De Moivre's theorem and its applications | 棣莫弗定理及其应用

📚 De Moivre's theorem and its applications | 棣莫弗定理及其应用

De Moivre's theorem is one of the most important tools in A-Level Further Mathematics for working with complex numbers in polar form. It links powers of complex numbers on the unit circle to multiples of an angle, and it underpins methods for proving trigonometric identities, finding roots of complex numbers, and solving polynomial equations.

棣莫弗定理是 A-Level 进阶数学中处理极坐标形式复数最重要的工具之一。它把单位圆上复数的幂与角度的倍数联系起来，是证明三角恒等式、求复数根以及解多项式方程等技巧的基础。

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1. Statement of De Moivre's theorem | 棣莫弗定理的表述

If a complex number has modulus 1 and argument θ, it can be written as cos θ + i sin θ. De Moivre's theorem states that raising this complex number to any integer power n is equivalent to multiplying the argument by n while keeping the modulus equal to 1.

如果一个复数的模为 1、辐角为 θ，那么它可以写成 cos θ + i sin θ。棣莫弗定理指出，把这个复数进行任意整数次幂运算，等价于把辐角乘以 n，而模保持为 1。

(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ for all integers n

ger n. This follows because the modulus r is raised to the power n and the argument θ is multiplied by n.

更一般地，如果 z = r(cos θ + i sin θ)，那么对于整数 n，有 zⁿ = rⁿ(cos nθ + i sin nθ)。这是因为模 r 乘方 n 次，而辐角 θ 乘以 n。

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2. Proof by induction for positive integers | 正整数情形的数学归纳法证明

For AQA exam purposes, it is useful to know how De Moivre' ositive integers. The base case n = 1 is immediate because both sides are cos θ + i sin θ.

从 AQA 考试的角度看，了解如何用数学归纳法证明正整数情形的棣莫弗定理是很有用的。基础情形 n = 1 显然成立，因为两边都是 cos θ + i sin θ。

eir moduli multiply and their arguments add, so the result has argument kθ + θ = (k + 1)θ.

假设结论对 n = k 成立，即 (cos θ + i sin θ)ᵏ = cos kθ + i sin kθ。现在两边同乘 cos θ + i sin θ。两个极坐标形式的复数相乘时，模相乘、辐角相加，因此结果的辐角为 kθ + θ = (k + 1)θ。

+ 1)θ + i sin(k + 1)θ. By mathematical induction, the theorem holds for all positive integers n.

因此，(cos θ + i sin θ)ᵏ⁺¹ = cos(k + 1)θ + i sin(k + 1)θ。根据数学归纳法，定理对所有正整数 n 都成立。

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3. Extending to zero and negative integer powers | 推广到零指数和负整数指数

De Moivre' -hand side is 1, and the right-hand side is cos 0 + i sin 0 = 1, so the result is trivial.

棣莫弗定理对 n = 0 以及负整数指数也成立。如果 n = 0，左边为 1，右边为 cos 0 + i sin 0 = 1，因此结论显然成立。

its conjugate, this equals cos mθ − i sin mθ = cos(−mθ) + i sin(−mθ) = cos nθ + i sin nθ.

对于负整数 n，令 n = −m，其中 m 为正整数。那么 (cos θ + i sin θ)ⁿ = 1 / (cos θ + i sin θ)ᵐ = 1 / (cos mθ + i sin mθ)。由于单位圆上复数的倒数等于它的共轭，所以结果为 cos mθ − i sin mθ = cos(−mθ) + i sin(−mθ) = cos nθ + i sin nθ。

Thus De Moivre's theorem is valid for all integers n, which is the version required at A-Level.

因此，棣莫弗定理对所有整数 n 都成立，这正是 A-Level 所要求的版本。

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4. Deriving multiple-angle identities | 推导倍角恒等式

One of the most common applications of De Moivre's theorem is deriving identities for cos nθ and sin nθ in terms of powers of sin θ and cos θ. For example, expand (cos θ + i sin θ)³ using the binomial theorem.

棣莫弗定理最常见的应用之一，是用 sin θ 和 cos θ 的幂来表示 cos nθ 和 sin nθ。例如，用二项式定理展开 (cos θ + i sin θ)³。

(cos θ + i sin θ)³ = cos³θ + 3i cos²θ sin θ − 3 cos θ sin²θ − i sin³θ

By De Moivre' d imaginary parts gives cos 3θ = cos³θ − 3 cos θ sin²θ and sin 3θ = 3 cos²θ sin θ − sin³θ.

根据棣莫弗定理，这个展开式也等于 cos 3θ + i sin 3θ。比较实部和虚部可得 cos 3θ = cos³θ − 3 cos θ sin²θ，以及 sin 3θ = 3 cos²θ sin θ − sin³θ。

. For larger n, equating real and imaginary parts after binomial expansion is a standard exam technique.

对于 n = 2，同样的方法可以得到熟悉的倍角公式 cos 2θ = cos²θ − sin²θ 和 sin 2θ = 2 sin θ cos θ。对于更大的 n，先进行二项式展开，再比较实部和虚部，是标准的考试技巧。

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5. Expressing powers of sin θ and cos θ as multiple angles | 把 sin θ 和 cos θ 的幂表示为倍角之和

De Moivre's theorem can also be used in reverse to express powers of sin θ and cos θ as sums of cosines or sines of multiple angles. This is especially useful in integration.

棣莫弗定理还可以反过来使用，把 sin θ 和 cos θ 的幂表示为倍角的余弦或正弦之和。这在积分中尤其有用。

θ. These two identities allow powers of cos θ or sin θ to be rewritten in terms of z and z⁻¹.

令 z = cos θ + i sin θ，则 z⁻¹ = cos θ − i sin θ。于是 z + z⁻¹ = 2 cos θ，z − z⁻¹ = 2i sin θ。这两个恒等式可以把 cos θ 或 sin θ 的幂改写成 z 和 z⁻¹ 的形式。

rouping z³ + z⁻³ as 2 cos 3θ and z + z⁻¹ as 2 cos θ gives cos³θ = (1/4)(cos 3θ + 3 cos θ).

例如，cos³θ = ((z + z⁻¹) / 2)³ = (1/8)(z³ + 3z + 3z⁻¹ + z⁻³)。把 z³ + z⁻³ 合并为 2 cos 3θ，把 z + z⁻¹ 合并为 2 cos θ，就得到 cos³θ = (1/4)(cos 3θ + 3 cos θ)。

such as ∫ cos³θ dθ easy to evaluate, because cos 3θ and cos θ can be integrated directly.

这样就可以直接计算 ∫ cos³θ dθ 这样的积分，因为 cos 3θ 和 cos θ 都可以直接积分。

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6. Finding nth roots of a complex number | 求复数的 n 次方根

De Moivre' n the equation zⁿ = w has exactly n distinct solutions for z when n is a positive integer.

棣莫弗定理是求复数 n 次方根的关键工具。如果 w = r(cos θ + i sin θ)，那么当 n 为正整数时，方程 zⁿ = w 恰好有 n 个不同的解。

te w = r(cos(θ + 2kπ) + i sin(θ + 2kπ)), where k is an integer. Taking the nth root gives:

由于复数的辐角只能确定到 2π 的整数倍，因此写成 w = r(cos(θ + 2kπ) + i sin(θ + 2kπ))，其中 k 为整数。开 n 次方根得到：

z = r^1/n ( cos((θ + 2kπ) / n) + i sin((θ + 2kπ) / n) )

all n distinct roots. Any other integer value of k simply repeats one of these roots because the arguments differ by an integer multiple of 2π.

取 k = 0, 1, 2, ..., n − 1 可以得到所有 n 个不同的根。任何其他整数值 k 都只是重复其中一个根，因为辐角相差 2π 的整数倍。

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7. Roots of unity and their geometric meaning | 单位根及其几何意义

unity. They are given by z = cos(2kπ / n) + i sin(2kπ / n), where k = 0, 1, 2, ..., n − 1.

当 w = 1 时，方程 zⁿ = 1 的解称为 n 次单位根。它们由 z = cos(2kπ / n) + i sin(2kπ / n) 给出，其中 k = 0, 1, 2, ..., n − 1。

Geometrically, the nth roots of unity lie on the unit circle centred at the origin and are equally spaced around it. They form the vertices of a regular n-gon with one vertex at 1.

从几何上看，n 次单位根位于以原点为圆心的单位圆上，并且均匀分布。它们构成一个正 n 边形的顶点，其中一个顶点在 1 处。

e roots can be written as 1, ω, ω², ..., ωⁿ⁻¹. The sum of all nth roots of unity is 0 for n > 1, which follows from the symmetry of the regular polygon or from the geometric series formula.

如果令 ω = cos(2π / n) + i sin(2π / n)，那么所有根可以写成 1, ω, ω², ..., ωⁿ⁻¹。当 n > 1 时，所有 n 次单位根的和为 0，这一点可以从正多边形的对称性看出，也可以由等比数列求和公式得到。

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8. Solving polynomial equations using De Moivre's theorem | 利用棣莫弗定理求解多项式方程

method is always the same: write the right-hand side in polar form, add 2kπ to the argument, and then apply De Moivre's theorem to take the nth root.

许多 A-Level 题目要求解 zⁿ = a + bi 形式的方程。方法始终相同：把右边写成极坐标形式，给辐角加上 2kπ，然后应用棣莫弗定理求 n 次方根。

) + i sin(π/2 + 2kπ)), and the roots are z = 2(cos((π/2 + 2kπ)/3) + i sin((π/2 + 2kπ)/3)).

例如，解 z³ = 8i，首先注意 8i 的模为 8、辐角为 π/2。因此 z³ = 8(cos(π/2 + 2kπ) + i sin(π/2 + 2kπ))，所以根为 z = 2(cos((π/2 + 2kπ)/3) + i sin((π/2 + 2kπ)/3))。

Substituting k = 0, 1, 2 gives z = √3 + i, z = −√3 + i, and z = −2i. Always give all distinct roots, not just the principal root.

代入 k = 0, 1, 2 得到 z = √3 + i，z = −√3 + i，以及 z = −2i。答题时一定要给出所有不同的根，而不能只给出

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