📚 Using de Moivre's Theorem to Evaluate Powers of Complex Numbers | 使用棣莫弗定理计算复数的幂
In AQA A-Level Mathematics, de Moivre's theorem is a standard tool for raising complex numbers in polar form to integer powers. This article explains how to convert a complex number to modulus-argument form, apply the theorem, and turn the result back into Cartesian form using exact trigonometric values.
在 AQA A-Level 数学中，棣莫弗定理是将极形式复数进行整数次幂运算的标准工具。本文讲解如何将复数转换为模-辐角形式、应用定理，以及利用精确三角值将结果转换回笛卡尔形式。
1. Complex Numbers in Modulus-Argument Form | 复数的模-辐角形式
ex number z = x + iy can be represented in polar form as z = r(cos θ + i sin θ), where r = |z| = √x² + y² nce from the origin to the point (x, y), and the argument θ is the angle measured anticlockwise from the positive real axis.
复数 z = x + iy 可以表示为极形式 z = r(cos θ + i sin θ)，其中 r = |z| = √x² + y²，θ = arg z。模 r 是原点到点 (x, y) 的距离，辐角 θ 是从正实轴逆时针测量的角度。
argument is usually chosen in the interval (-π, π]. For example, z = 1 + i has modulus r = √1² + 1² = √2 and argument θ = π/4, so z = √2(cos π/4 + i sin π/4).
在 AQA 考试题中，主辐角通常选取在区间 (-π, π] 内。例如，z = 1 + i 的模为 r = √1² + 1² = √2，辐角为 θ = π/4，因此 z = √2(cos π/4 + i sin π/4)。
z = r(cos θ + i sin θ)
2. Statement of de Moivre's Theorem | 棣莫弗定理的表述
For any real angle θ and any integer n, de Moivre' nθ + i sin nθ. More generally, if z = r(cos θ + i sin θ), then zⁿ = rⁿ(cos nθ + i sin nθ).
对于任意实数角 θ 和任意整数 n，棣莫弗定理指出 (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ。更一般地，如果 z = r(cos θ + i sin θ)，则 zⁿ = rⁿ(cos nθ + i sin nθ)。
The theorem is powerful because the exponent moves inside the trigonometric functions as a multiplier. This avoids expanding brackets repeatedly and is valid for all integer powers, including negative integers.
该定理非常有用，因为指数会以乘法因子的形式进入三角函数内部。这样就不需要反复展开括号，并且该定理对所有整数次幂成立，包括负整数次幂。
(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ
3. Deriving the Theorem from Euler's Formula | 从欧拉公式推导定理
Euler's formula gives eiθ = cos θ + i sin θ. Therefore, (cos θ + i sin θ)ⁿ = (eiθ)ⁿ = einθ = cos nθ + i sin nθ. This is a quick way to understand why de Moivre's theorem works, although AQA also expects the proof by induction for positive integer n.
欧拉公式给出 eiθ = cos θ + i sin θ。因此，(cos θ + i sin θ)ⁿ = (eiθ)ⁿ = einθ = cos nθ + i sin nθ。这是快速理解棣莫弗定理为何成立的方法，不过 AQA 也要求掌握正整数 n 情况下的归纳法证明。
of, the base case n = 1 is true. Assuming the result for n = k, multiply (cos θ + i sin θ)k by (cos θ + i sin θ) and use the compound angle formulae to obtain cos(k + 1)θ + i sin(k + 1)θ.
对于形式化证明，基础情形 n = 1 显然成立。假设 n = k 时结果成立，将 (cos θ + i sin θ)k 乘以 (cos θ + i sin θ)，并利用和角公式即可得到 cos(k + 1)θ + i sin(k + 1)θ。
eiθ = cos θ + i sin θ
4. Evaluating Positive Integer Powers | 求正整数次幂
To evaluate a positive integer power, first write the complex number in modulus-argument form, then raise the modulus to the required power and multiply the argument by that power.
要计算正整数次幂，首先将复数写成模-辐角形式，然后将模提升到所需次幂，并将辐角乘以该次幂。
2⁵ = 32 and the argument becomes 5 × π/6 = 5π/6. Therefore z⁵ = 32(cos 5π/6 + i sin 5π/6).
例 1：如果 z = 2(cos π/6 + i sin π/6)，求 z⁵。模变为 2⁵ = 32，辐角变为 5 × π/6 = 5π/6。因此 z⁵ = 32(cos 5π/6 + i sin 5π/6)。
Using exact values, cos 5π/6 = -√3/2 and sin 5π/6 = 1/2, so z⁵ = 32(-√3/2 + i/2) = -16√3 + 16i.
利用精确值，cos 5π/6 = -√3/2，sin 5π/6 = 1/2，所以 z⁵ = 32(-√3/2 + i/2) = -16√3 + 16i。
Example 2: Evaluate (1 + i)⁴. First, 1 + i = √2(cos π/4 + i sin π/4). Then (1 + i)⁴ = (√2)⁴(cos 4π/4 + i sin 4π/4) = 4(cos π + i sin π) = 4(-1 + 0i) = -4.
例 2：计算 (1 + i)⁴。首先，1 + i = √2(cos π/4 + i sin π/4)。然后 (1 + i)⁴ = (√2)⁴(cos 4π/4 + i sin 4π/4) = 4(cos π + i sin π) = 4(-1 + 0i) = -4。
zⁿ = rⁿ(cos nθ + i sin nθ)
5. Using Exact Trigonometric Values | 使用精确三角函数值
AQA frequently requires final answers in exact Cartesian form, so you must know the exact sine and cosine values for common angles such as π/6, π/4, π/3, π/2 and their multiples.
AQA 经常要求最终答案写成精确笛卡尔形式，因此你必须熟悉常见角（如 π/6、π/4、π/3、π/2 及其倍数）的精确正弦和余弦值。
| θ | cos θ | sin θ |
|---|---|---|
| 0 | 1 | 0 |
| π/6 | √3/2 | 1/2 |
| π/4 | √2/2 | √2/2 |
| π/3 | 1/2 | √3/2 |
| π/2 | 0 | 1 |
| π | -1 | 0 |
⁶. Here r = 1 and θ = π/3, so z⁶ = cos 6π/3 + i sin 6π/3 = cos 2π + i sin 2π = 1 + 0i = 1.
例：如果 z = cos π/3 + i sin π/3，求 z⁶。这里 r = 1，θ = π/3，所以 z⁶ = cos 6π/3 + i sin 6π/3 = cos 2π + i sin 2π = 1 + 0i = 1。
8 and the argument is 3π/6 = π/2, so the result is 8(cos π/2 + i sin π/2) = 8(0 + i) = 8i.
另一个例子：计算 [2(cos π/6 + i sin π/6)]³。模为 2³ = 8，辐角为 3π/6 = π/2，所以结果为 8(cos π/2 + i sin π/2) = 8(0 + i) = 8i。
6. Negative Integer Powers | 负整数次幂
De Moivre' ⁿ = r⁻ⁿ(cos(-nθ) + i sin(-nθ)). This is sometimes written as z⁻ⁿ = r⁻ⁿ(cos nθ - i sin nθ).
棣莫弗定理同样适用于负整数次幂。如果 n 为正整数且 z = r(cos θ + i sin θ)，那么 z⁻ⁿ = r⁻ⁿ(cos(-nθ) + i sin(-nθ))。有时也写作 z⁻ⁿ = r⁻ⁿ(cos nθ - i sin nθ)。
+ i sin π/4), find z⁻³. First, r = 2 and θ = π/4, so z⁻³ = 2⁻³(cos(-3π/4) + i sin(-3π/4)).
例：如果 z = 2(cos π/4 + i sin π/4)，求 z⁻³。首先，r = 2，θ = π/4，所以 z⁻³ = 2⁻³(cos(-3π/4) + i sin(-3π/4))。
Using exact values, cos(-3π/4) = -√2/2 and sin(-3π/4) = -√2/2. Therefore z⁻³ = 1/8(-√2/2 - i√2/2) = -√2/16 - i√2/16.
利用精确值，cos(-3π/4) = -√2/2，sin(-3π/4) = -√2/2。因此 z⁻³ = 1/8(-√2/2 - i√2/2) = -√2/16 - i√2/16。
z⁻ⁿ = r⁻ⁿ(cos(-nθ) + i sin(-nθ))
7. Fractional Powers and Roots | 分数次幂与方根
De Moivre' ), the n-th roots are given by adding multiples of 2π to the argument before dividing by n.
棣莫弗定理是针对整数次幂提出的，但可以推广用于求方根。对于复数 z = r(cos θ + i sin θ)，其 n 次方根可以通过在辐角上加上 2π 的整数倍，再除以 n 来得到。
The n distinct roots are zk = r1/n portant because a fractional power does not give a unique value unless a specific root is chosen.
n 个不同的方根为 zk = r1/n[cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)]，其中 k = 0, 1, 2, ..., n - 1。这一点很重要，因为分数次幂通常不会给出唯一值，除非指定了某一个方根。
ample: Find the cube roots of 8i. First, 8i = 8(cos π/2 + i sin π/2). The cube roots are 81/3[cos((π/2 + 2πk)/3) + i sin((π/2 + 2πk)/3)] for k = 0, 1, 2.
例：求 8i 的立方根。首先，8i = 8(cos π/2 + i sin π/2)。其立方根为 81/3[cos((π/2 + 2πk)/3) + i sin((π/2 + 2πk)/3)]，其中 k = 0, 1, 2。
For k = 0, the root is 2(cos π/6 + i sin π/6) = √3 + i. For k = 1, the root is 2(cos 5π/6 + i sin 5π/6) = -√3 + i. For k = 2, the root is 2(cos 3π/2 + i sin 3π/2) = -2i.
当 k = 0 时，方根为 2(cos π/6 + i sin π/6) = √3 + i。当 k = 1 时，方根为 2(cos 5π/6 + i sin 5π/6) = -√3 + i。当 k = 2 时，方根为 2(cos 3π/2 + i sin 3π/2) = -2i。
zk = r1/n[cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)]
8. Simplifying Expressions of the Form (cos θ + i sin θ)ⁿ
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