📚 The nth Roots of Unity | 复数单位根
The nth roots of unity are the complex numbers that satisfy the equation zn = 1. They form a beautiful link between algebra, geometry and trigonometry, and are a core topic in AQA A-level further mathematics.
复数单位根是满足方程 zn = 1 的复数。它们将代数、几何与三角学优美地联系起来，是 AQA A-level 进阶数学的核心内容。
1. Definition | 定义
For a positive integer n, an nth root of unity is any complex number z such that zn = 1. Because the modulus of 1 is 1, every such root must have modulus 1.
对于正整数 n，若复数 z 满足 zn = 1，则称 z 为 n 次单位根。由于 1 的模为 1，每个这样的根都必须满足模为 1。
By the Fundamental Theorem of Algebra, the equation zn = 1 has exactly n complex roots, counting multiplicity. These roots are distinct for positive integer n.
根据代数基本定理，方程 zn = 1 恰好有 n 个复数根（按重数计）。当 n 为正整数时，这些根互不相同。
Each root lies on the unit circle in the Argand diagram, because the product of a complex number and its conjugate gives its squared modulus.
每个根都位于阿甘图（复平面）中的单位圆上，因为复数与其共轭相乘得到其模的平方。
2. Solving zn = 1 | 解方程 zn = 1
We write z in modulus-argument form: z = r(cos θ + i sin θ). Since zn = rn(cos nθ + i sin nθ) by de Moivre’s theorem, we require rn = 1 and nθ ≡ 0 (mod 2π).
我们将 z 写成模辐角形式：z = r(cos θ + i sin θ)。根据棣莫弗定理，zn = rn(cos nθ + i sin nθ)，因此需要 rn = 1 且 nθ ≡ 0 (mod 2π)。
Since r is a positive real number, r = 1. The argument equation gives θ = 2kπ/n for k = 0, 1, …, n−1.
因为 r 为正实数，所以 r = 1。辐角方程给出 θ = 2kπ/n，其中 k = 0, 1, …, n−1。
z = cos(2kπ/n) + i sin(2kπ/n) = e2kπi/n, k = 0, 1, …, n−1
This explicit formula is the basis of every problem involving roots of unity.
这个显式公式是处理一切单位根问题的基础。
3. Unit Circle and Regular Polygons | 单位圆与正多边形
The n roots of unity all have modulus 1, so they lie on the unit circle with equally spaced arguments 2π/n apart.
n 次单位根的模都为 1，因此它们位于单位圆上，相邻辐角间隔均为 2π/n。
Connecting the points in order produces a regular n-gon inscribed in the unit circle. One vertex is always at (1, 0).
按顺序连接这些点，就得到单位圆的内接正 n 边形，其中一个顶点始终在 (1, 0) 处。
For example, the fourth roots of unity are 1, i, −1, −i, which form a square. The sixth roots form a regular hexagon.
例如，四次单位根为 1, i, −1, −i，构成一个正方形；六次单位根构成正六边形。
This geometric view helps you remember the roots quickly, instead of memorising separate values.
这种几何观点能帮助你快速记住各根，而不必单独记忆每个数值。
4. Sum and Product of the Roots | 根的和与积
For the equation zn = 1, the coefficient of zn−1 is zero. By Vieta’s formulas, the sum of all n roots is 0.
对方程 zn = 1，zn−1 的系数为 0。由韦达定理，所有 n 个根之和为 0。
The product of the roots is (−1)n+1 times the constant term. For zn − 1 = 0, the product equals (−1)n+1(−1) = (−1)n−1.
根的乘积等于常数项乘以 (−1)n+1。对于 zn − 1 = 0，乘积等于 (−1)n+1(−1) = (−1)n−1。
Sum = 0, Product = (−1)n−1
When n is odd, the product is 1; when n is even, the product is −1. These facts are often tested in AQA exam questions.
当 n 为奇数时，乘积为 1；当 n 为偶数时，乘积为 −1。这些结论在 AQA 考试中常被考查。
5. Primitive Roots of Unity | 本原单位根
An nth root of unity ω is called primitive if ωk ≠ 1 for every positive integer k < n. In other words, no smaller power of ω equals 1.
若 n 次单位根 ω 满足对所有小于 n 的正整数 k 都有 ωk ≠ 1，则称 ω 为 n 次本原单位根。换句话说，ω 的任何更低次幂都不等于 1。
Using the exponential form ω = e2kπi/n, the root is primitive exactly when gcd(k, n) = 1.
利用指数形式 ω = e2kπi/n，该根是本原的当且仅当 gcd(k, n) = 1。
The number of primitive nth roots equals φ(n), where φ is Euler’s totient function. For example, when n = 6, the primitive roots are eπi/3 and e5πi/3, since 1 and −1 satisfy x2 = 1 or x3 = 1 already.
本原 n 次单位根的个数等于 φ(n)，其中 φ 是欧拉函数。例如 n = 6 时，本原根为 eπi/3 和 e5πi/3，因为 1 和 −1 已经满足 x2 = 1 或 x3 = 1。
6. Conjugate Pairs and Real Coefficients | 共轭对与实系数
If z is a root of zn = 1, then its complex conjugate z̄ is also a root, because the coefficients of zn − 1 are real.
若 z 是 zn = 1 的一个根，则其共轭复数 z̄ 也是根，因为 zn − 1 的系数为实数。
For n > 2, roots come in conjugate pairs except for real roots 1 and possibly −1. This helps in factorising real polynomials.
当 n > 2 时，除实根 1 以及可能的 −1 外，其余根总是成对共轭出现。这有助于分解实系数多项式。
For n even: roots are 1, −1 and (n−2)/2 conjugate pairs; for n odd: roots are 1 and (n−1)/2 conjugate pairs
For n even, the non-real roots pair up as e2kπi/n and e−2kπi/n = e2(n−k)πi/n.
当 n 为偶数时，非实根按 e2kπi/n 与 e−2kπi/n = e2(n−k)πi/n 成对出现。
7. Cube Roots of Unity | 三次单位根
The cube roots of unity are the solutions of z3 = 1. By the formula, they are 1, cos(2π/3) + i sin(2π/3), and cos(4π/3) + i sin(4π/3).
三次单位根是方程 z3 = 1 的解。由公式，它们是 1、cos(2π/3) + i sin(2π/3)、cos(4π/3) + i sin(4π/3)。
The second root is usually denoted ω; then the third is ω² because ω³ = 1 and ω ≠ 1.
第二个根通常记为 ω，第三个根为 ω²，因为 ω³ = 1 且 ω ≠ 1。
ω = −½ + (√3/2)i, ω² = −½ − (√3/2)i
Key properties are ω³ = 1, 1 + ω + ω² = 0, and ω̄ = ω². These are used to simplify algebraic expressions.
关键性质为 ω³ = 1、1 + ω + ω² = 0，以及 ω̄ = ω²。这些性质可用于化简代数式。
For example, (1 + ω − ω²)³ = (−2ω²)³ = −8ω⁶ = −8.
例如，(1 + ω − ω²)³ = (−2ω²)³ = −8ω⁶ = −8。
8. Fourth and Sixth Roots | 四次与六次单位根
The fourth roots of unity are the solutions of z⁴ = 1: they are 1, i, −1, −i. Their sum is 0 and their product is −1.
四次单位根是方程 z⁴ = 1 的解：它们是 1, i, −1, −i。它们的和为 0，乘积为 −1。
The sixth roots are the angles 0, π/3, 2π/3, π, 4π/3, 5π/3 on the unit circle. In exponential form they are ekπi/3 for k = 0, …, 5.
六次单位根对应单位圆上的辐角 0, π/3, 2π/3, π, 4π/3, 5π/3。用指数形式写作 ekπi/3，k = 0, …, 5。
Notice that the fourth roots are a subset of the eighth roots, and the cube roots are a subset of the sixth roots. In general, if d divides n, the dth roots are included among the nth roots.
注意四次根是八次根的子集，三次根是六次根的子集。一般地，若 d 整除 n，则 d 次单位根都是 n 次单位根的子集。
This subset relationship follows because if zd = 1, then zn = (zd)n/d = 1.
这个子集关系是因为若 zd = 1，则 zn = (zd)n/d = 1。
9. Factorising Polynomials | 因式分解
Knowing the roots of unity allows us to factorise zn − 1 over different number systems.
知道单位根使我们能在不同数系中分解 zn − 1。
Over the complex numbers, zn − 1 = (z − 1)(z − ω)(z − ω²)⋯(z − ωn−1), where ω = e2πi/n.
在复数范围内，zn − 1 = (z − 1)(z − ω)(z − ω²)⋯(z − ωn−1)，其中 ω = e2πi/n。
Over the real numbers, non-real conjugate pairs combine into quadratic factors. For example, z⁴ − 1 = (z − 1)(z + 1)(z² + 1).
在实数范围内，共轭非实根对合并为二次因式。例如，z⁴ − 1 = (z − 1)(z + 1)(z² + 1)。
z⁶ − 1 = (z − 1)(z + 1)(z² + z + 1)(z² − z + 1)
This factorisation follows by grouping conjugate roots: e±πi/3 and e±2πi/3.
该分解通过对共轭根 e±πi/3 和 e±2πi/3 配对而得到。
10. Geometric Transformations | 几何变换
Multiplication by an nth root of unity rotates the complex plane by 2π/n. Repeated multiplication cycles through all the roots.
乘以 n 次单位根相当于将复平面旋转 2π/n。不断重复乘法会在所有根之间循环。
If z is any non-zero complex number, then z, zω, zω², …, zωn−1 form the vertices of a regular n-gon centred at the origin.
若 z 是任意非零复数，则 z, zω, zω², …, zωn−1 构成以原点为中心的正 n 边形的顶点。
This idea is used in problems about roots of polynomial equations: if a polynomial has coefficients that are invariant under rotation, its roots often include roots of unity.
这个思想常用于多项式方程根的问题：若多项式的系数在旋转下保持不变，其根常包含单位根。
For instance, the roots of z⁵ = 32 are five equally spaced points on the circle radius 2: 2e2kπi/5.
例如，z⁵ = 32 的根是半径为 2 的圆上五个等距点：2e2kπi/5。
11. Exam Pitfalls and Key Facts | 考试易错点与关键事实
Common mistake: forgetting that 1 is always a root. The root for k = 0 is 1, never omit it.
常见错误：忘记 1 始终是一个根。k = 0 对应的根就是 1，不要遗漏。
Common mistake: using θ = 2kπ/n instead of 2kπ/n in degrees? Always use radians in A-level complex number work.
常见错误：误用 θ = 2kπ/n 时忘记本题使用弧度制。A-level 复数运算中一律使用弧度。
Key facts: the sum of all roots is 0; the product is (−1)n−1; every primitive root generates all roots as powers.
关键事实：所有根之和为 0；乘积为 (−1)n−1；每个本原根的幂可以生成所有根。
Also remember: if ω is a primitive nth root, then the full set of roots is 1, ω, ω², …, ωn−1.
还要记住：若 ω 是 n 次本原单位根，则全部根为 1, ω, ω², …, ωn−1。
12. Conclusion | 总结
The nth roots of unity tie together complex numbers, trigonometry and geometry. Mastering the formula z = e2kπi/n and the two key sums is essential.
复数单位根将复数、三角学与几何紧密联系起来。掌握公式 z = e2kπi/n 以及两个关键求和至关重要。
Practice sketching regular polygons, computing primitive roots, and factorising zn − 1 to build confidence for AQA exam questions.
通过练习绘制正多边形、计算本原根以及分解 zn − 1，可以为 AQA 考试建立信心。
With these tools, problems that initially look abstract become quick, visual and predictable.
掌握这些工具后，原本抽象的题目会变得快速、直观且可预测。
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