The nth Roots of Unity | n 次单位根

📚 The nth Roots of Unity | n 次单位根

The nth roots of unity are one of the central ideas in the complex numbers topic of AQA A-Level Mathematics. They connect algebraic equations with geometric symmetry, and they appear in factorisation, trigonometry, and polynomial problems. This article explains how to find, interpret and use these roots step by step.

n 次单位根是 AQA A-Level 数学复数部分的核心概念之一。它们将代数方程与几何对称联系起来，并出现在因式分解、三角学和多项式问题中。本文将逐步解释如何求解、解释和运用这些根。

1. What Are the nth Roots of Unity? | 什么是 n 次单位根

eger n, the nth roots of unity are the complex numbers z that satisfy the equation zⁿ = 1.

对于正整数 n，n 次单位根是满足方程 zⁿ = 1 的复数 z。

y n roots when counted with multiplicity. In the case of zⁿ = 1, all n roots are distinct.

由于这是 n 次多项式方程，代数基本定理保证按重数计算恰好有 n 个根。在 zⁿ = 1 的情形中，所有 n 个根都互不相同。

For example, when n = 2 the roots are z = 1 and z = −1, since 1² = 1 and (−1)² = 1.

例如，当 n = 2 时，根为 z = 1 和 z = −1，因为 1² = 1 且 (−1)² = 1。

2. Solving zⁿ = 1 Using de Moivre's Theorem | 用棣莫弗定理求解 zⁿ = 1

side 1 in exponential polar form. Since cosine and sine have period 2π, we can write 1 = e^2πik for any integer k.

最可靠的方法是将右边的 1 写成指数极坐标形式。由于余弦和正弦的周期为 2π，因此对任意整数 k，可以写成 1 = e^2πik。

zⁿ = e^2πik ⇒ z = e^2πik/n

Taking the nth root gives z = e^2πik/n. To obtain distinct solutions, we choose k = 0, 1, 2, ..., n − 1.

开 n 次方得到 z = e^2πik/n。为了得到互不相同的解，我们取 k = 0, 1, 2, ..., n − 1。

Using Euler's formula, each root can also be written as z = cos(2πk/n) + i sin(2πk/n).

利用欧拉公式，每个根也可以写成 z = cos(2πk/n) + i sin(2πk/n)。

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3. The Standard Formula and Notation | 标准公式与记号

Let ω (omega) denote the principal nth root of unity, defined by ω = e^2πi/n. Then the complete set of nth roots of unity is {1, ω, ω², ..., ωⁿ⁻¹}.

令 ω（欧米伽）表示主 n 次单位根，定义为 ω = e^2πi/n。那么 n 次单位根的完整集合为 {1, ω, ω², ..., ωⁿ⁻¹}。

This notation is useful because every root is a power of one number, and multiplication of roots corresponds to adding exponents.

这种记号很有用，因为每个根都是一个数的幂，并且根的乘法对应于指数相加。

Since ωⁿ = e^2πi = 1, the powers repeat cyclically: ωⁿ = 1, ωⁿ⁺¹ = ω, and so on.

因为 ωⁿ = e^2πi = 1，这些幂循环重复：ωⁿ = 1，ωⁿ⁺¹ = ω，依此类推。

4. Geometric Interpretation: Regular n-gon | 几何解释：正 n 边形

gand diagram, all nth roots of unity lie on the unit circle centred at the origin, because |z| = 1 for every root.

在阿尔冈图上，所有 n 次单位根都位于以原点为圆心的单位圆上，因为每个根的模 |z| = 1。

of 2π/n radians. Joining them in order produces a regular n-gon with one vertex at z = 1.Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

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