📚 Exercise H.4: De Moivre's Theorem and Applications | 练习H.4：棣莫弗定理及其应用
Exercise H.4 in the IB Mathematics Higher Level course is a pivotal set of problems that deepens your understanding of complex numbers. This article unpacks the core ideas behind De Moivre's theorem — a bridge between trigonometry and algebra — and shows you how to tackle typical exam questions on powers, roots, and identities with confidence.
IB 数学高等级课程中的练习 H.4 是一组关键问题，能帮你加深对复数的理解。本文拆解了棣莫弗定理的核心思想——它是三角与代数之间的桥梁——并展示如何自信地应对关于幂、根与恒等式的典型考题。
1. Polar Form and the Foundation | 极坐标形式与基础
expressed in polar form as z = r(cos θ + i sin θ), often abbreviated as r cis θ, where r = |z| makes multiplication and exponentiation far more manageable than working with a + bi form.
每个复数 z = x + iy 都可以用极坐标形式表示为 z = r(cos θ + i sin θ)，常缩写为 r cis θ，其中 r = |z| 是模，θ = arg(z) 是辐角。这种表示法使乘法和乘方运算远比 a + bi 形式更易处理。
θ₂ is simply r₁r₂ cis(θ₁ + θ₂). This multiplicative property is the intuitive bedrock of De Moivre's theorem.
例如，两个复数 z₁ = r₁ cis θ₁ 与 z₂ = r₂ cis θ₂ 的乘积就是 r₁r₂ cis(θ₁ + θ₂)。这个乘法性质是棣莫弗定理的直观根基。
2. Stating De Moivre's Theorem | 棣莫弗定理的陈述
For any real number θ and any integer n, the theorem states: (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ. If we include the modulus r, the full form is [r(cos θ + i sin θ)]ⁿ = rⁿ (cos nθ + i sin nθ). The proof for positive integers relies on mathematical induction, and the result extends to all integers (and even rational powers with care).
对任意实数 θ 和整数 n，定理指出：(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ。若包含模 r，完整形式为 [r(cos θ + i sin θ)]ⁿ = rⁿ (cos nθ + i sin nθ)。对正整数的证明依赖于数学归纳法，该结果可推广到所有整数（甚至在谨慎对待下可推广到有理数次幂）。
z = r(cos θ + i sin θ) ⟹ zⁿ = rⁿ (cos nθ + i sin nθ)
This compact statement unlocks a vast range of calculations, from finding (1 + i√3)⁵ to deriving triple-angle formulas.
这个紧凑的陈述解锁了大量计算，从求 (1 + i√3)⁵ 到推导三倍角公式。
3. Computing Powers of Complex Numbers | 计算复数的乘方
To evaluate a high power such as (1 + i)⁸, convert to polar form first. Here, 1 + i has modulus √2 and argument π/4, so (√2 cis π/4)⁸ = (√2 /4) = 16 cis 2π = 16. Without polar form, expanding binomially would be extremely tedious.
要计算像 (1 + i)⁸ 这样的高次幂，应先将它转换为极坐标形式。这里 1 + i 的模为 √2，辐角为 π/4，因此 (√2 cis π/4)⁸ = (√2)⁸ cis (8 × π/4) = 16 cis 2π = 16。若不用极坐标形式，用二项式展开将极其繁琐。
usually a + bi. In the example above, 16 cis 2π = 16(cos 2π + i sin 2π) = 16(1 + 0i) = 16.
请记住最后要将答案表达为题目要求的形式，通常是 a + bi。上例中 16 cis 2π = 16(cos 2π + i sin 2π) = 16(1 + 0i) = 16。
4. Roots of Complex Numbers | 复数的根
De Moivre' e w is a given complex number. If w = r cis θ, then the n distinct nth roots are given by:
棣莫弗定理在解方程 zⁿ = w（w 为给定复数）时同样强大。若 w = r cis θ，则 n 个不同的 n 次方根由以下公式给出：
zₖ = r1/n cis ((θ + 2kπ)/n) for k = 0, 1, 2, …, n−1
Each root lies on a circle of radius r1/n centred at the origin, spaced by an angle of 2π/n. This gives a regular n-gon pattern on the Argand diagram.
每个根都位于以原点为圆心、半径为 r1/n 的圆上，相邻两根的辐角差为 2π/n。这在阿尔冈图上构成了正 n 边形的图案。
5. Roots of Unity | 单位根
e roots of unity are cis θ with θ = 2kπ/3 for k = 0,1,2. This yields 1, cis 2π/3 = −½ + i(√3/2), cis 4π/3 = −½ − i(√3/2). Notice that the non-real roots are complex conjugates.
当 w = 1 时，我们得到 n 次单位根。例如，三次单位根为 cis θ，其中 θ = 2kπ/3，k = 0,1,2。这就得到 1、cis 2π/3 = −½ + i(√3/2)、cis 4π/3 = −½ − i(√3/2)。注意非实数根互为共轭。
ity is 0, and the product is (−1)ⁿ⁻¹. These can be proven using the polynomial zⁿ − 1 = 0.
单位根的性质在 IB 考题中经常出现：所有 n 次单位根之和为 0，其乘积为 (−1)ⁿ⁻¹。这些性质可利用多项式 zⁿ − 1 = 0 来证明。
6. Proving Trigonometric Identities | 证明三角恒等式
One of the most elegant applications of De Moivre's theorem is deriving trigonometric identities. By equating (cos θ + i sin θ)ⁿ expanded via the binomial theorem with cos nθ + i sin nθ, we can extract real and imaginary parts.
棣莫弗定理最优雅的应用之一就是推导三角恒等式。将 (cos θ + i sin θ)ⁿ 用二项式定理展开，并与 cos nθ + i sin nθ 的实部和虚部分别相等，即可提取恒等式。
Substituting s² = 1 − c² yields cos 3θ = 4 cos³θ − 3 cos θ and sin 3θ = 3 sin θ − 4 sin³θ.
以 n = 3 为例：(c + i s)³ = c³ + 3c²(i s) + 3c(i s)² + (i s)³ = c³ + 3i c² s − 3c s² − i s³。实部 c³ − 3c s² 必然等于 cos 3θ；虚部 3c² s − s³ = sin 3θ。代入 s² = 1 − c² 得到 cos 3θ = 4 cos³θ − 3 cos θ 与 sin 3θ = 3 sin θ − 4 sin³θ。
7. Finding Complex Conjugates and Symmetry | 共轭复数与对称性
ar in conjugate pairs. For instance, if z = r cis θ is a root of a real polynomial, then z̅ = r cis(−θ) must also be a root.
一个常见的考试技巧是：若多项式系数为实数，则复数根必定成共轭对出现。例如，若 z = r cis θ 是某实系数多项式的根，则 z̅ = r cis(−θ) 也必定是其根。
This observation often reduces the number of equations you need to solve in Exercise H.4 questions, as you can immediately deduce the second root once one complex root is found.
这一观察常能减少练习 H.4 题中需解的方程数量，因为你一旦找到一个复数根，就可以立即推出第二个根。
8. Recognizing the Principal Argument and Multiple Forms | 识别主辐角与多种表达形式
When using De Moivre' y −π < θ ≤ π) or to add integer multiples of 2π as needed. For example, −1 − i has modulus √2, but its argument could be −3π/4 or 5π/4. Using the wrong argument leads to incorrect power or root values.
使用棣莫弗定理时，必须使用主辐角（通常是 −π < θ ≤ π）或根据需要加上 2π 的整数倍。例如，−1 − i 的模为 √2，但其辐角可以是 −3π/4 或 5π/4。若使用错误的辐角，会导致幂或根的值错误。
In root problems, the general solution formula automatically adds 2kπ, so you are free to choose any convenient representation of θ, as long as consistency is maintained through all k values.
在求根问题中，通解公式已经自动加上了 2kπ，因此你可以自由选择任一方便的 θ 表示形式，只要在所有 k 值中保持一致即可。
9. Common Pitfalls and How to Avoid Them | 常见陷阱与如何避免
| Pitfall | How to avoid |
|---|---|
| Forgetting to multiply the argument by n in the power rule | Write the expression in full: rⁿ cis(nθ) and double-check. |
| Using De Moivre for non-integer n without considering multi-valuedness | For fractional powers, always use the roots formula. |
| Losing precision when converting arguments from radians to degrees | Stick to radians and keep exact values with π. |
| Omitting the modulus when raising to a power | Treat the modulus separately: rⁿ. |
中文对照：
| 陷阱 | 避免方法 |
|---|---|
| 在乘方法则中忘记将辐角乘以 n | 写出完整表达式：rⁿ cis(nθ) 并再次检查。 |
| 对非整数 n 使用棣莫弗定理而未考虑多值性 | 分数次幂时务必使用求根公式。 |
| 将弧度转换为角度时丢失精度 | 坚持使用弧度并以 π 保留精确值。 |
| 求幂时遗漏模 | 单独处理模：rⁿ。 |
10. Exam-Style Example with Full Solution | 考试风格例题及完整解答
Question: Given z = √3 + i, find z⁵ and determine all fourth roots of z, expressing answers in exact Cartesian form a + bi.
问题：已知 z = √3 + i，求 z⁵ 并确定 z 的所有四次方根，将答案表示为精确的笛卡尔形式 a + bi。
First, |z| = √((√3)² + 1²) = √3+1 = 2. Argument θ = arctan(1/√3) = π/6. Thus z = 2 cis(π/6).
首先，|z| = √((√3)² + 1²) = √3+1 = 2。辐角 θ = arctan(1/√3) = π/6。因此 z = 2 cis(π/6)。
Part (a): z⁵ = 2⁵ cis(5 × π/6) = 32 cis(5π/6) = 32(cos 5π/6 + i sin 5π/6) = 32(−√3/2 + i(1/2)) = −16√3 + 16i.
(a) 部分：z⁵ = 2⁵ cis(5 × π/6) = 32 cis(5π/6) = 32(cos 5π/6 + i sin 5π/6) = 32(−√3/2 + i(1/2)) = −16√3 + 16i。
Part (b): We need the four 4th roots of z, i.e., solve w⁴ = 2 cis(π/6). Modulus of each root = 21/4 = ⁴√2 for k = 0,1,2,3. This gives angles: π/24, 13π/24, 25π/24, 37π/24. Express each in a + bi:
(b) 部分：我们需要 z 的四个四次方根，即解 w⁴ = 2 cis(π/6)。每个根的模 = 21/4 = ⁴√2。辐角为：(π/6 + 2kπ)/4，k = 0,1,2,3。得到角度：π/24、13π/24、25π/24、37π/24。分别表示为 a + bi：
- w₀ = ⁴√2 (cos π/24 + i sin π/24)
- w₁ = ⁴√2 (cos 13π/24 + i sin 13π/24) = ⁴√2 (cos(π − 11π/24) + i sin(...)) but better to keep exact;
- w₂ = ⁴√2 (cos 25π/24 + i sin 25π/24) = ⁴√2 (−cos π/24 − i sin π/24)
- w₃ = ⁴√2 (cos 37π/24 + i sin 37π/24) = ⁴√2 (cos(2π − 11π/24) + i sin(...)) = ⁴√2 (cos 11π/24 − i sin 11π/24)
Often the question accepts the polar form, but you can apply half-angle formulas if exact Cartesian values are demanded. The key is showing the method clearly.
通常题目接受极坐标形式，但若要求精确的笛卡尔值，你可以应用半角公式。关键在于清晰展示解题过程。
11. Linking to Other IB Topics | 与其它 IB 主题的联系
De Moivre's theorem is not an isolated topic. It appears in proofs of trigonometric limits, in summing series involving cosine and sine, in vector rotations (multiplying by i gives a 90° rotation), and in the study of complex mappings. Exercise H.4 often combines these ideas to test deeper understanding.
棣莫弗定理并非孤立主题。它出现在三角极限的证明中、含余弦和正弦的级数求和中、向量旋转（乘以 i 相当于旋转 90°）以及复变映射的研究中。练习 H.4 常将这些思路结合起来，检验更深入的理解。
.. + cos(2n−1)θ as a closed form using the real part of a geometric series of complex numbers.
例如，你可能被要求将和式 S = cos θ + cos 3θ + cos 5θ + ... + cos(2n−1)θ 表达为一个闭合形式，方法是利用复数等比级数的实部。
12. Tips for Success in Exercise H.4 | 成功攻克练习 H.4 的技巧
Always sketch the number on an Argand diagram to visualise the argument and modulus. Check that your answers make sense geometrically — the roots should be symmetrically spaced. Practise converting between forms rapidly: from x + iy to r cis θ and back. Memorise the exact trig values for common angles (0, π/6, π/4, π/3, π/2, etc.) to save time in exams.
始终在阿尔冈图上草绘复数以直观把握辐角与模。检查答案在几何上是否合理——根应当均匀对称分布。练习在形式间快速转换：从 x + iy 到 r cis θ 再转回。熟记常见角度（0、π/6、π/4、π/3、π/2 等）的精确三角值，以便在考试中节省时间。
Finally, when dealing with identities, isolate real and imaginary parts carefully and don' atic practice, Exercise H.4 becomes one of the most rewarding parts of the IB HL syllabus.
最后，在处理恒等式时，要仔细分离实部和虚部，不要忘记 i² = −1。通过系统练习，练习 H.4 将成为 IB 高等级课程中最有收获的部分之一。
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