📚 De Moivre's Theorem and its Applications | 棣莫弗定理及其应用
De Moivre's theorem is a central tool in AQA Further Mathematics for working with complex numbers in polar form. It provides a fast, reliable way to compute powers and roots, derive trigonometric identities, and sum trigonometric series. This article explains the theorem, its proof, and the main applications you should master for the exam.
棣莫弗定理是 AQA 进阶数学中处理极坐标形式复数的核心工具。它为计算复数的幂与根、推导三角恒等式以及求解三角级数提供了一种快速而可靠的方法。本文将解释该定理、其证明，以及你在考试中需要掌握的主要应用。
1. Polar Form and the Key Idea | 极坐标形式与核心思想
Any non-zero complex number z = x + iy can be written in modulus-argument form:
z = r(cos θ + i sin θ)
where r = |z| = √x² + y² rval (−π, π]. Multiplication of two complex numbers multiplies their moduli and adds their arguments. De Moivre's theorem turns this geometric rule into a practical formula for powers.
任意非零复数 z = x + iy 都可以写成模-辐角形式 z = r(cos θ + i sin θ)，其中 r = |z| = √x² + y²，θ = arg z，通常取在区间 (−π, π] 内。两个复数相乘时，模相乘、辐角相加。棣莫弗定理正是把这一几何规律转化为实用的求幂公式。
2. Statement of De Moivre's Theorem | 棣莫弗定理的表述
For any integer n, De Moivre's theorem states:
(r(cos θ + i sin θ))ⁿ = rⁿ(cos nθ + i sin nθ)
When r = 1, this reduces to the simpler and very common form:
(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ
It is essential to remember that this standard single-valued form holds for integer n. Rational powers can produce more than one value, as discussed in Section 5.
对于任意整数 n，棣莫弗定理指出：(r(cos θ + i sin θ))ⁿ = rⁿ(cos nθ + i sin nθ)。当 r = 1 时，它简化为更常见的形式：(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ。必须记住，这一标准单值形式要求 n 为整数；有理数幂可能会产生多个值，详见第 5 节。
3. Proof by Induction for Positive Integers | 正整数情形的归纳法证明
For n = 1, the statement is obvious because both sides equal cos θ + i sin θ. Assume that:
(cos θ + i sin θ)ᵏ = cos kθ + i sin kθ
Then multiply both sides by (cos θ + i sin θ):
(cos θ + i sin θ)ᵏ⁺¹ = (cos kθ + i sin kθ)(cos θ + i sin θ)
Therefore the statement holds for k + 1, so by induction it holds for all positive integers n.
该定理可通过数学归纳法对正整数证明。n = 1 时命题显然成立，因为两边都等于 cos θ + i sin θ。假设 (cos θ + i sin θ)ᵏ = cos kθ + i sin kθ，然后两边乘以 (cos θ + i sin θ)。利用和角公式，实部为 cos kθ cos θ − sin kθ sin θ = cos(k+1)θ，虚部为 sin kθ cos θ + cos kθ sin θ = sin(k+1)θ。因此命题对 k+1 成立，由归纳法可知它对所有正整数 n 都成立。
4. Extending to Negative Powers | 扩展到负整数幂
For negative powers, write:
(cos θ + i sin θ)⁻ⁿ = 1 / (cos θ + i sin θ)ⁿ
Since dividing by a complex number is equivalent to multiplying by its conjugate over the modulus squared, we have:
1 / (cos θ + i sin θ) = cos θ − i sin θ = cos(−θ) + i sin(−θ)
Applying the positive-integer result then gives:
(cos θ + i sin θ)⁻ⁿ = cos(−nθ) + i sin(−nθ)
So De Moivre's theorem holds for all integers n. For a modulus r, we also use r⁻ⁿ = 1 / rⁿ.
对于负整数幂，可写成 (cos θ + i sin θ)⁻ⁿ = 1 / (cos θ + i sin θ)ⁿ。由于除以一个复数等价于乘以它的共轭复数再除以模的平方，我们有 1 / (cos θ + i sin θ) = cos θ − i sin θ = cos(−θ) + i sin(−θ)。应用正整数情形的结果可得 (cos θ + i sin θ)⁻ⁿ = cos(−nθ) + i sin(−nθ)。因此棣莫弗定理对所有整数 n 均成立。对于模 r，还需使用 r⁻ⁿ = 1 / rⁿ。
5. Rational Powers and Multiple Values | 有理数幂与多值性
If n = p/q, where p and q are coprime integers, then zp/q is not unique: there are exactly q distinct values. To find them, write z in polar form and use:
z1/q = r1/q [cos((θ + 2kπ)/q) + i sin((θ + 2kπ)/q)]
, 1, 2, ..., q − 1. Then raise each of these q roots to the power p. This is how De Moivre's theorem locates the q distinct qth roots of a complex number.
如果 n = p/q，其中 p 与 q 为互质整数，那么 zp/q 并不唯一：它恰好有 q 个不同的值。求这些值时，先将 z 写成极坐标形式，并使用公式 z1/q = r1/q [cos((θ + 2kπ)/q) + i sin((θ + 2kπ)/q)]，其中 k = 0, 1, 2, ..., q−1。然后将这 q 个根分别取 p 次幂。棣莫弗定理正是这样找到复数的 q 个不同的 q 次根。
6. Worked Example: Finding a Power | 例题：求复数的幂
Example: compute (1 + i)⁸. First write 1 + i in modulus-argument form:
1 + i = √2(cos π/4 + i sin π/4)
Then apply De Moivre's theorem:
(1 + i)⁸ = (√2)⁸(cos 8 × π/4 + i sin 8 × π/4) = 16(cos 2π + i sin 2π) = 16
Notice that 8 × π/4 = 2π
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