📚 Exercise H.2: Complex Numbers in Polar Form and de Moivre’s Theorem | 练习 H.2：复数的极形式和棣莫弗定理
This set of IB HL exercises explores the power of polar representation for complex numbers, focusing on multiplication, division, integer powers, and the extraction of roots. de Moivre’s theorem is applied not just for computation but also to derive trigonometric identities and sum series. The problems in Exercise H.2 typically extend beyond routine manipulation, challenging you to connect algebraic, geometric, and trigonometric perspectives of complex numbers.
这组 IB HL 练习探索复数极坐标表示的力量，重点关注乘除运算、整数次幂以及求根。棣莫弗定理不仅用于计算，还应用于推导三角恒等式和级数求和。练习 H.2 中的题目通常超越常规操作，挑战你将复数的代数、几何和三角视角联系起来。
1. Review of Polar Form and Modulus-Argument | 极形式与模-辐角复习
Any non‑zero complex number z = x + iy can be written in polar form as z = r(cos θ + i sin θ), where r = |z| = √(x² + y²) and θ = arg(z), with −π < θ ≤ π as the principal argument. The polar form is particularly convenient for multiplying and dividing complex numbers, because the arguments add or subtract while the moduli are multiplied or divided.
任何非零复数 z = x + iy 都可以写成极形式 z = r(cos θ + i sin θ)，其中 r = |z| = √(x² + y²)，θ = arg(z)，主辐角满足 −π < θ ≤ π。极形式特别便于复数的乘除运算，因为辐角相加或相减而模相乘或相除。
For example, z₁ = √3 + i has modulus r₁ = 2 and argument θ₁ = π/6, so z₁ = 2(cos(π/6) + i sin(π/6)). Similarly, z₂ = 1 − i√3 gives r₂ = 2, θ₂ = −π/3. Writing numbers in this way quickly reveals the effect of operations on the Argand diagram.
例如，z₁ = √3 + i 的模为 2，辐角为 π/6，因此 z₁ = 2(cos(π/6) + i sin(π/6))。类似地，z₂ = 1 − i√3 有 r₂ = 2，θ₂ = −π/3。用这种方式表示能快速揭示运算在阿尔冈图上的效果。
2. Multiplication and Division in Polar Form | 极形式的乘除
If z₁ = r₁(cos θ₁ + i sin θ₁) and z₂ = r₂(cos θ₂ + i sin θ₂), then z₁z₂ = r₁r₂[cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)]. Division follows naturally: z₁ / z₂ = (r₁/r₂)[cos(θ₁ − θ₂) + i sin(θ₁ − θ₂)], provided r₂ ≠ 0. These rules make repetitive multiplication and even exponentiation much simpler than using Cartesian form.
若 z₁ = r₁(cos θ₁ + i sin θ₁)，z₂ = r₂(cos θ₂ + i sin θ₂)，则 z₁z₂ = r₁r₂[cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)]。除法也自然得出：z₁ / z₂ = (r₁/r₂)[cos(θ₁ − θ₂) + i sin(θ₁ − θ₂)]，只要 r₂ ≠ 0。这些规则使重复乘法和指数运算比用笛卡尔形式简单得多。
Exercise H.2 often asks you to express products like (1 + i√3)³(1 − i)² by first converting each factor to polar form. The modulus‑argument product rule then yields the resulting modulus and argument directly, after which you convert back to a + bi form if required.
练习 H.2 经常要求通过首先将每个因子转换为极形式来计算如 (1 + i√3)³(1 − i)² 的乘积。模‑辐角乘积法则直接给出结果的模和辐角，之后如有需要再转回 a + bi 形式。
3. de Moivre’s Theorem for Integer Powers | 整数次幂的棣莫弗定理
de Moivre’s theorem states that (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ) for any integer n. For a complex number in polar form z = r(cos θ + i sin θ), raising to the power n gives zⁿ = rⁿ[cos(nθ) + i sin(nθ)]. This holds for both positive and negative integers; for negative n the modulus becomes rⁿ and the argument is nθ.
棣莫弗定理指出，对于任意整数 n，有 (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)。对极形式的复数 z = r(cos θ + i sin θ)，求 n 次幂得 zⁿ = rⁿ[cos(nθ) + i sin(nθ)]。这对正负整数都成立；n 为负时模为 rⁿ，辐角为 nθ。
Proving the theorem for natural numbers uses induction and the compound angle formulas. The result is extended to negative integers by writing z⁻ⁿ = 1 / zⁿ and using the division rule, confirming the same pattern. Many IB questions test your ability to handle negative powers correctly, especially when arguments go outside (−π, π] and need adjustment.
对自然数的证明使用归纳法和和角公式。然后通过写 z⁻ⁿ = 1 / zⁿ 并运用除法法则将其推广到负整数，证实了相同的模式。许多 IB 问题考察你是否能正确处理负次幂，尤其是当辐角超出 (−π, π] 并需要调整时。
4. Finding Powers of Complex Numbers | 求复数的幂
A typical Exercise H.2 problem might be: Express (1 + i)⁸ in the form a + bi. Using de Moivre, first write 1 + i = √2(cos(π/4) + i sin(π/4)). Then (1 + i)⁸ = (√2)⁸[cos(8×π/4) + i sin(8×π/4)] = 16(cos 2π + i sin 2π) = 16(1 + 0i) = 16. The technique avoids binomial expansion entirely.
典型的练习 H.2 问题可能是：将 (1 + i)⁸ 表示为 a + bi 形式。运用棣莫弗定理，首先写出 1 + i = √2(cos(π/4) + i sin(π/4))。那么 (1 + i)⁸ = (√2)⁸[cos(8×π/4) + i sin(8×π/4)] = 16(cos 2π + i sin 2π) = 16(1 + 0i) = 16。该方法完全避免了二项式展开。
For more complicated bases, such as (√3 − i)⁻⁵, the same steps apply: find modulus and argument, apply the exponent to each, and then reduce the argument modulo 2π. In IB exams, the final answer is often required in exact Cartesian form, so be prepared to recognise sine and cosine values of standard angles.
对于更复杂的底数，如 (√3 − i)⁻⁵，同样步骤：求模和辐角，将指数分别作用于两者，然后将辐角模 2π 化简。在 IB 考试中，最终答案通常要求以精确的笛卡尔形式给出，因此要准备好识别标准角的正弦和余弦值。
5. Roots of Complex Numbers Using de Moivre | 使用棣莫弗定理求复数根
To find the n distinct nth roots of a complex number w = s(cos φ + i sin φ), we set z = r(cos θ + i sin θ) such that zⁿ = w. By de Moivre, rⁿ = s and nθ = φ + 2kπ, where k ∈ {0, 1, 2, …, n−1}. Hence the roots are given by zₖ = s^(1/n)[cos((φ + 2kπ)/n) + i sin((φ + 2kπ)/n)].
为求复数 w = s(cos φ + i sin φ) 的 n 个不同的 n 次方根，设 z = r(cos θ + i sin θ) 使得 zⁿ = w。由棣莫弗定理，rⁿ = s 且 nθ = φ + 2kπ，其中 k ∈ {0, 1, 2, …, n−1}。因此根为 zₖ = s^(1/n)[cos((φ + 2kπ)/n) + i sin((φ + 2kπ)/n)]。
Exercise H.2 frequently includes questions such as: Find all cube roots of 8i. First, 8i = 8(cos(π/2) + i sin(π/2)). Then cube roots have modulus 2 and arguments (π/2 + 2kπ)/3 = π/6, 5π/6, 3π/2. Converting to Cartesian form gives √3 + i, −√3 + i, and −2i. These three points are equally spaced around the circle of radius 2, forming an equilateral triangle.
练习 H.2 经常包含类似问题：求 8i 的所有立方根。首先，8i = 8(cos(π/2) + i sin(π/2))。立方根的模为 2，辐角为 (π/2 + 2kπ)/3 = π/6、5π/6、3π/2。转为笛卡尔形式得 √3 + i、−√3 + i 和 −2i。这三个点均匀分布在半径为 2 的圆上，构成等边三角形。
6. Solving Equations zⁿ = w | 解方程 zⁿ = w
Solving polynomial equations of the form zⁿ − w = 0, where w is a complex constant, reduces directly to finding the nth roots of w. For instance, z⁴ + 16 = 0 → z⁴ = −16 = 16(cos π + i sin π). The four fourth roots are spaced by π/2, yielding the four solutions z = 2 cis(π/4 + kπ/2) for k = 0, 1, 2, 3.
解形如 zⁿ − w = 0（w 为复数常数）的多项式方程可直接归结为求 w 的 n 次方根。例如，z⁴ + 16 = 0 → z⁴ = −16 = 16(cos π + i sin π)。四个四次方根间隔 π/2，得到四个解 z = 2 cis(π/4 + kπ/2)，k = 0, 1, 2, 3。
In Exercise H.2, you may meet equations like z⁵ = 1 − i√3. The right‑hand side must be expressed in polar form with exact modulus and argument. Then the five solutions are written systematically using a table showing the angle increments. A common mistake is to forget that the principal argument of w can be negative; you can still use the same formula by starting with that negative value and adding 2kπ.
在练习 H.2 中，你可能会遇到像 z⁵ = 1 − i√3 这样的方程。需要将右侧表示为具有精确模和辐角的极形式。然后通过表格系统地写出五个解，展示角度增量。常见错误是忘记 w 的主辐角可以是负值；你仍然可以使用相同公式，从该负值开始加上 2kπ。
7. Geometric Interpretation of Complex Roots | 复数根的几何解释
The nth roots of a non‑zero complex number lie at the vertices of a regular n‑gon centred at the origin. For example, the five fifth roots of any number all have the same modulus, so they sit on a circle. Their arguments differ by 2π/5, creating a symmetric star or pentagon. This geometric understanding helps check the algebraic solutions and answers questions about distance between roots.
非零复数的 n 次方根位于以原点为中心的正 n 边形的顶点上。例如，任何一个数的五个五次方根都有相同的模，因此它们位于一个圆上。它们的辐角相差 2π/5，形成对称的星形或五边形。这种几何理解有助于检查代数解并回答关于根之间距离的问题。
Exercise H.2 sometimes asks for the product or sum of all roots. By symmetry, the sum of all roots of zⁿ − w = 0 is zero (unless n = 1), because the polygonal vertices sum vectorially to the origin. The product is (−1)ⁿ⁻¹ w for the equation zⁿ − w = 0, a useful shortcut that can be proved using Vieta’s formulas and also confirmed by multiplying the polar forms.
练习 H.2 有时会问所有根的乘积或和。由对称性，方程 zⁿ − w = 0 的所有根之和为零（除非 n = 1），因为多边形顶点向量和指向原点。对于该方程，乘积为 (−1)ⁿ⁻¹ w，这是一个有用的捷径，可用韦达定理证明，也可通过极形式相乘来确认。
8. Proving Trigonometric Identities | 证明三角恒等式
de Moivre’s theorem is a powerful tool for proving identities like cos 3θ = 4 cos³θ − 3 cos θ. By expanding (cos θ + i sin θ)³ using the binomial theorem, then equating real and imaginary parts to cos 3θ and sin 3θ from de Moivre, you can derive double and triple angle formulas. In Exercise H.2, you may be asked to express cos 5θ entirely in terms of cos θ.
棣莫弗定理是证明恒等式如 cos 3θ = 4 cos³θ − 3 cos θ 的有力工具。通过利用二项式定理展开 (cos θ + i sin θ)³，再将实部与虚部分别与棣莫弗给出的 cos 3θ 和 sin 3θ 等同，你可以推导出二倍角和三倍角公式。在练习 H.2 中，可能要求将 cos 5θ 完全用 cos θ 来表示。
The approach: (cos θ + i sin θ)⁵ = cos 5θ + i sin 5θ. Expand the left side: cos⁵θ + 5i cos⁴θ sin θ − 10 cos³θ sin²θ − 10i cos²θ sin³θ + 5 cos θ sin⁴θ + i sin⁵θ. Collect real parts and replace sin²θ with 1−cos²θ. After simplification, you obtain cos 5θ = 16 cos⁵θ − 20 cos³θ + 5 cos θ. Similar methods can be used to express sin nθ in terms of sin θ, or to find powers of trig functions.
方法是：(cos θ + i sin θ)⁵ = cos 5θ + i sin 5θ。展开左边：cos⁵θ + 5i cos⁴θ sin θ − 10 cos³θ sin²θ − 10i cos²θ sin³θ + 5 cos θ sin⁴θ + i sin⁵θ。收集实部并用 sin²θ = 1−cos²θ 替换。化简后得到 cos 5θ = 16 cos⁵θ − 20 cos³θ + 5 cos θ。类似方法可用于将 sin nθ 表示为 sin θ 的函数，或求三角函数的幂。
9. Applications in Sums of Series | 在级数求和中的应用
Complex numbers and de Moivre allow compact summation of trigonometric series. For example, the sum C = Σ cos kθ (k=0 to n) can be found by considering the geometric series Σ cis(kθ) whose sum is (1 − cis((n+1)θ))/(1 − cis θ). The real part of this expression yields a closed form for C. Such problems appear in Exercise H.2 to link geometric series with trigonometry.
复数和棣莫弗定理使得三角级数可以紧凑求和。例如，和 C = Σ cos kθ (k=0 到 n) 可通过考虑几何级数 Σ cis(kθ) 求得，其和为 (1 − cis((n+1)θ))/(1 − cis θ)。该表达式的实部给出 C 的封闭形式。这类问题出现在练习 H.2 中，以联系几何级数与三角学。
A typical task: Use the sum of a geometric series to prove that Σ_{k=0}^{n} 2⁻ᵏ cos(kθ) = (4 − 2 cos θ) / (5 − 4 cos θ) as n → ∞. By letting z = (1/2) cis θ, the sum is a geometric progression with ratio z. Computing Σ zᵏ and taking the real part, then letting n → ∞, yields the result. This demonstrates the interplay of limits, series, and complex numbers.
一个典型任务：利用几何级数之和证明当 n → ∞ 时 Σ_{k=0}^{n} 2⁻ᵏ cos(kθ) = (4 − 2 cos θ) / (5 − 4 cos θ)。设 z = (1/2) cis θ，该和为公比 z 的几何级数。计算 Σ zᵏ 并取实部，再令 n → ∞，即可得到结果。这展示了极限、级数和复数之间的相互作用。
10. Challenging Mixed Problems from Exercise H.2 | 练习 H.2 中的挑战性综合题
Near the end of Exercise H.2, you find problems that combine several ideas: solving an equation like z⁵ = z̅, where z̅ is the complex conjugate. Writing z = r cis θ and z̅ = r cis(−θ), the equation becomes r⁵ cis(5θ) = r cis(−θ). Equating moduli gives r⁵ = r ⇒ r = 0 or r = 1. Equating arguments yields 5θ = −θ + 2kπ ⇒ θ = kπ/3. So the solutions are 0 and six points on the unit circle.
在练习 H.2 接近尾声处，你会遇到结合多个知识点的题目：解方程如 z⁵ = z̅，其中 z̅ 是共轭复数。令 z = r cis θ，z̅ = r cis(−θ)，方程变为 r⁵ cis(5θ) = r cis(−θ)。比较模得 r⁵ = r ⇒ r = 0 或 r = 1。比较辐角得 5θ = −θ + 2kπ ⇒ θ = kπ/3。因此解为 0 以及单位圆上的六个点。
Another challenging type involves finding arguments of sums or products without explicit Cartesian conversion. For instance, given |z₁| = |z₂| = 1 and arg(z₁) = 2 arg(z₂), determine arg(z₁ + z₂). Using geometry or factorisation z₁ + z₂ = z₂ (z₁/z₂ + 1), the problem reduces to understanding rotation and the sum of two unit vectors. These questions test deep comprehension of polar representation.
另一种挑战类型涉及在无须显式笛卡尔转换的情况下求复数之和或积的辐角。例如，已知 |z₁| = |z₂| = 1 且 arg(z₁) = 2 arg(z₂)，求 arg(z₁ + z₂)。利用几何或因式分解 z₁ + z₂ = z₂ (z₁/z₂ + 1)，问题归结为理解旋转及两个单位向量的和。这类题测试对极坐标表示的深度理解。
11. Avoiding Common Pitfalls and IB Exam Tips | 避免常见误区与 IB 考试技巧
Many students lose marks by not expressing the final answer in exact Cartesian form when required, or by giving arguments outside the principal range. Always check that your principal argument satisfies −π < θ ≤ π. Also, when taking roots, ensure you list all n distinct values – a common error is to stop after finding the first one or two.
许多学生因未按要求用精确笛卡尔形式表达最终答案而丢分，或给出的辐角超出主值范围。始终检查主辐角是否满足 −π < θ ≤ π。此外，求根时要确保列出了所有 n 个不同的值——常见错误是找出一两个后就停止了。
When using de Moivre for negative powers, carefully convert the complex number to polar form, apply the negative exponent to the modulus, and multiply the argument by the exponent. Then, if needed, add or subtract multiples of 2π to bring the argument into the standard range. Practice with a variety of modulus values (rational, irrational) and arguments that are not special angles to build confidence.
当对负次幂使用棣莫弗定理时，仔细将复数转换为极形式，将负指数应用于模，并将辐角乘以指数。然后如有需要，通过加减 2π 的整数倍将辐角调整到标准范围。通过练习各种模值（有理数、无理数）和非特殊角的辐角来建立信心。
12. Summary and Further Practice | 总结与进一步练习
Exercise H.2 consolidates the core IB HL content on complex numbers in polar form, de Moivre’s theorem, and root extraction. Mastery of these techniques enables you to tackle any power or root problem efficiently, prove trigonometric identities elegantly, and sum series that would otherwise be intractable. Revisit the harder problems, especially those linking geometry and algebra, to refine your problem‑solving skills.
练习 H.2 巩固了 IB HL 关于复数极形式、棣莫弗定理和求根的核心内容。掌握这些技巧使你能高效地解决任何幂或根的问题，优雅地证明三角恒等式，并对原本难处理的级数求和。重温较难的题目，尤其是那些联系几何与代数的题目，以提升你的解题技巧。
For further revision, derive the identities for sin 5θ and cos 5θ independently, solve equations like z⁶ = −1 + i, and prove that the sum of the nth roots of unity is zero. Use online graphing tools to visualize roots on the complex plane and deepen your geometric intuition. Regular timed practice with exam‑style questions will prepare you for both Paper 1 and Paper 2 requirements.
为进一步复习，独立推导 sin 5θ 和 cos 5θ 的恒等式，解如 z⁶ = −1 + i 的方程，并证明 n 次单位根之和为零。使用在线绘图工具在复平面上可视化根以加深几何直觉。定期限时练习考试型题目将为 Paper 1 和 Paper 2 的要求做好准备。
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