📚 The nth Roots of Complex Numbers | 复数的n次方根求法
In IB Mathematics Analysis and Approaches HL, complex numbers form a key topic, and one of the most elegant results is the method for finding the n-th roots of a complex number. This article gives a complete guide: the polar form, De Moivre’s Theorem, the derivation of the n-th root formula, worked examples, geometric insight, and common exam pitfalls.
在IB数学分析与方法HL课程中，复数是核心主题之一，而其中最精彩的结论之一就是求复数n次方根的方法。本文将提供完整指南：极坐标形式、棣莫弗定理、n次方根公式的推导、典型例题、几何意义以及常见考试陷阱。
1. Review: Polar Form of Complex Numbers | 回顾：复数的极坐标形式
Any complex number z = a + bi can be written in polar form as z = r(cos θ + i sin θ), where r = |z| = √(a² + b²) is the modulus, and θ = arg(z) is the argument, the angle measured counterclockwise from the positive real axis.
任何复数 z = a + bi 都可以写成极坐标形式 z = r(cos θ + i sin θ)，其中 r = |z| = √(a² + b²) 是模长，θ = arg(z) 是辐角，即从正实轴逆时针测量的角度。
The argument is not unique: adding any integer multiple of 2π gives the same complex number. The principal argument is usually chosen in the interval (-π, π] or [0, 2π), depending on convention. IB typically uses (-π, π].
辐角不唯一：加上任意整数倍的 2π 都表示同一个复数。主辐角通常取在区间 (-π, π] 或 [0, 2π) 内，具体看约定。IB课程通常使用 (-π, π]。
Using Euler’s formula, the same number can be written compactly as z = re^(iθ), which is extremely useful when multiplying, dividing, and raising complex numbers to powers.
借助欧拉公式，同一复数可简洁地写成 z = re^(iθ)，这在复数的乘法、除法以及幂运算中极为方便。
2. De Moivre’s Theorem | 棣莫弗定理
De Moivre’s Theorem is the engine that powers root extraction. It states that for any integer n and any real angle θ:
棣莫弗定理是复数的开方运算的核心工具。它指出：对任意整数 n 和任意实数角度 θ：
(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)
Equivalently, in exponential form: (e^(iθ))ⁿ = e^(i nθ). This theorem holds for all integers n, and it is the reason that multiplying complex numbers corresponds to adding their arguments while raising to a power corresponds to multiplying the argument.
等价地，在指数形式下可写为：(e^(iθ))ⁿ = e^(i nθ)。该定理对所有整数 n 均成立，也正是因为这一性质，复数的乘法对应辐角相加，而幂运算对应辐角相乘。
For example, (cos 20° + i sin 20°)⁶ = cos 120° + i sin 120° = -½ + i√3⁄2. This theorem is the bridge between powers of complex numbers and their roots.
例如，(cos 20° + i sin 20°)⁶ = cos 120° + i sin 120° = -½ + i√3⁄2。棣莫弗定理正是连接复数的幂与复数的方根之间的桥梁。
3. Deriving the nth Root Formula | 推导n次方根公式
Suppose we are given a complex number w = r(cos θ + i sin θ) and we wish to solve zⁿ = w, i.e. find all n-th roots of w.
假设给定复数 w = r(cos θ + i sin θ)，我们希望求解 zⁿ = w，即求 w 的全部n次方根。
Write the unknown root as z = ρ(cos φ + i sin φ), where ρ = |z| and φ = arg(z). By De Moivre’s Theorem:
将未知根写成 z = ρ(cos φ + i sin φ)，其中 ρ = |z|，φ = arg(z)。根据棣莫弗定理：
zⁿ = ρⁿ (cos(nφ) + i sin(nφ)) = r(cos θ + i sin θ)
Matching moduli gives ρⁿ = r, so ρ = r^(1/n), the positive real n-th root of r. Matching arguments requires that nφ and θ differ by an integer multiple of 2π:
比较模长可得 ρⁿ = r，因此 ρ = r^(1/n)，即 r 的正实数n次方根。比较辐角则要求 nφ 与 θ 相差整数倍的 2π：
nφ = θ + 2kπ ⇒ φ = (θ + 2kπ)/n, k = 0, ±1, ±2, …
Thus the n-th roots of w are given by the formula:
因此 w 的n次方根由如下公式给出：
zₖ = r1/n [cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)]
Taking k = 0, 1, 2, …, n – 1 produces n distinct roots; taking any other integer k simply repeats one of these n roots, because the roots are 2π/n apart in angle.
取 k = 0, 1, 2, …, n – 1 恰好得到 n 个不同的根；取其他任何整数 k 只会重复这 n 个根，因为它们之间的角度间隔为 2π/n。
An equivalent compact form uses the cis notation: zₖ = r1/n cis((θ + 2kπ)/n). In exponential form: zₖ = r1/n e^(i(θ + 2kπ)/n).
等价的简洁形式可用 cis 记号：zₖ = r1/n cis((θ + 2kπ)/n)。指数形式为：zₖ = r1/n e^(i(θ + 2kπ)/n)。
4. Special Case: Roots of Unity | 特殊情况：单位根
When w = 1, we have r = 1 and θ = 0, so the n-th roots of unity are:
当 w = 1 时，r = 1，θ = 0，因此n次单位根为：
zₖ = cos(2kπ/n) + i sin(2kπ/n) = cis(2kπ/n), k = 0, 1, …, n – 1
For example, the cube roots of unity are z₀ = 1, z₁ = -½ + i√3⁄2, and z₂ = -½ – i√3⁄2. A key property is that the sum of all n-th roots of unity is zero for n ≥ 2:
例如，三次单位根为 z₀ = 1，z₁ = -½ + i√3⁄2，z₂ = -½ – i√3⁄2。一个重要性质是：当 n ≥ 2 时，所有n次单位根之和为零：
1 + ω + ω² + … + ωⁿ⁻¹ = 0
where ω = cis(2π/n). This property is frequently tested in IB exam questions involving polynomial equations and geometric series.
其中 ω = cis(2π/n)。这一性质在IB考试中经常与多项式方程和等比数列结合考查。
5. Worked Example: Cube Roots of 8 | 例题：求8的立方根
Find all cube roots of 8, giving answers in Cartesian form a + bi.
求 8 的所有立方根，并以代数形式 a + bi 作答。
Step 1: Write 8 in polar form. Since 8 is a positive real number, r = 8 and θ = 0, so 8 = 8(cos 0 + i sin 0).
步骤1：将 8 写成极坐标形式。由于 8 是正实数，r = 8，θ = 0，故 8 = 8(cos 0 + i sin 0)。
Step 2: Apply the formula with n = 3. The modulus of each root is ρ = 8^(1/3) = 2, and the angles are φₖ = (0 + 2kπ)/3 = 2kπ/3 for k = 0, 1, 2.
步骤2：代入 n = 3 的公式。每个根的模长为 ρ = 8^(1/3) = 2，辐角为 φₖ = (0 + 2kπ)/3 = 2kπ/3，其中 k = 0, 1, 2。
Step 3: Compute each root.
步骤3：逐一计算每个根。
| k | φₖ = 2kπ/3 | zₖ = 2 cis(2kπ/3) |
| 0 | 0 | 2 |
| 1 | 2π/3 | 2(-½ + i√3⁄2) = -1 + i√3 |
| 2 | 4π/3 | 2(-½ – i√3⁄2) = -1 – i√3 |
Check: (-1 + i√3)³ = 8? Expanding (or using De Moivre’s Theorem) confirms each root satisfies z³ = 8. The three roots form an equilateral triangle on the circle of radius 2.
验证：(-1 + i√3)³ = 8 吗？通过展开（或棣莫弗定理）可确认每个根都满足 z³ = 8。这三个根在半径为 2 的圆上构成一个等边三角形。
6. Worked Example: Fourth Roots of -16 | 例题：求-16的四次方根
Find all solutions to z⁴ = -16.
求方程 z⁴ = -16 的全部解。
Step 1: Polar form. The modulus is 16 and the principal argument is π, so -16 = 16(cos π + i sin π).
步骤1：极坐标形式。模长为 16，主辐角为 π，故 -16 = 16(cos π + i sin π)。
Step 2: Apply the formula with n = 4. Each root has modulus ρ = 16^(1/4) = 2, and angles:
步骤2：代入 n = 4 的公式。每个根的模长为 ρ = 16^(1/4) = 2，辐角为：
φₖ = (π + 2kπ)/4 = π/4 + kπ/2, k = 0, 1, 2, 3
Step 3: List the roots.
步骤3：列出所有根。
- k = 0: z₀ = 2(cos π/4 + i sin π/4) = 2(√2⁄2 + i√2⁄2) = √2 + i√2
- k = 1: z₁ = 2(cos 3π/4 + i sin 3π/4) = 2(-√2⁄2 + i√2⁄2) = -√2 + i√2
- k = 2: z₂ = 2(cos 5π/4 + i sin 5π/4) = 2(-√2⁄2 – i√2⁄2) = -√2 – i√2
- k = 3: z₃ = 2(cos 7π/4 + i sin 7π/4) = 2(√2⁄2 – i√2⁄2) = √2 – i√2
These four roots are equally spaced on the circle of radius 2 at angles π/4, 3π/4, 5π/4, and 7π/4, forming a square centered at the origin.
这四个根在半径 2 的圆上以 π/4、3π/4、5π/4、7π/4 的角度等
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