Exercise 21J.2: De Moivre’s Theorem & Complex Roots | 练习21J.2：棣莫弗定理与复数根

📚 Exercise 21J.2: De Moivre’s Theorem & Complex Roots | 练习21J.2：棣莫弗定理与复数根

In IB Mathematics HL, Exercise 21J.2 typically challenges students to apply De Moivre’s theorem to compute powers and roots of complex numbers, prove trigonometric identities, and interpret results geometrically. This article revisits the core concepts and walks through typical problem-solving strategies.

在IB数学高水平课程中，练习21J.2通常要求学生运用棣莫弗定理计算复数的幂与根、证明三角恒等式，并从几何角度诠释结果。本文回顾相关核心概念，并带您梳理典型的解题思路。

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1. Review: Polar Form of Complex Numbers | 复习：复数的极坐标形式

A complex number z = a + bi can be expressed in polar form z = r(cos θ + i sin θ) where r = |z| = √(a² + b²) is the modulus and θ = arg(z) is the argument. The principal argument often lies in (−π, π]. Writing it as r cis θ is also common.

复数 z = a + bi 可写为极坐标形式 z = r(cos θ + i sin θ)，其中模 r = |z| = √(a² + b²)，辐角 θ = arg(z)。通常主辐角取值在 (−π, π] 范围内，也常简记为 r cis θ。

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

De Moivre’s theorem states that for any real number θ and integer n, (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ. This elegant formula extends to rational exponents with care taken for multiple values.

棣莫弗定理指出，对任何实数 θ 和整数 n，有 (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ。该公式极其优美，推广到有理指数时需留意多值性。

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3. Computing Powers of Complex Numbers | 计算复数的幂

To find zⁿ, express z in polar form, raise the modulus to the power n, and multiply the argument by n: zⁿ = rⁿ (cos nθ + i sin nθ). Example: (1 + i√3)⁶. First, 1 + i√3 has modulus 2 and argument π/3; thus (2 cis π/3)⁶ = 64 cis 2π = 64.

计算 zⁿ 时，先将 z 化为极坐标式，模求 n 次幂，辐角乘以 n：zⁿ = rⁿ (cos nθ + i sin nθ)。例如 (1 + i√3)⁶：模为 2，辐角为 π/3，故 (2 cis π/3)⁶ = 64 cis 2π = 64。

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4. Roots of Complex Numbers: The General Principle | 复数根的一般原理

Solving zⁿ = w for a given complex w yields n distinct roots. If w = R cis Φ, then the roots are zₖ = R^(1/n) cis ( (Φ + 2πk)/n ) for k = 0, 1, 2, …, n−1.

解方程 zⁿ = w 时，对于给定的复数 w，恰有 n 个不同的根。若 w = R cis Φ，则根为 zₖ = R^(1/n) cis ( (Φ + 2πk)/n )，k = 0, 1, 2, …, n−1。

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5. Finding Cube Roots of a Specific Complex Number | 求具体复数的立方根

Take w = 8i. Modulus R = 8, argument Φ = π/2. Then z = ∛(8) cis ( (π/2 + 2πk)/3 ) = 2 cis (π/6 + 2πk/3). The three roots: z₀ = 2 cis π/6 = √3 + i, z₁ = 2 cis 5π/6 = −√3 + i, z₂ = 2 cis 3π/2 = −2i. Plotting them shows an equilateral triangle.

以 w = 8i 为例，模 R = 8，辐角 Φ = π/2。则 z = ∛(8) cis ( (π/2 + 2πk)/3 ) = 2 cis (π/6 + 2πk/3)。三个根为：z₀ = 2 cis π/6 = √3 + i，z₁ = 2 cis 5π/6 = −√3 + i，z₂ = 2 cis 3π/2 = −2i。它们在复平面上构成等边三角形。

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6. Roots of Unity | 单位根

When w = 1, the solutions of zⁿ = 1 are called the nth roots of unity. They are given by ωₖ = cis (2πk/n), k = 0,1,…,n−1. Their sum is 0, and they are symmetrically spaced around the unit circle.

当 w = 1 时，方程 zⁿ = 1 的解称为 n 次单位根，记作 ωₖ = cis (2πk/n)，k = 0,1,…,n−1。它们之和为零，且在单位圆上对称分布。

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7. Using De Moivre to Prove Trigonometric Identities | 利用棣莫弗定理证明三角恒等式

By expanding (cos θ + i sin θ)ⁿ via binomial theorem and equating real/imaginary parts, we obtain cos nθ and sin nθ in terms of powers of sin θ and cos θ. For n = 3: cos 3θ = 4cos³θ − 3cos θ, sin 3θ = 3sin θ − 4sin³θ.

将 (cos θ + i sin θ)ⁿ 用二项式定理展开，并比较实部和虚部，即可将 cos nθ 和 sin nθ 表达为 sin θ、cos θ 的幂。例如 n=3 时：cos 3θ = 4cos³θ − 3cos θ，sin 3θ = 3sin θ − 4sin³θ。

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8. Applying Multiple-Angle Identities | 多倍角公式的应用

Typical Exercise 21J.2 problems ask to express sin 5θ in terms of sin θ, or to prove that tan 3θ equals a rational function of tan θ. De Moivre and binomial expansion is the key.

练习21J.2中的典型题目要求将 sin 5θ 用 sin θ 表示，或证明 tan 3θ 等于关于 tan θ 的有理分式。棣莫弗定理与二项式展开是核心。

Example: (cos θ + i sin θ)⁵ = cos 5θ + i sin 5θ. Expanding (c + i s)⁵ gives c⁵ + 5i c⁴s − 10c³s² − 10i c²s³ + 5c s⁴ + i s⁵. Then equate imaginary part: sin 5θ = 5c⁴s − 10c²s³ + s⁵. Substitute c² = 1 − s² to get sin 5θ = 16 sin⁵θ − 20 sin³θ + 5 sin θ.

例如：(cos θ + i sin θ)⁵ = cos 5θ + i sin 5θ。展开 (c + i s)⁵ 得 c⁵ + 5i c⁴s − 10c³s² − 10i c²s³ + 5c s⁴ + i s⁵。比较虚部：sin 5θ = 5c⁴s − 10c²s³ + s⁵。代入 c² = 1 − s²，得 sin 5θ = 16 sin⁵θ − 20 sin³θ + 5 sin θ。

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9. Geometric Interpretation: Roots on the Complex Plane | 几何解释：复平面上的根

The n nth roots of a non-zero complex number form a regular n-gon centred at the origin. This geometric insight helps check answers and solve locus problems.

非零复数的 n 个 n 次根在复平面上构成以原点为中心的正 n 边形。这一几何直观有助于验证答案并解决轨迹问题。

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10. Common Pitfalls and Tips | 常见错误与解题技巧

• English: Forgetting to add 2πk when finding arguments for roots. Always use zₖ = R^(1/n) cis((Φ+2πk)/n).
• 中文：求根时忘记在辐角上加上 2πk，务必使用 zₖ = R^(1/n) cis((Φ+2πk)/n)。
• English: Using degrees instead of radians, especially in calculus contexts, stick to radians.
• 中文：使用角度制而非弧度制，尤其在微积分背景下，应坚持用弧度。
• English: Confusing principal root with all roots; state all n values explicitly.
• 中文：混淆主根与所有根；必须明确写出全部 n 个值。
• English: Algebraic mistakes when expanding (cos θ + i sin θ)ⁿ; double-check binomial coefficients.
• 中文：二项式展开时易出代数错误，仔细核对二项式系数。

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11. Further Practice and Extension | 进一步练习与拓展

To master Exercise 21J.2, try these: (a) Find the 4th roots of −16. (b) Prove cos 4θ = 8cos⁴θ − 8cos²θ + 1. (c) Solve z⁵ + 32 = 0. (d) Show the 5th roots of unity sum to 0.

为掌握练习21J.2，请尝试以下题目：(a) 求 −16 的四次方根；(b) 证明 cos 4θ = 8cos⁴θ − 8cos²θ + 1；(c) 解方程 z⁵ + 32 = 0；(d) 证明 5 次单位根之和为 0。

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12. Summary | 总结

De Moivre’s theorem is a powerful tool linking algebra, trigonometry, and geometry. By confidently converting between Cartesian and polar forms, applying the theorem, and interpreting roots geometrically, you will ace IB exercises like 21J.2.

棣莫弗定理是连接代数、三角与几何的强有力工具。通过熟练进行笛卡尔与极坐标形式间的转换、运用该定理并从几何角度理解根的意义，你将轻松应对类似练习21J.2的IB题目。

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