De Moivre's Theorem and Complex Roots | 棣莫弗定理与复数根

📚 De Moivre’s Theorem and Complex Roots | 棣莫弗定理与复数根

The third support pack for Edexcel International AS/A Level Further Mathematics (9665) takes your understanding of complex numbers to the next level. At its heart lies De Moivre’s theorem, a powerful tool that links trigonometry, algebra, and geometry, enabling you to compute powers, roots, and intricate identities with elegance.

Edexcel 国际 AS/A Level 进阶数学（9665）的第三套学习支持包将你对复数的理解推向更高层次。其核心是棣莫弗定理——一个将三角学、代数和几何巧妙联系起来的强大工具，让你能够优雅地计算幂、方根以及复杂的三角恒等式。

1. Revisiting the Polar Form | 重温极坐标形式

A complex number z = x + yi can be located on an Argand diagram using its modulus r = √(x² + y²) and argument θ = arctan(y/x). The polar form z = r (cos θ + i sin θ) is essential because it turns multiplication into a combination of moduli and arguments: multiply the moduli, add the arguments.

复数 z = x + yi 可在阿根图上用模 r = √(x² + y²) 和辐角 θ = arctan(y/x) 定位。极坐标形式 z = r (cos θ + i sin θ) 必不可少，因为它将乘法转化为模的乘积与辐角的相加。

2. Statement of De Moivre’s Theorem | 棣莫弗定理的陈述

For any real number θ and any integer n, De Moivre’s theorem asserts that (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ). When the complex number is also multiplied by a modulus r, the full expression becomes [r (cos θ + i sin θ)]ⁿ = rⁿ [cos(nθ) + i sin(nθ)].

对于任意实数 θ 和任意整数 n，棣莫弗定理断言 (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)。当复数还带有模 r 时，完整的表达式为 [r (cos θ + i sin θ)]ⁿ = rⁿ [cos(nθ) + i sin(nθ)]。

3. Proof by Induction for Positive n | 正整数幂的归纳法证明

For n = 1 the statement is trivial. Assume the theorem holds for n = k, i.e., (cos θ + i sin θ)ᵏ = cos(kθ) + i sin(kθ). Multiply both sides by (cos θ + i sin θ) and use the compound angle formulas: cos(kθ)cos θ – sin(kθ)sin θ = cos[(k+1)θ] and sin(kθ)cos θ + cos(kθ)sin θ = sin[(k+1)θ]. This gives the result for n = k+1, completing the induction for positive integers.

当 n = 1 时结论显然成立。假设定理对 n = k 成立，即 (cos θ + i sin θ)ᵏ = cos(kθ) + i sin(kθ)。两边同乘 (cos θ + i sin θ) 并利用复合角公式：cos(kθ)cos θ – sin(kθ)sin θ = cos[(k+1)θ] 和 sin(kθ)cos θ + cos(kθ)sin θ = sin[(k+1)θ]，即得到 n = k+1 的结果，完成对正整数的归纳证明。

4. Extending to Negative Integers | 扩展到负整数指数

For a negative integer n = –m (m > 0), write (cos θ + i sin θ)⁻ᵐ = 1 / (cos θ + i sin θ)ᵐ. By the positive-integer version, the denominator becomes cos(mθ) + i sin(mθ). Multiply numerator and denominator by the complex conjugate cos(mθ) – i sin(mθ) to obtain cos(–mθ) + i sin(–mθ). Thus the theorem holds for all integer n.

对于负整数 n = –m (m > 0)，将 (cos θ + i sin θ)⁻ᵐ 写为 1 / (cos θ + i sin θ)ᵐ。根据正整数情况，分母变为 cos(mθ) + i sin(mθ)。分子分母同乘共轭复数 cos(mθ) – i sin(mθ) 即可得到 cos(–mθ) + i sin(–mθ)。因此定理对所有整数 n 均成立。

5. Applying De Moivre’s Theorem to Powers | 应用棣莫弗定理求复数的幂

To evaluate (1 + i√3)⁵, first express the base in polar form: modulus r = √(1² + (√3)²) = 2, argument θ = π/3. Then by the theorem, (2[cos(π/3) + i sin(π/3)])⁵ = 2⁵ [cos(5π/3) + i sin(5π/3)] = 32(1/2 – i√3/2) = 16 – 16i√3. The modulus is raised to the power, the argument is multiplied.

要计算 (1 + i√3)⁵，先将底数化为极坐标形式：模 r = √(1² + (√3)²) = 2，辐角 θ = π/3。然后根据定理，(2[cos(π/3) + i sin(π/3)])⁵ = 2⁵ [cos(5π/3) + i sin(5π/3)] = 32(1/2 – i√3/2) = 16 – 16i√3。模进行乘方，辐角被倍数化。

6. Deriving Trigonometric Identities | 推导三角恒等式

By expanding (cos θ + i sin θ)³ using the binomial theorem and equating real and imaginary parts with cos(3θ) + i sin(3θ), we obtain cos(3θ) = cos³θ – 3 cos θ sin²θ and sin(3θ) = 3 cos²θ sin θ – sin³θ. These can be rewritten using sin²θ = 1 – cos²θ to give cos(3θ) = 4 cos³θ – 3 cos θ and sin(3θ) = 3 sin θ – 4 sin³θ, which are standard triple-angle formulas.

通过二项式定理展开 (cos θ + i sin θ)³，并将其实部与虚部与 cos(3θ) + i sin(3θ) 分别对应，我们得到 cos(3θ) = cos³θ – 3 cos θ sin²θ 和 sin(3θ) = 3 cos²θ sin θ – sin³θ。再利用 sin²θ = 1 – cos²θ 可改写为 cos(3θ) = 4 cos³θ – 3 cos θ 和 sin(3θ) = 3 sin θ – 4 sin³θ，这正是标准的三倍角公式。

7. The Formula for nth Roots | n 次方根公式

If zⁿ = w, where w = R(cos φ + i sin φ), then z = w^(1/n) = R^(1/n)[cos((φ + 2kπ)/n) + i sin((φ + 2kπ)/n)] for k = 0, 1, 2, …, n–1. The n roots lie equally spaced on a circle of radius R^(1/n) centred at the origin.

若 zⁿ = w，且 w = R(cos φ + i sin φ)，则 z = w^(1/n) = R^(1/n)[cos((φ + 2kπ)/n) + i sin((φ + 2kπ)/n)]，其中 k = 0, 1, 2, …, n–1。这 n 个根等距分布在以原点为圆心、半径为 R^(1/n) 的圆周上。

8. Solving Equations in the Form z^n = a + bi | 解形如 z^n = a + bi 的方程

To solve z⁴ = –8 – 8i√3, first put –8 – 8i√3 into polar form: the modulus is 16, the argument is 4π/3 (or –2π/3, but we use the principal +2π adjustment). Then z = 16^(1/4) × [cos((4π/3 + 2kπ)/4) + i sin((4π/3 + 2kπ)/4)] = 2[cos(π/3 + kπ/2) + i sin(π/3 + kπ/2)], giving four distinct roots for k = 0,1,2,3.

求解 z⁴ = –8 – 8i√3 时，先将 –8 – 8i√3 化为极坐标形式：模为 16，辐角为 4π/3（或 –2π/3，但通过加 2π 调整即可）。于是 z = 16^(1/4) × [cos((4π/3 + 2kπ)/4) + i sin((4π/3 + 2kπ)/4)] = 2[cos(π/3 + kπ/2) + i sin(π/3 + kπ/2)]，k = 0,1,2,3 给出四个不同的根。

9. Properties of the nth Roots of Unity | n 次单位根的性质

The equation zⁿ = 1 has n solutions given by e^(2kπi/n) for k = 0,1,…,n–1. Their sum is zero, and they form the vertices of a regular n-gon on the unit circle. If ω is a primitive nth root, then 1 + ω + ω² + … + ωⁿ⁻¹ = 0 is a crucial identity for factorising polynomials and summing series.

方程 zⁿ = 1 有 n 个解，由 e^(2kπi/n)，k = 0,1,…,n–1 给出。它们的和为零，且构成单位圆上正 n 边形的顶点。若 ω 为本原 n 次根，则 1 + ω + ω² + … + ωⁿ⁻¹ = 0 是一个关键恒等式，用于因式分解多项式和级数求和。

10. Geometric Interpretation on the Argand Diagram | 阿根图上的几何解释

Multiplying by (cos θ + i sin θ) rotates a complex number anticlockwise by θ, while raising to a power n scales the modulus by rⁿ and multiplies the argument by n. The nth roots of a complex number are the vertices of a regular n-gon, illustrating how algebra and geometry merge seamlessly.

乘以 (cos θ + i sin θ) 会将复数逆时针旋转 θ，而求 n 次幂则将模缩放至 rⁿ 并将辐角乘以 n。复数的 n 次方根对应正 n 边形的顶点，完美展现了代数与几何的融合。

11. Common Pitfalls and How to Avoid Them | 常见误区与规避方法

Forgetting the ±2kπ term: Always write the argument as φ + 2kπ before dividing by n. Confusing radians and degrees: Stick to radians in further maths unless the question states otherwise. Ignoring the principal argument range: When giving final answers, express the argument within (–π, π] or [0, 2π) as required. Misapplying De Moivre for non-integer n: The theorem in its basic form requires integer n; for rational powers, use the root formula with multiple values.

遗漏 ±2kπ 项：务必在除以 n 之前将辐角写为 φ + 2kπ。弧度与角度混淆：在进阶数学中坚持使用弧度，除非题目特别说明。忽视主辐角范围：给出最终答案时，按题目要求将辐角控制在 (–π, π] 或 [0, 2π) 内。将棣莫弗定理误用于非整数指数：定理的基本形式要求 n 为整数；对于有理数幂，应使用带多值的方根公式。

12. Summary and Further Practice | 总结与拓展练习

De Moivre’s theorem transforms intricate trigonometric manipulations into systematic algebraic steps. Mastering it means you can effortlessly find powers, extract roots, prove identities, and solve polynomial equations in the complex domain. Regular practice with past-paper questions from the 9665 specification will solidify your fluency.

棣莫弗定理将复杂的三角运算转化为系统的代数步骤。掌握它意味着你能轻松地求幂、开方、证明恒等式并在复数域求解多项式方程。定期用 9665 大纲的历年真题进行练习，将强化你的熟练度。

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De Moivre’s Theorem and Complex Roots | 棣莫弗定理与复数根