📚 Example 3.4.3 – Application to de Moivre's Theorem for Positive Integers | 例 3.4.3 – 德莫弗定理对正整数的应用
De Moivre's theorem is one of the most powerful tools in A-Level further mathematics. It connects complex numbers with trigonometry, allowing us to compute powers of complex numbers and derive trigonometric identities in a compact and elegant way. In this article we focus on its application when the exponent is a positive integer.
德莫弗定理是 A-Level 进阶数学中最强大的工具之一。它将复数与三角学联系起来，使我们能够简洁优美地计算复数的幂并推导三角恒等式。本文重点关注指数为正整数时的应用。
1. Statement of de Moivre's Theorem | 德莫弗定理的表述
For any positive integer n, the theorem states:
对于任意正整数 n，定理表述如下：
(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)
Here θ is a real number, i is the imaginary unit, and n is a positive integer. The theorem also holds for all real n, but the positive integer case is the most frequently tested in exams.
其中 θ 是实数，i 是虚数单位，n 是正整数。该定理对所有实数 n 也成立，但正整数情形是考试中最常考查的。
2. Proof by Mathematical Induction | 用数学归纳法证明
eorem for positive integers using induction. First, the base case n = 1 is trivially true:
我们可以用归纳法证明该定理对正整数成立。首先，基例 n = 1 显然成立：
(cos θ + i sin θ)¹ = cos θ + i sin θ
Assume the result holds for some positive integer k. Then for k + 1:
假设结果对某个正整数 k 成立。那么对于 k + 1：
(cos θ + i sin θ)ᵏ⁺¹ = (cos θ + i sin θ)ᵏ (cos θ + i sin θ)
= (cos kθ + i sin kθ)(cos θ + i sin θ)
Expanding using trigonometric addition formulas gives cos((k+1)θ) + i sin((k+1)θ). Hence the theorem holds for all positive integers by induction.
利用三角加法公式展开，得到 cos((k+1)θ) + i sin((k+1)θ)。因此，由归纳法可知定理对所有正整数 n 成立。
3. Basic Application: Computing Powers | 基本应用：计算幂
The most straightforward application is computing powers of complex numbers. Write the complex number in modulus-argument form, apply de Moivre's theorem, then convert back to Cartesian form if required.
最直接的应用是计算复数的幂。先将复数写成模-辐角形式，应用德莫弗定理，必要时再转换回直角坐标形式。
Example: Compute (1 + i)¹⁰.
例如：计算 (1 + i)¹⁰。
First, write 1 + i in modulus-argument form:
首先，将 1 + i 写成模-辐角形式：
1 + i = √2 (cos π/4 + i sin π/4)
Then by de Moivre's theorem:
然后由德莫弗定理：
(1 + i)¹⁰ = (√2)¹⁰ (cos(10π/4) + i sin(10π/4))
Simplifying: (√2)¹⁰ = 2⁵ = 32, and 10π/4 = 5π/2, so cos(5π/2) = 0 and sin(5π/2) = 1.
化简：(√2)¹⁰ = 2⁵ = 32，并且 10π/4 = 5π/2，所以 cos(5π/2) = 0，sin(5π/2) = 1。
(1 + i)¹⁰ = 32i
4. Deriving Multiple Angle Formulas | 推导倍角公式
De Moivre's theorem provides a systematic way to derive formulas for cos(nθ) and sin(nθ) in terms of powers of cos θ and sin θ. This is one of the most important applications for exam problems.
德莫弗定理为推导 cos(nθ) 和 sin(nθ) 关于 cos θ 和 sin θ 幂次的公式提供了系统方法。这是考试题目中最重要的应用之一。
For n = 2:
对于 n = 2：
(cos θ + i sin θ)² = cos 2θ + i sin 2θ
Expanding the left side:
展开左边：
cos²θ + 2i sin θ cos θ − sin²θ = cos 2θ + i sin 2θ
Equating real and imaginary parts gives the familiar double angle formulas.
比较实部和虚部，即得到常见的二倍角公式。
5. Worked Example: cos 3θ and sin 3θ | 实例：cos 3θ 与 sin 3θ
Let us derive formulas for cos 3θ and sin 3θ. Using de Moivre's theorem with n = 3:
我们来推导 cos 3θ 和 sin 3θ 的公式。在德莫弗定理中取 n = 3：
(cos θ + i sin θ)³ = cos 3θ + i sin 3θ
Expanding the left side using the binomial theorem:
利用二项式定理展开左边：
cos³θ + 3i cos²θ sin θ − 3 cos θ sin²θ − i sin³θ
Grouping real and imaginary parts:
将实部和虚部分组：
cos 3θ = cos³θ − 3 cos θ sin²θ
sin 3θ = 3 cos²θ sin θ − sin³θ
e expressions are exact and can be further transformed using sin²θ = 1 − cos²θ if desired.
这些表达式是精确的，如果需要，还可以利用 sin²θ = 1 − cos²θ 进一步变形。
6. Expressing cos(nθ) as a Polynomial in cos θ | 将 cos(nθ) 表示为 cos θ 的多项式
olely as a polynomial in cos θ. For example, using cos²θ = 1 − sin²θ in the formula above:
当 n 为奇数时，cos(nθ) 可以仅表示为 cos θ 的多项式。例如，在上述公式中利用 cos²θ = 1 − sin²θ：
cos 3θ = 4 cos³θ − 3 cos θ
Similarly, sin(nθ) can be written as sin θ times a polynomial in cos θ:
类似地，sin(nθ) 可以写成 sin θ 乘以 cos θ 的多项式：
sin 3θ = 3 sin θ − 4 sin³θ
These forms are particularly useful when solving trigonometric equations or evaluating integrals.
这些形式在解三角方程或计算积分时特别有用。
7. Application to Binomial Expansion | 在二项式展开中的应用
De Moivre's theorem can be combined with the binomial theorem to find powers of complex numbers more systematically. For a complex number in polar form:
德莫弗定理可以与二项式定理结合，更系统地计算复数的幂。对于极坐标形式的复数：
(r(cos θ + i sin θ))ⁿ = rⁿ(cos nθ + i sin nθ)
This is useful when r is a simple number, such as 1, 2, or √2. In exams, you may need to expand (a + bi)ⁿ directly and then compare with the polar form.
当 r 是简单数，如 1、2 或 √2 时，这很有用。考试中，你可能需要直接展开 (a + bi)ⁿ，然后与极坐标形式进行比较。
8. Finding Roots of Complex Numbers (Positive Integer Connection) | 求复数的根（正整数联系）
Although the focus is on positive integer powers, de Moivre' process of finding nth roots. If z = r(cos θ + i sin θ), then the nth roots are given by:
虽然我们关注的是正整数幂，但德莫弗定理也是求 n 次方根的基础。如果 z = r(cos θ + i sin θ)，那么其 n 次方根为：
r1/n (cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)), k = 0, 1, ..., n−1
This formula uses the theorem with fractional exponents, but for integer n it connects powers to roots in a natural way.
该公式使用了分数指数情形的定理，但对于整数 n，它自然地联系了幂与根。
9. Worked Example: (√3 + i)⁶ | 实例：(√3 + i)⁶
Compute (√3 + i)⁶.
计算 (√3 + i)⁶。
The modulus is:
模为：
|√3 + i| = √3 + 1 = 2
The argument is:
辐角为：
tan⁻¹(1/√3) = π/6
Thus:
因此：
√3 + i = 2(cos π/6 + i sin π/6)
Applying de Moivre:
应用德莫弗定理：
(√3 + i)⁶ = 2⁶(cos(6π/6) + i sin(6π/6)) = 64(cos π + i sin π) = −64
10. Trigonometric Identities via de Moivre | 利用德莫弗定理推导三角恒等式
De Moivre's theorem can prove many identities without lengthy geometric arguments. For example, to express cos 5θ:
德莫弗定理可以在没有冗长几何论证的情况下证明许多恒等式。例如，要表示 cos 5θ：
cos 5θ = Re[(cos θ + i sin θ)⁵]
Expanding with the binomial theorem and taking the real part yields:
用二项式定理展开并取实部，得到：
cos 5θ = cos⁵θ − 10 cos³θ sin²θ + 5 cos θ sin⁴θ
This method is clean and avoids memorising dozens of trigonometric identities.
这种方法简洁明了，避免了死记硬背大量的三角恒等式。
11. Common Pitfalls and Exam Tips | 常见错误与考试技巧
r n. Always write (r(cos θ + i sin θ))ⁿ = rⁿ(cos nθ + i sin nθ), not r(cos nθ + i sin nθ).
学生经常忘记将模 r 变为 rⁿ。务必写成 (r(cos θ + i sin θ))ⁿ = rⁿ(cos nθ + i sin nθ)，而不是 r(cos nθ + i sin nθ)。
Another common mistake is not simplifying the angle nθ modulo 2π. For example, cos(10π/3) should be simplified to cos(4π/3) before evaluating.
另一个常见错误是未将角度 nθ 化到 0 到 2π 之间。例如，cos(10π/3) 应先化简为 cos(4π/3) 再求值。
carried out correctly, especially alternating signs. Practice with n = 3 and n = 4 first.
在比较实部和虚部时，确保二项式展开正确，特别注意符号交替。先练习 n = 3 和 n = 4 的情形。
12. Summary and Further Practice | 总结与进一步练习
De Moivre's theorem for positive integers is a foundational tool in complex numbers. It allows us to compute powers, derive multiple-angle formulas, and simplify trigonometric expressions. Mastery of this theorem requires familiarity with binomial expansion and the ability to switch between Cartesian and polar forms.
德莫弗定理对正整数是复数部分的基础工具。它使我们能够计算幂、推导倍角公式并简化三角表达式。掌握该定理需要熟悉二项式展开，并能够在直角坐标与极坐标形式之间转换。
ession for cos 4θ in terms of cos θ. (c) Show that sin 5θ = 16 sin⁵θ − 20 sin³θ + 5 sin θ.
建议练习：(a) 计算 (2(cos 20° + i sin 20°))⁵。(b) 推导 cos 4θ 关于 cos θ 的表达式。(c) 证明 sin 5θ = 16 sin⁵θ − 20 sin³θ + 5 sin θ。
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