IB Math Exercise 18H: Mastering De Moivre’s Theorem and Complex Roots | IB数学练习18H：掌握棣莫弗定理与复数根

📚 IB Math Exercise 18H: Mastering De Moivre’s Theorem and Complex Roots | IB数学练习18H：掌握棣莫弗定理与复数根

In IB Mathematics Analysis and Approaches HL, Exercise 18H is a collection of challenging problems centred on the polar representation of complex numbers, De Moivre's theorem, and the extraction of roots. Mastering these techniques is essential for tackling Paper 1 and Paper 2 questions on complex numbers, and for deepening your understanding of the algebraic structure of complex numbers.

在IB数学分析与方法HL课程中，练习18H汇集了一系列围绕复数极坐标表示、棣莫弗定理以及求复数根的难题。掌握这些技巧对于解答试卷一和试卷二中有关复数的题目至关重要，也能加深你对复数代数结构的理解。

1. Polar Form and Euler's Formula | 极坐标形式与欧拉公式

Any non-zero complex number z = a + bi can be expressed in polar form as z = r(cos θ + i sin θ), where r = |z| is the modulus and θ = arg(z) is the argument, usually chosen in (−π, π] or [0, 2π). Euler's formula states that e^iθ = cos θ + i sin θ, giving the compact exponential form z = r e^iθ.

任何一个非零复数 z = a + bi 都可以用极坐标形式表示为 z = r(cos θ + i sin θ)，其中 r = |z| 是模，θ = arg(z) 是辐角，通常选取在 (−π, π] 或 [0, 2π) 内。欧拉公式 e^iθ = cos θ + i sin θ 将极坐标形式进一步简化为指数形式 z = r e^iθ。

This representation is particularly useful because multiplication of complex numbers becomes a simple operation: add the arguments and multiply the moduli.

这一表示方法格外有用，因为复数的乘法运算变得非常简单：辐角相加，模长相乘。

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2. Multiplication and Division in Polar Form | 极坐标形式下的乘法与除法

Given two complex numbers z₁ = r₁(cos θ₁ + i sin θ₁) and z₂ = r₂(cos θ₂ + i sin θ₂), their product is z₁z₂ = r₁r₂ [cos(θ₁+θ₂) + i sin(θ₁+θ₂)].

给定两个复数 z₁ = r₁(cos θ₁ + i sin θ₁) 和 z₂ = r₂(cos θ₂ + i sin θ₂)，它们的乘积为 z₁z₂ = r₁r₂ [cos(θ₁+θ₂) + i sin(θ₁+θ₂)]。

Similarly, division yields z₁/z₂ = (r₁/r₂) [cos(θ₁−θ₂) + i sin(θ₁−θ₂)]. These formulas form the basis for De Moivre's theorem.

类似地，除法给出 z₁/z₂ = (r₁/r₂) [cos(θ₁−θ₂) + i sin(θ₁−θ₂)]。这些公式构成了棣莫弗定理的基础。

In the exponential form, z₁z₂ = r₁r₂ e^{i(θ₁+θ₂)} and z₁/z₂ = (r₁/r₂) e^{i(θ₁−θ₂)}, which makes the extension to integer powers straightforward.

在指数形式下，z₁z₂ = r₁r₂ e^{i(θ₁+θ₂)}，z₁/z₂ = (r₁/r₂) e^{i(θ₁−θ₂)}，这使得向整数次幂的推广变得顺理成章。

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3. De Moivre's Theorem | 棣莫弗定理

De Moivre's theorem states that for any real number θ and any integer n, (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ).

棣莫弗定理指出，对于任意实数 θ 和任意整数 n，有 (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)。

If z = r(cos θ + i sin θ), then zⁿ = rⁿ (cos nθ + i sin nθ). The theorem can be proved by induction for positive integers and extended to negative integers using the reciprocal relationship.

若 z = r(cos θ + i sin θ)，则 zⁿ = rⁿ (cos nθ + i sin nθ)。该定理可以通过数学归纳法对正整数进行证明，并利用倒数关系推广到负整数。

zⁿ = rⁿ e^inθ = rⁿ (cos nθ + i sin nθ)

Many Exercise 18H questions ask you to compute high powers of a complex number efficiently without expanding the binomial directly.

练习18H中的许多题目都要求你高效地计算复数的高次幂，而不必直接展开二项式。

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4. Finding Complex Roots Using De Moivre's Theorem | 利用棣莫弗定理求复数根

To find the n distinct nth roots of a complex number w = r(cos θ + i sin θ), we set zⁿ = w and obtain z_k = r^1/n [cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)], for k = 0, 1, 2, …, n−1.

要求复数 w = r(cos θ + i sin θ) 的 n 个不同 n 次方根，我们令 zⁿ = w，得到 z_k = r^1/n [cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)]，其中 k = 0, 1, 2, …, n−1。

The addition of multiples of 2π reflects the periodicity of the argument; all n roots are equally spaced around a circle of radius r^1/n in the complex plane.

加上 2π 的整数倍体现了辐角的周期性；所有 n 个根在复平面上均匀分布在半径为 r^1/n 的圆周上。

be roots of 8 can be found by writing 8 = 8(cos 0 + i sin 0) and applying the formula with n=3, leading to the three roots: 2, −1 + i√3, −1 − i√3.

例如，8 的立方根可以通过将 8 写为 8(cos 0 + i sin 0) 并代入 n=3 的公式来求得，得到三个根：2，−1 + i√3，−1 − i√3。

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5. Roots of Unity and Their Properties | 单位根及其性质

The nth roots of unity are the solutions of zⁿ = 1. They are given by ω_k = cos(2kπ/n) + i sin(2kπ/n) for k = 0, 1, …, n−1. The principal root is ω = cos(2π/n) + i sin(2π/n).

单位根是方程 zⁿ = 1 的解，由 ω_k = cos(2kπ/n) + i sin(2kπ/n) 给出，其中 k = 0, 1, …, n−1。原始根通常取 ω = cos(2π/n) + i sin(2π/n)。

Key properties include: the sum of all nth roots of unity is 0, and the product is (−1)ⁿ⁻¹. If ω is a primitive root, then the set {1, ω, ω², …, ωⁿ⁻¹} generates all roots.

关键性质包括：所有 n 次单位根之和为 0，乘积为 (−1)ⁿ⁻¹。若 ω 是本原根，则集合 {1, ω, ω², …, ωⁿ⁻¹} 生成全部根。

n	Roots of Unity (first few)	Sum
2	1, −1	0
3	1, −½ + i(√3/2), −½ − i(√3/2)	0
4	1, i, −1, −i	0

Exercise 18H often includes problems that exploit these symmetry properties to factorise polynomials or simplify sums.

练习18H中常包含利用这些对称性质来因式分解多项式或简化求和的题目。

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6. Solving Polynomial Equations with Complex Coefficients | 求解复数系数多项式方程

De Moivre's theorem is invaluable when solving equations like zⁿ = c, but it also helps with more elaborate polynomials. By substituting z = r e^iθ and comparing moduli and arguments, you can systematically find all solutions.

棣莫弗定理在解诸如 zⁿ = c 这类方程时非常宝贵，它也同样适用于更复杂的多项式。通过代换 z = r e^iθ 并比较模与辐角，你可以系统地求出所有解。

For instance, to solve z⁴ + 16 = 0, rewrite it as z⁴ = −16 = 16(cos π + i sin π). Then the four roots are z_k = 2 [cos((π+2kπ)/4) + i sin((π+2kπ)/4)], k=0,1,2,3.

例如，求解 z⁴ + 16 = 0，将其改写为 z⁴ = −16 = 16(cos π + i sin π)。由此得到四个根 z_k = 2 [cos((π+2kπ)/4) + i sin((π+2kπ)/4)]，k=0,1,2,3。

Always remember to express the complex number on the right-hand side in exact polar form before extracting roots.

一定要记住，在开方之前将右边的复数用精确的极坐标形式表示出来。

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7. Trigonometric Identities via De Moivre | 借助棣莫弗证明三角恒等式

Another common application found in Exercise 18H is deriving trigonometric identities. By writing cos nθ + i sin nθ = (cos θ + i sin θ)ⁿ and expanding the right-hand side using the binomial theorem, you can equate real and imaginary parts to obtain expressions for cos nθ and sin nθ in terms of powers of cos θ and sin θ.

练习18H中另一个常见应用是推导三角恒等式。通过写出 cos nθ + i sin nθ = (cos θ + i sin θ)ⁿ 并用二项式定理展开右边，令实部和虚部分别相等，即可得到用 cos θ 和 sin θ 的幂表示的 cos nθ 和 sin nθ 的表达式。

For n=3: (cos θ + i sin θ)³ = cos³θ + 3i cos²θ sin θ − 3 cos θ sin²θ − i sin³θ. Hence, cos 3θ = 4 cos³θ − 3 cos θ and sin 3θ = 3 sin θ − 4 sin³θ.

以 n=3 为例：(cos θ + i sin θ)³ = cos³θ + 3i cos²θ sin θ − 3 cos θ sin²θ − i sin³θ。因此，cos 3θ = 4 cos³θ − 3 cos θ，sin 3θ = 3 sin θ − 4 sin³θ。

This technique can also be used to find sums such as Σ cos kθ by summing geometric series in complex exponential form.

这一技巧还可以通过对复数指数形式的几何级数求和来得到诸如 Σ cos kθ 的和式。

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8. Typical Exercise 18H Problem Walkthrough | 练习18H典型题目详解

Problem: Find all four fourth roots of −4 − 4i√3 and plot them on an Argand diagram.

题目：求 −4 − 4i√3 的全部四个四次方根，并绘在阿尔冈图上。

Step 1: Compute the modulus: |−4 − 4i√3| = √16 + 48 = √64 = 8.

步骤1：计算模：|−4 − 4i√3| = √16 + 48 = √64 = 8。

Step 2: Determine the argument θ. The point lies in the third quadrant, so θ = −π + arctan(√3) = −2π/3 (or equivalently 4π/3). Choose θ = −2π/3 for the principal argument.

步骤2：确定辐角 θ。该点位于第三象限，故 θ = −π + arctan(√3) = −2π/3（或等价地 4π/3）。选择主辐角 θ = −2π/3。

Thus, −4 − 4i√3 = 8[cos(−2π/3) + i sin(−2π/3)].

于是，−4 − 4i√3 = 8[cos(−2π/3) + i sin(−2π/3)]。

Step 3: The fourth roots are z_k = 8^1/4 [cos((−2π/3 + 2kπ)/4) + i sin((−2π/3 + 2kπ)/4)], with k = 0, 1, 2, 3; 8^1/4 = (2³)^1/4 = 2^3/4 = ∜(8).

步骤3：四次方根为 z_k = 8^1/4 [cos((−2π/3 + 2kπ)/4) + i sin((−2π/3 + 2kπ)/4)]，其中 k = 0,1,2,3；8^1/4 = 2^3/4 ≈ 1.682。

Step 4: List the four roots by evaluating for each k and simplify arguments if required.

步骤4：分别代入每个 k 并化简辐角，列出四个根。

This systematic approach handles any similar Exercise 18H item.

这种系统的方法可以处理任何类似的练习18H题目。

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9. Common Mistakes and Tips | 常见错误与高分技巧

Mistake 1: Forgetting that the argument is not unique and failing to add 2kπ when finding roots, which results in missing roots.

错误1：忘记辐角的多值性，在求根时没有加上 2kπ，导致遗漏其他根。

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