📚 De Moivre's Theorem and Its Applications | 棣莫弗定理及其应用
De Moivre's theorem is a powerful tool in A-Level mathematics that links complex numbers, trigonometry, and algebra. It states a simple formula for raising complex numbers in polar form to any integer power, yet it leads to elegant results for finding roots of equations, deriving trigonometric identities, and solving polynomial problems.
棣莫弗定理是 A-Level 数学中连接复数、三角与代数的强大工具。它以简洁公式给出极坐标形式复数的整数次幂，进而衍生出求方程根、推导三角恒等式及求解多项式问题等优雅结论。
1. Complex Numbers in Polar Form | 复数的极坐标形式
omplex number z = x + yi can be written in polar form as z = r(cos θ + i sin θ), where r = |z| = √x² + y² is the modulus and θ = arg(z) is the argument, usually chosen so that −π < θ ≤ π.
任意复数 z = x + yi 可写成极坐标形式 z = r(cos θ + i sin θ)，其中 r = |z| = √x² + y² 为模长，θ = arg(z) 为辐角，通常取 −π < θ ≤ π。
For example, the complex number 1 + i has modulus √2 and argument π/4, so its polar form is √2(cos(π/4) + i sin(π/4)).
例如，复数 1 + i 的模为 √2，辐角为 π/4，故其极坐标形式为 √2(cos(π/4) + i sin(π/4))。
2. Statement of De Moivre's Theorem | 棣莫弗定理的表述
For any real number θ and any integer n,
对于任意实数 θ 和任意整数 n，
(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)
is not 1, because multiplying by rⁿ gives (r(cos θ + i sin θ))ⁿ = rⁿ(cos(nθ) + i sin(nθ)).
当模长不为 1 时定理同样成立，乘以 rⁿ 得 (r(cos θ + i sin θ))ⁿ = rⁿ(cos(nθ) + i sin(nθ))。
3. Proof for Positive Integers | 正整数情形的证明
rivial. Assume the theorem holds for n = k, i.e. (cos θ + i sin θ)ᵏ = cos(kθ) + i sin(kθ).
我们用数学归纳法证明。基底情形 n = 1 显然成立。假设 n = k 时定理成立，即 (cos θ + i sin θ)ᵏ = cos(kθ) + i sin(kθ)。
Multiply both sides by (cos θ + i sin θ):
两边同乘以 (cos θ + i sin θ)：
(cos θ + i sin θ)ᵏ⁺¹ = (cos(kθ) + i sin(kθ))(cos θ + i sin θ)
nce the statement is true for n = k+1, so by induction it holds for all positive integers.
展开并利用和角公式可得 cos((k+1)θ) + i sin((k+1)θ)。因此当 n = k+1 时命题成立，由归纳法知对所有正整数成立。
4. Extension to Negative and Rational Powers | 负整数与有理数幂的推广
θ − i sin θ, we obtain (cos θ + i sin θ)⁻ᵐ = cos(mθ) − i sin(mθ) = cos(−mθ) + i sin(−mθ).
对负整数 −m，利用正整数情形以及 (cos θ + i sin θ)⁻¹ = cos θ − i sin θ，可得 (cos θ + i sin θ)⁻ᵐ = cos(mθ) − i sin(mθ) = cos(−mθ) + i sin(−mθ)。
For rational powers p/q, the expression (cos θ + i sin θ)p/q of the principal result. This subtlety is essential when solving equations such as zⁿ = a.
对于有理数幂 p/q，表达式 (cos θ + i sin θ)p/q 可取 q 个不同值，对应主结果的所有 q 次根。这个细微之处在解 zⁿ = a 这类方程时至关重要。
5. Powers of Complex Numbers | 复数的幂运算
De Moivre' a direct way to compute powers of complex numbers. Given z = r(cos θ + i sin θ), we have:
棣莫弗定理提供了计算复数幂的直接方法。若 z = r(cos θ + i sin θ)，则：
zⁿ = rⁿ(cos(nθ) + i sin(nθ))
nce, to compute (1 + i)⁶, first write 1 + i = √2(cos(π/4) + i sin(π/4)). Then (1 + i)⁶ = (√2)⁶(cos(6·π/4) + i sin(6·π/4)) = 8(cos(3π/2) + i sin(3π/2)) = 8(0 − i) = −8i.
例如，计算 (1 + i)⁶：先写 1 + i = √2(cos(π/4) + i sin(π/4))，则 (1 + i)⁶ = (√2)⁶(cos(6·π/4) + i sin(6·π/4)) = 8(cos(3π/2) + i sin(3π/2)) = 8(0 − i) = −8i。
6. Finding nth Roots of a Complex Number | 求复数的 n 次方根
To solve wⁿ = z, where z = r(cos θ + i sin θ), the n solutions are:
求解 wⁿ = z，其中 z = r(cos θ + i sin θ)，其 n 个解为：
wₖ = r1/n [cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)], k = 0, 1, 2, …, n−1
The parameter k produces n distinct roots equally spaced around a circle of radius r1/n. For example, the cube roots of unity solve w³ = 1 = cos 0 + i sin 0, giving:
参数 k 给出 n 个不同根，均匀分布在半径 r1/n 的圆上。例如，解 w³ = 1 = cos 0 + i sin 0 得到单位根的三个值：
w₀ = 1, w₁ = cos(2π/3) + i sin(2π/3) = −1/2 + i√3/2, w₂ = cos(4π/3) + i sin(4π/3) = −1/2 − i√3/2.
7. Application to Trigonometric Identities | 应用于三角恒等式
De Moivre' ields identities for cos(nθ) and sin(nθ) in terms of powers of cos θ and sin θ. For n = 3:
棣莫弗定理结合二项式展开可导出 cos(nθ) 和 sin(nθ) 关于 cos θ 与 sin θ 幂次的恒等式。对 n = 3：
os(3θ) + i sin(3θ) = (cos θ + i sin θ)³ = cos³θ + 3i cos²θ sin θ − 3 cos θ sin²θ − i sin³θ
cos θ sin²θ = 4 cos³θ − 3 cos θ, and sin(3θ) = 3 cos²θ sin θ − sin³θ = 3 sin θ − 4 sin³θ.
比较实部和虚部，得 cos(3θ) = cos³θ − 3 cos θ sin²θ = 4 cos³θ − 3 cos θ，以及 sin(3θ) = 3 cos²θ sin θ − sin³θ = 3 sin θ − 4 sin³θ。
8. Expressing cos nθ and sin nθ in Powers of cosθ and sinθ | 用 cosθ 和 sinθ 的幂表示 cos nθ 与 sin nθ
For larger n, use the binomial expansion:
对更大的 n，使用二项式展开：
(cos θ + i sin θ)ⁿ = Σk=0n C(n,k) (cos θ)ⁿ⁻ᵏ (i sin θ)ᵏ
is the real part of this sum, and sin(nθ) is the imaginary part. For example, when n = 4:
则 cos(nθ) 为该和的实部，sin(nθ) 为虚部。例如当 n = 4 时：
cos(4θ) = cos⁴θ − 6 cos²θ sin²θ + sin⁴θ = 8 cos⁴θ − 8 cos²θ + 1, and sin(4θ) = 4 cos³θ sin θ − 4 cos θ sin³θ.
9. Solving Polynomial Equations Using Roots | 利用根求解多项式方程
rm solutions to the polynomial equation wⁿ − z = 0. For example, the roots of w⁴ = 16 are:
任意复数 z 的 n 次方根构成多项式方程 wⁿ − z = 0 的解。例如，w⁴ = 16 的根为：
- w₀ = 2(cos 0 + i sin 0) = 2
- w₁ = 2(cos(π/2) + i sin(π/2)) = 2i
- w₂ = 2(cos π + i sin π) = −2
- w₃ = 2(cos(3π/2) + i sin(3π/2)) = −2i
These four roots form a square in the complex plane, symmetrically placed about the origin.
这四个根在复平面上构成一个关于原点对称的正方形。
10. Exam Tips and Common Pitfalls | 应试技巧与常见陷阱
In AQA A-Level examinations, always convert to polar form before applying De Moivre's theorem. Remember to adjust the argument to the principal range when finding roots.
在 AQA A-Level 考试中，应用棣莫弗定理前务必转换为极坐标形式。求根时记得将辐角调整到主值范围。
- ncluded: (r(cos θ + i sin θ))ⁿ = rⁿ(cos(nθ) + i sin(nθ)), not just the trigonometric part.
- When taking nth roots, the argument must be divided by n, and add 2πk/n for k = 0, 1, …, n−1. Do not forget the k range.
- refully separate real and imaginary terms. The imaginary unit follows i² = −1 and i³ = −i.
- 注意包含模长： (r(cos θ + i sin θ))ⁿ = rⁿ(cos(nθ) + i sin(nθ))，而不只是三角部分。
- 求 n 次方根时，辐角要除以 n，并加上 2πk/n，其中 k = 0, 1, …, n−1，切勿遗漏 k 的范围。
- 二项式展开时仔细分离实部和虚部，注意虚数单位满足 i² = −1、i³ = −i。
Practice with past paper questions involving √3 cos 20° + i sin 20°)⁶, and solving equations like z³ = −8. These are typical AQA contexts.
建议多练习涉及 √3 + i、类似 (cos 20° + i sin 20°)⁶ 的幂运算，以及求解 z³ = −8 等 AQA 典型题型。
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