📚 Exercise 15K: Complex Numbers, De Moivre's Theorem and Roots | 练习15K：复数、棣莫弗定理与根
Exercise 15K in the IB Mathematics course is a turning point where complex numbers move from cartesian addition to polar multiplication, unlocking powerful tools like De Moivre's theorem and the ability to extract nth roots. This article provides a complete walkthrough of the topic, covering polar form conversion, the statement and proof of De Moivre's theorem, computing powers, finding roots of unity and general roots, geometric interpretation, and worked examples that mirror typical exam questions.
在IB数学课程中，练习15K是一个转折点，复数从直角坐标的加法运算转向极坐标的乘法运算，棣莫弗定理与求n次方根等强大工具随之开启。本文全程详解这一主题，涵盖极坐标转换、棣莫弗定理的表述与证明、幂的计算、单位根与一般根的求解、几何解释以及贴近真题的范例，帮助你彻底吃透练习15K。
1. Cartesian Form vs Polar Form | 直角坐标形式与极坐标形式
the complex plane. Its polar form expresses z in terms of the distance from the origin r = |z| = √x² + y² al axis. The polar form is z = r (cos θ + i sin θ). The argument is usually chosen so that –π < θ ≤ π for the principal value, but any θ + 2kπ is equally valid.
复数 z = x + iy 可以表示为复平面上的点 (x, y)。极坐标形式则用点到原点的距离 r = |z| = √x² + y² 以及从正实轴量起的辐角 θ = arg(z) 来表示，即 z = r (cos θ + i sin θ)。辐角通常取主值 –π < θ ≤ π，但所有 θ + 2kπ 均等价。
Converting between forms: Given x and y, r = √x² + y² quadrant. To convert from polar to cartesian, simply compute x = r cos θ and y = r sin θ.
形式互化：已知 x, y，则 r = √x² + y²，θ = arctan(y/x) 并根据象限调整。由极坐标化为直角坐标时，只需计算 x = r cos θ，y = r sin θ。
2. Multiplication and Division in Polar Form | 极坐标形式下的乘法与除法
+ θ₂)]. The moduli multiply and the arguments add. For division, z₁ ÷ z₂ = (r₁/r₂) [cos(θ₁ – θ₂) + i sin(θ₁ – our makes polar form far more efficient for multiplication and powers than cartesian form.
若 z₁ = r₁(cos θ₁ + i sin θ₁)，z₂ = r₂(cos θ₂ + i sin θ₂)，则乘积为 z₁z₂ = r₁r₂ [cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)]。模长相乘，辐角相加。除法为 z₁ ÷ z₂ = (r₁/r₂) [cos(θ₁ – θ₂) + i sin(θ₁ – θ₂)]，其中 r₂ ≠ 0。这一性质使得极坐标形式在乘法与乘方运算中远比直角坐标高效。
3. De Moivre's Theorem Statement | 棣莫弗定理的表述
De Moivre' rⁿ [cos(nθ) + i sin(nθ)]. The theorem holds for all real n, although in Exercise 15K we usually work with integer and rational exponents.
棣莫弗定理指出，对于任意实数 n，(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)。加入模长之后，完整形式为 [r(cos θ + i sin θ)]ⁿ = rⁿ [cos(nθ) + i sin(nθ)]。该定理对所有实数 n 成立，不过在练习15K中我们主要处理整数指数与有理指数。
4. Proof of De Moivre's Theorem for Integer Exponents | 整数指数下棣莫弗定理的证明
using the polar multiplication rule gives cos(kθ + θ) + i sin(kθ + θ) = cos((k+1)θ) + i sin((k+1)θ). The inductive step holds, so the theorem is valid for all positive integers n.
我们借助数学归纳法证明正整数指数的情形。奠基步 n = 1：(cos θ + i sin θ)¹ = cos θ + i sin θ，成立。假设 n = k 时成立：(cos θ + i sin θ)ᵏ = cos(kθ) + i sin(kθ)。则 n = k+1 时： sin θ)ᵏ⁺¹ = (cos θ + i sin θ)ᵏ (cos θ + i sin θ) = [cos(kθ) + i sin(kθ)](cos θ + i sin θ)，利用极坐标乘法规则得到 cos(kθ + θ) + i sin(kθ + θ) = cos((k+1)θ) + i sin((k+1)θ)。归纳步成立，定理对所有正整数 n 成立。
For negative integers, let n = – m > 0: (cos θ + i sin θ)⁻ᵐ = 1 / (cos θ + i sin θ)ᵐ = 1 / [cos(mθ) + i sin(mθ)] = cos(mθ) – i sin(mθ) = cos(–mθ) + i sin(–mθ). Thus the formula holds.
对于负整数，令 n = –m (m > 0)：(cos θ + i sin θ)⁻ᵐ = 1 / (cos θ + i sin θ)ᵐ = 1 / [cos(mθ) + i sin(mθ)] = cos(mθ) – i sin(mθ) = cos(–mθ) + i sin(–mθ)，公式依然成立。
5. Computing Powers of Complex Numbers | 计算复数的幂
De Moivre's theorem turns cumbersome binomial expansions into simple multiplications of the argument. For example, to evaluate (1 + i√3)⁶, first rewrite 1 + i√3 in polar form: modulus r = √(1² + (√3)²) = 2, and argument θ = arctan(√3/1) = π/3. Thus (1 + i√3 i sin(π/3))]⁶ = 2⁶ [cos(6·π/3) + i sin(6·π/3)] = 64 (cos 2π + i sin 2π) = 64(1 + 0i) = 64.
棣莫弗定理将笨拙的二项式展开转化为辐角的简单乘法。例如，计算 (1 + i√3)⁶ ，先将 1 + i√3 写成极坐标形式：模 r = √(1² + (√3)²) = 2，辐角 θ = arctan(√3/1) = π/3。因此 (1 + i√3 [2(cos(π/3) + i sin(π/3))]⁶ = 2⁶ [cos(6·π/3) + i sin(6·π/3)] = 64 (cos 2π + i sin 2π) = 64。
Without De Moivre's theorem, expanding (1 + i√3)⁶ using the binomial theorem and simplifying powers of i would be far more time‑consuming and prone to error. This highlights why the polar form is essential in IB exams.
若不用棣莫弗定理，用二项式定理展开 (1 + i√3)⁶ 再化简 i 的幂将极为耗时且易错，这凸显了极坐标形式在IB考试中的重要性。
6. Introduction to nth Roots | n次方根引入
, the equation rⁿ (cos nθ + i sin nθ) = w = R (cos φ + i sin φ) yields rⁿ = R, hence r = R1/n k is an integer. Thus θ = (φ + 2kπ)/n. Distinct roots are obtained by taking k = 0, 1, 2, …, n–1.
若 zⁿ = w，则 z 称为 w 的一个 n 次方根。由于辐角具有 2π 周期性，方程 rⁿ (cos nθ + i sin nθ) = w = R (cos φ + i sin φ) 给出 rⁿ = R，于是 r = R1/n（正实数的n次主根），且 nθ = φ + 2kπ，其中 k 为整数。故 θ = (φ + 2kπ)/n。取 k = 0, 1, 2, …, n–1 便得到 n 个互异的根。
7. Roots of Unity | 单位根
r = 1, θ = (0 + 2kπ)/n = 2kπ/n. The roots are zₖ = cos(2kπ/n) + i sin(2kπ/n) for k = 0, 1, …, n– ts of unity. For example, the cube roots of unity (n = 3) are 1, cos(2π/3) + i sin(2π/3) = –½ + i(√3/2), and cos(4π/3) + i sin(4π/3) = –½ – i(√3/2).
当 w = 1 时，我们解 zⁿ = 1。在极坐标中 1 = 1·(cos 0 + i sin 0)。由 n 次方根公式，r = 1，θ = (0 + 2kπ)/n = 2kπ/n。根为 zₖ = cos(2kπ/n) + i sin(2kπ/n)，k = 0, 1, …, n–1。这些数称为 n 次单位根。例如三次单位根 (n=3) 为 1，cos(2π/3) + i sin(2π/3) = –½ + i(√3/2)，以及 cos(4π/3) + i sin(4π/3) = –½ – i(√3/2)。
is can be seen from the geometric symmetry or by summing the geometric series 1 + ω + ω² + … + ωⁿ⁻¹ = 0 where ω is a primitive root.
最精妙的性质之一是，当 n ≥ 2 时所有 n 次单位根之和为零。这一点既可以从几何对称性看出，也可以通过等比数列求和 1 + ω + ω² + … + ωⁿ⁻¹ = 0（ω为本原根）加以证明。
8. General Formula for nth Roots | n次方根的一般公式
To find all n distinct nth roots of a general complex number w = R(cos φ + i sin φ), use:
zₖ = R1/n [cos((φ + 2kπ)/n) + i sin((φ + 2kπ)/n)], k = 0, 1, ..., n–1
为求一般复数 w = R(cos φ + i sin φ) 的所有 n 个 n 次方根，使用：
zₖ = R1/n [cos((φ + 2kπ)/n) + i sin((φ + 2kπ)/n)], k = 0, 1, ..., n–1
Note that R1/n epeats the same angle modulo 2π. This is the complete solution set of the equation zⁿ = w.
注意 R1/n 表示模长 R 的唯一正实数 n 次方根。该公式恰好给出 n 个不同的值，因为 k 增加 n 会使辐角增加 2π，回到同一点。这正是方程 zⁿ = w 的完整解集。
9. Geometric Representation on the Complex Plane | 复平面上的几何表示
The n nth roots of w lie on a circle centered at the origin with radius R1/n qually spaced by an angle of 2π/n radians. Thus the roots form the vertices of a regular n‑ polygon is inscribed in the unit circle; for general w, it is rotated by φ/n relative to the positive real axis.
w 的 n 个 n 次方根分布在以原点为圆心、半径为 R1/n 的圆上，彼此间隔 2π/n 弧度。因此这些根构成复平面上一个正 n 边形的顶点。对于 w = 1，该正多边形内接于单位圆；对于一般 w，整体图形绕正实轴旋转 φ/n。
o sketch the roots without evaluating all trigonometric values. For instance, solving z⁴ = –16: modulus of –16 is 16, so radius = 161/4 = 2. Argument of – es (π + 2kπ)/4 = π/4, 3π/4, 5π/4, 7π/4. They sit on a circle of radius 2 at 45° intervals.
利用几何视角，无需逐一求值便可快速画出根的分布。例如求解 z⁴ = –16：–16 的模为 16，故半径 = 161/4 = 2；辐角为 π，因此根的角度为 (π+2kπ)/4 = π/4，3π/4，5π/4，7π/4。它们分布在半径为2的圆上，相隔45°。
10. Worked Example: Solving z³ = –8 | 实例解析：解 z³ = –8
We solve z³ = –8 following the steps expected in Exercise 15K. First, express –8 in polar form: modulus = 8, argument = π (since –8 lies on the negative real axis). So – i sin θ), then z³ = r³ [cos(3θ) + i sin(3θ)] = 8 (cos π + i sin π). Equate moduli: r³ = 8 ⇒ r = 2. Equate arguments: 3θ = π + 2kπ, so θ = π/3 + 2kπ/3.
按练习15K的标准步骤求解 z³ = –8。首先将 –8 写成极坐标形式：模 = 8，辐角 = π（因为 –8 在负实轴上）。于是 –8 = 8 (cos π + i sin π)。设 z = r(cos θ + i sin θ)，则 z³ = r³ [cos(3θ) + i sin(3θ)] = 8 (cos π + i sin π)。比较模：r³ = 8 ⇒ r = 2。比较辐角：3θ = π + 2kπ，故 θ = π/3 + 2kπ/3。
Take k = 0, 1, 2 to get the three cube roots:
- k = 0: z₀ = 2 (cos(π/3) + i sin(π/3)) = 2(½ + i √3/2) = 1 + i√3
- k = 1: z₁ = 2 (cos π + i sin π) = 2(–1 + i·0) = –2
- k = 2: z₂ = 2 (cos(5π/3) + i sin(5π/3)) = 2(½ – i √3/2) = 1 – i√3
取 k = 0, 1, 2 得三个立方根：
- k = 0: z₀ = 2 (cos(π/3) + i sin(π/3)) = 2(½ + i √3/2) = 1 + i√3
- k = 1: z₁ = 2 (cos π + i sin π) = 2(–1 + i·0) = –2
- k = 2: z₂ = 2 (cos(5π/3) + i sin(5π/3)) = 2(½ – i √3/2) = 1 – i√3
These three points form an equilateral triangle on the circle of radius 2, confirming the geometric prediction.
这三个点构成半径为2的圆上的等边三角形，印证了几何预期。
11. Linking De Moivre's Theorem with Trigonometric Identities | 棣莫弗定理与三角恒等式的联系
De Moivre's theorem provides a neat way to derive multiple‑ ties. Expanding (cos θ + i sin θ)³ using the binomial theorem gives cos³θ + 3i cos²θ sin θ – 3 cos θ sin²θ – i sin³θ. Meanwhile, by De Moivre' θ + i sin θ)³ = cos 3θ + i sin 3θ. Equating real and imaginary parts yields cos 3θ = cos³θ – 3 cos θ sin²θ and sin 3θ = 3 cos²θ sin θ – sin³θ. Using sin²θ = 1 – cos²θ, these can be written purely in terms of cos θ or sin θ, giving the well‑known triple‑angle formulas.
棣莫弗定理为推导倍角公式提供了简洁途径。用二项式定理展开 (cos θ + i sin θ)³ 得 cos³θ + 3i cos²θ sin θ – 3 cos θ sin²θ – i sin³θ。另一方面，由棣莫弗定理，(cos θ + i sin θ)³ = cos 3θ + i sin 3θ。比较实部和虚部即得 cos 3θ = cos³θ – 3 cos θ sin²θ，sin 3θ = 3 cos²θ sin θ – sin³θ。利用 sin²θ = 1 – cos²θ 可将其化为只含 cos θ 或 sin θ 的形式，得到熟知的三倍角公式。
Similarly, expressions for cos(nθ) and sin(nθ) as polynomials in cos θ and sin θ can be derived, and exam questions frequently ask to use De Moivre to find tan(nθ) or to express cos⁵θ in terms of multiple angles.
类似地，可将 cos(nθ) 和 sin(nθ) 表示为 cos θ 和 sin θ 的多项式，考试常要求用棣莫弗定理求 tan(nθ)，或将 cos⁵θ 表示为多倍角的线性组合。
12. Common Exam Pitfalls and Tips | 常见考试陷阱与技巧
- Forgetting the modulus when applying De Moivre: Always write [r(cos θ + i sin θ)]ⁿ = rⁿ[cos(nθ) + i sin(nθ)]. Missing rⁿ will cost marks.
- 忽略模长：运用定理时务必写成 [r(cos θ + i sin θ)]ⁿ = rⁿ[cos(nθ) + i sin(nθ)]。漏掉 rⁿ 会严重失分。
- Incorrect argument ranges: When required to give the principal argument, make sure final roots are expressed with –π < θ ≤ π. Adjust when necessary.
- 辐角范围错误：若要求给出辐角主值，应确保最终的根满足 –π < θ ≤ π，必要时进行加减 2π 调整。
- Overlooking the number of roots: An nth root equation yields exactly n distinct solutions. Forgetting one or adding extra non‑distinct ones is a common mistake.
- 遗漏或重复根：n 次方根方程恰有 n 个互异解，漏解或加入非互异解是常见失误。
- Decimal approximations: Unless instructed otherwise, leave roots in exact polar or surd form. Decimal answers often lose the exactness required at IB level.
- 小数近似：除非题目明确要求，否则应将根保留为精确极坐标或根式形式。小数近似往往不符合IB对精确性的要求。
- Verification: Check your answers by raising one of the roots back to the original power; a quick mental check of the modulus and argument can catch sign errors.
- 验证：将所求的某个根乘回原方程进行检验；快速心算模与辐角能及时发现符号错误。
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