. Expanding the product and applying the compound angle identities gives cos((k+1)θ) + i sin((k+1)θ), completing the induction.
假设 (cos θ + i sin θ)ᵏ = cos(kθ) + i sin(kθ)。两边同乘 (cos θ + i sin θ)，得到 (cos θ + i sin θ)ᵏ⁺¹ = [cos(kθ) + i sin(kθ)][cos θ + i sin θ]。展开乘积并应用和角公式，得到 cos((k+1)θ) + i sin((k+1)θ)，归纳完成。
For negative integers, we use the fact that (cos θ + i sin θ)⁻¹ = cos θ – i sin θ = cos(-θ) + i sin(-θ), which is simply the theorem with n = -1. Combining this with the positive case establishes the theorem for all integers.
对于负整数，我们利用 (cos θ + i sin θ)⁻¹ = cos θ – i sin θ = cos(-θ) + i sin(-θ)，这正是 n = -1 时的定理。将此与正整数情形结合，即可确立定理对所有整数成立。
4. Evaluating Integer Powers | 计算整数次幂
To raise a complex number to an integer power, we first convert it to polar form and then apply de Moivre’s theorem. Consider z = 1 + i. We find r = √(1² + 1²) = √2 and θ = arctan(1) = π/4, so z = √2(cos(π/4) + i sin(π/4)).
要将复数提高到整数次幂，我们首先将其转换为极坐标形式，然后应用棣莫弗定理。考虑 z = 1 + i。我们求得 r = √(1² + 1²) = √2，θ = arctan(1) = π/4，因此 z = √2(cos(π/4) + i sin(π/4))。
To compute z⁶, we apply the theorem: z⁶ = (√2)⁶[cos(6 × π/4) + i sin(6 × π/4)] = 8[cos(3π/2) + i sin(3π/2)] = 8(0 – i) = -8i. Without de Moivre’s theorem, expanding (1 + i)⁶ would require the binomial theorem with complex terms — a far more laborious calculation.
要计算 z⁶，我们应用定理：z⁶ = (√2)⁶[cos(6 × π/4) + i sin(6 × π/4)] = 8[cos(3π/2) + i sin(3π/2)] = 8(0 – i) = -8i。如果没有棣莫弗定理，展开 (1 + i)⁶ 需要使用含复数项的二项式定理——计算量要大得多。
When the argument is large, we simplify by subtracting multiples of 2π before applying the theorem. For example, to compute (cos(7π/3) + i sin(7π/3))⁵, we note that 5 × 7π/3 = 35π/3 = 11π + 2π/3, so the result is cos(2π/3) + i sin(2π/3) = -1/2 + i√3/2.
当辐角较大时，我们先减去 2π 的整数倍再应用定理。例如，计算 (cos(7π/3) + i sin(7π/3))⁵ 时，我们注意到 5 × 7π/3 = 35π/3 = 11π + 2π/3，因此结果为 cos(2π/3) + i sin(2π/3) = -1/2 + i√3/2。
5. Negative Powers and Reciprocals | 负指数幂与倒数
De Moivre’s theorem handles negative powers seamlessly. Since (cos θ + i sin θ)⁻ⁿ = cos(-nθ) + i sin(-nθ) = cos(nθ) – i sin(nθ), we can compute reciprocals directly. For a complex number z = r(cos θ + i sin θ), the reciprocal is z⁻¹ = r⁻¹(cos(-θ) + i sin(-θ)).
棣莫弗定理可以完美地处理负指数。由于 (cos θ + i sin θ)⁻ⁿ = cos(-nθ) + i sin(-nθ) = cos(nθ) – i sin(nθ)，我们可以直接计算倒数。对于复数 z = r(cos θ + i sin θ)，其倒数为 z⁻¹ = r⁻¹(cos(-θ) + i sin(-θ))。
As an example, consider z = 2(cos(π/3) + i sin(π/3)). Then z⁻¹ = (1/2)(cos(-π/3) + i sin(-π/3)) = (1/2)(1/2 – i√3/2) = 1/4 – i√3/4. We can verify this: 2(cos(π/3) + i sin(π/3)) × (1/2)(cos(-π/3) + i sin(-π/3)) = 1(cos 0 + i sin 0) = 1. ✓
例如，考虑 z = 2(cos(π/3) + i sin(π/3))。则 z⁻¹ = (1/2)(cos(-π/3) + i sin(-π/3)) = (1/2)(1/2 – i√3/2) = 1/4 – i√3/4。我们可以验证：2(cos(π/3) + i sin(π/3)) × (1/2)(cos(-π/3) + i sin(-π/3)) = 1(cos 0 + i sin 0) = 1。✓
For higher negative powers, such as z⁻³, we compute z⁻³ = r⁻³(cos(-3θ) + i sin(-3θ)). This avoids the need to calculate the reciprocal in Cartesian form first, which often involves rationalising a denominator with complex terms.
对于更高的负指数，如 z⁻³，我们计算 z⁻³ = r⁻³(cos(-3θ) + i sin(-3θ))。这避免了我们先在笛卡尔形式下计算倒数（通常需要对含复数项的分母进行有理化）的麻烦。
6. Fractional Powers and Roots | 分数指数幂与方根
De Moivre’s theorem extends to fractional powers, but with an important caveat: the result is multi-valued. To find the n-th roots of a complex number z = r(cos θ + i sin θ), we use the formula z^(1/n) = r^(1/n)(cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)) for k = 0, 1, 2, …, n-1.
棣莫弗定理可以扩展到分数指数，但有一个重要的注意事项：结果是多值的。要求复数 z = r(cos θ + i sin θ) 的 n 次方根，我们使用公式 z^(1/n) = r^(1/n)(cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n))，其中 k = 0, 1, 2, …, n-1。
For example, the cube roots of 8 are found by writing 8 = 8(cos 0 + i sin 0). Then the cube roots are 8^(1/3)(cos(2kπ/3) + i sin(2kπ/3)) for k = 0, 1, 2. This gives: 2(cos 0 + i sin 0) = 2; 2(cos(2π/3) + i sin(2π/3)) = -1 + i√3; and 2(cos(4π/3) + i sin(4π/3)) = -1 – i√3.
例如，8 的立方根可以通过将 8 写成 8(cos 0 + i sin 0) 来求得。然后立方根为 8^(1/3)(cos(2kπ/3) + i sin(2kπ/3))，其中 k = 0, 1, 2。这给出：2(cos 0 + i sin 0) = 2；2(cos(2π/3) + i sin(2π/3)) = -1 + i√3；以及 2(cos(4π/3) + i sin(4π/3)) = -1 – i√3。
The n-th roots of any complex number always lie on a circle of radius r^(1/n) in the complex plane, equally spaced at angles of 2π/n. This geometric fact makes finding roots a visual and intuitive process.
任何复数的 n 次方根总是位于复平面上半径为 r^(1/n) 的圆上，以 2π/n 的角度等间距分布。这一几何事实使得求根过程变得直观。
7. Roots of Unity | 单位根
The n-th roots of unity are the solutions to the equation zⁿ = 1. Using de Moivre’s theorem, we write 1 = 1(cos 0 + i sin 0), and the roots are cos(2kπ/n) + i sin(2kπ/n) for k = 0, 1, …, n-1. These roots have remarkable algebraic and geometric properties.
单位根是方程 zⁿ = 1 的解。利用棣莫弗定理，我们将 1 写成 1(cos 0 + i sin 0)，根为 cos(2kπ/n) + i sin(2kπ/n)，其中 k = 0, 1, …, n-1。这些根具有显著的代数和几何性质。
For n = 3, the cube roots of unity are 1, ω = -1/2 + i√3/2, and ω² = -1/2 – i√3/2, where ω satisfies ω³ = 1 and 1 + ω + ω² = 0. These properties are frequently tested in A-Level questions and are invaluable for simplifying complex algebraic expressions involving roots of unity.
对于 n = 3，单位立方根为 1、ω = -1/2 + i√3/2 和 ω² = -1/2 – i√3/2，其中 ω 满足 ω³ = 1 且 1 + ω + ω² = 0。这些性质在 A-Level 考试中经常出现，对于化简涉及单位根的复杂代数表达式非常有价值。
For n = 4, the fourth roots of unity are 1, i, -1, and -i, which form the vertices of a square in the complex plane. In general, the n-th roots of unity form a regular n-gon centred at the origin with one vertex at the point 1.
对于 n = 4，单位四次根为 1、i、-1 和 -i，它们在复平面上构成正方形的四个顶点。一般而言，n 次单位根在复平面上构成以原点为中心的正 n 边形，其中一个顶点位于点 1 处。
8. Trigonometric Identities from de Moivre | 由棣莫弗定理推导三角恒等式
One of the most important applications of de Moivre’s theorem is deriving trigonometric identities. By expanding (cos θ + i sin θ)ⁿ using the binomial theorem and equating real and imaginary parts, we obtain expressions for cos(nθ) and sin(nθ) in terms of powers of cos θ and sin θ.
棣莫弗定理最重要的应用之一就是推导三角恒等式。通过使用二项式定理展开 (cos θ + i sin θ)ⁿ，并比较实部和虚部，我们可以得到 cos(nθ) 和 sin(nθ) 关于 cos θ 和 sin θ 的幂的表达式。
For n = 2: (cos θ + i sin θ)² = cos(2θ) + i sin(2θ). Expanding the left side: cos²θ – sin²θ + 2i sin θ cos θ. Equating real parts: cos(2θ) = cos²θ – sin²θ. Equating imaginary parts: sin(2θ) = 2 sin θ cos θ. These are the familiar double-angle formulae.
对于 n = 2：(cos θ + i sin θ)² = cos(2θ) + i sin(2θ)。展开左边：cos²θ – sin²θ + 2i sin θ cos θ。比较实部：cos(2θ) = cos²θ – sin²θ。比较虚部：sin(2θ) = 2 sin θ cos θ。这就是熟悉的二倍角公式。
For n = 3: cos(3θ) = 4cos³θ – 3cos θ and sin(3θ) = 3sin θ – 4sin³θ. These triple-angle identities appear regularly in A-Level papers, and examiners often expect students to derive them directly from de Moivre’s theorem rather than memorising them.
对于 n = 3：cos(3θ) = 4cos³θ – 3cos θ，sin(3θ) = 3sin θ – 4sin³θ。这些三倍角恒等式在 A-Level 试卷中经常出现，考官通常期望学生直接从棣莫弗定理推导它们，而不是死记硬背。
9. Geometric Interpretation | 几何解释
Multiplying a complex number by another complex number of unit modulus corresponds to a rotation in the complex plane. De Moivre’s theorem makes this precise: multiplying z by (cos θ + i sin θ) rotates z counterclockwise by angle θ. Raising z to the power n therefore rotates z by nθ and scales its modulus by rⁿ.
将一个复数乘以一个单位模长的复数对应于复平面上的旋转。棣莫弗定理使得这一点变得精确：将 z 乘以 (cos θ + i sin θ) 相当于将 z 逆时针旋转角度 θ。因此，将 z 提高至 n 次幂相当于将 z 旋转 nθ 并将其模长缩放 rⁿ。
This perspective explains why the n-th roots of a complex number are equally spaced: the argument must be divided into n equal parts, and adding 2kπ to the original argument (which does not change the number) yields n distinct positions on the circle of radius r^(1/n).
这一视角解释了为什么复数的 n 次方根是等间距分布的：辐角必须被分成 n 等份，而将原始辐角加上 2kπ（这不改变该数）会在半径为 r^(1/n) 的圆上产生 n 个不同的位置。
For instance, the square roots of i are found by noting i = cos(π/2) + i sin(π/2). The square roots are cos(π/4) + i sin(π/4) = (1 + i)/√2 and cos(5π/4) + i sin(5π/4) = -(1 + i)/√2, which lie at opposite ends of a diameter.
例如，i 的平方根可通过注意到 i = cos(π/2) + i sin(π/2) 来求得。平方根为 cos(π/4) + i sin(π/4) = (1 + i)/√2 和 cos(5π/4) + i sin(5π/4) = -(1 + i)/√2，它们位于直径的两端。
10. Worked Exam Question | 考试真题演练
Let us work through a typical AQA examination question. Question: Given that z = √3 + i, find z⁵ in Cartesian form a + bi, where a and b are real numbers.
让我们完整地解一道典型的 AQA 考试题。题目：已知 z = √3 + i，求 z⁵ 的笛卡尔形式 a + bi，其中 a、b 为实数。
Step 1 — Find the modulus and argument. r = √((√3)² + 1²) = √(3 + 1) = 2. The argument θ satisfies tan θ = 1/√3, and since z is in the first quadrant, θ = π/6. Thus z = 2(cos(π/6) + i sin(π/6)).
第一步——求模长和辐角。r = √((√3)² + 1²) = √(3 + 1) = 2。辐角 θ 满足 tan θ = 1/√3，由于 z 在第一象限，θ = π/6。因此 z = 2(cos(π/6) + i sin(π/6))。
Step 2 — Apply de Moivre’s theorem. z⁵ = 2⁵(cos(5π/6) + i sin(5π/6)) = 32(-√3/2 + i(1/2)) = -16√3 + 16i.
第二步——应用棣莫弗定理。z⁵ = 2⁵(cos(5π/6) + i sin(5π/6)) = 32(-√3/2 + i(1/2)) = -16√3 + 16i。
Step 3 — Verify the answer. We can check the modulus: |z⁵| = √((-16√3)² + 16²) = √(768 + 256) = √1024 = 32 = 2⁵. The modulus is correct, confirming our calculation.
第三步——验证答案。我们可以检查模长：|z⁵| = √((-16√3)² + 16²) = √(768 + 256) = √1024 = 32 = 2⁵。模长正确，确认我们的计算无误。
11. Common Pitfalls | 常见错误
Students frequently make several recurring errors when applying de Moivre’s theorem. The first is forgetting to include the modulus r⁵ when raising to a power — the theorem states zⁿ = rⁿ(cos(nθ) + i sin(nθ)), and omitting the rⁿ factor leads to answers with an incorrect modulus.
学生在应用棣莫弗定理时常犯几个反复出现的错误。第一个是忘记在幂运算中包含模长的幂 rⁿ——定理指出 zⁿ = rⁿ(cos(nθ) + i sin(nθ))，省略 rⁿ 因子会导致答案模长不正确。
The second common mistake is using degrees instead of radians for the argument when applying the theorem, particularly when combining it with compound angle identities or when the question specifies radians. Always check whether the argument should be in radians and keep a consistent unit throughout.
第二个常见错误是在应用定理时辐角使用度数而非弧度，尤其是与和角公式结合使用或题目明确要求弧度时。始终检查辐角是否应以弧度为单位，并在整个计算过程中保持单位一致。
The third mistake occurs with negative or fractional powers: students forget that the argument must be manipulated carefully. For roots, it is essential to add 2kπ to the argument before dividing by n to obtain all distinct roots. Forgetting this step yields only one root instead of n distinct roots.
第三个错误出现在负指数或分数指数的情况：学生忘记辐角必须仔细处理。对于方根，必须在除以 n 之前将辐角加上 2kπ，才能得到所有不同的根。忘记这一步只会得到一个根，而非 n 个不同的根。
12. Summary | 总结
De Moivre’s theorem is a cornerstone of complex number manipulation at A-Level. The key results to remember are: (1) for integer n, (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ); (2) for roots, add 2kπ before dividing; (3) the theorem enables both power calculations and trigonometric identity derivation. Mastery of these techniques will serve you well across multiple examination topics.
棣莫弗定理是 A-Level 复数运算的基石。需要记住的关键结果有：(1) 对于整数 n，(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)；(2) 对于方根，先加 2kπ 再除以 n；(3) 该定理既能用于幂运算，也能用于推导三角恒等式。掌握这些技巧将使你在多个考试主题中游刃有余。
When preparing for examinations, practise converting between Cartesian and polar forms fluently, and always verify your final answer by checking that the modulus is correct and that the argument lies in the expected range. With consistent practice, de Moivre’s theorem becomes one of the most reliable tools in your mathematical toolkit.
备考时，请流畅地练习笛卡尔形式与极坐标形式之间的转换，并始终通过检查模长是否正确以及辐角是否在预期范围内来验证最终答案。通过持续练习，棣莫弗定理将成为你数学工具箱中最可靠的工具之一。
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