The Cube Roots of Unity | 单位立方根

📚 The Cube Roots of Unity | 单位立方根

The cube roots of unity are the three solutions of the equation z³ = 1 in the complex number system. This topic appears frequently in AQA A-Level Mathematics, especially in pure mathematics and further mathematics papers, where students are expected to manipulate ω and use its elegant properties to simplify expressions.

单位立方根是方程 z³ = 1 在复数范围内的三个解。这个主题在 AQA A-Level 数学中频繁出现，尤其是在纯数学和高等数学试卷中，学生需要熟练运用 ω 及其优美性质来化简表达式。

1. Solving z³ = 1 | 解方程 z³ = 1

We begin by solving the equation z³ = 1. One obvious solution is z = 1. To find the other two, we factorise or use De Moivre’s theorem. Since z³ – 1 = 0, we can write (z – 1)(z² + z + 1) = 0.

我们从解方程 z³ = 1 开始。一个显然的解是 z = 1。为了找到另外两个解，我们可以因式分解或使用棣莫弗定理。因为 z³ – 1 = 0，所以可以写成 (z – 1)(z² + z + 1) = 0。

Setting z – 1 = 0 gives z = 1. Solving z² + z + 1 = 0 using the quadratic formula yields the other two roots.

令 z – 1 = 0 得 z = 1。用二次方程求根公式解 z² + z + 1 = 0，即可得到另外两个根。

z = (-1 ± √(-3)) / 2 = -1/2 ± (√3/2)i

Thus the three roots are 1, -1/2 + (√3/2)i, and -1/2 – (√3/2)i.

因此三个根是 1、-1/2 + (√3/2)i 和 -1/2 – (√3/2)i。

2. The notation ω and ω² | 记号 ω 与 ω²

By convention, the non-real cube roots of unity are denoted by ω and ω². We define ω = -1/2 + (√3/2)i, so that ω² = -1/2 – (√3/2)i. The other choice simply swaps ω and ω².

按惯例，单位立方根中的非实数根用 ω 和 ω² 表示。我们定义 ω = -1/2 + (√3/2)i，那么 ω² = -1/2 – (√3/2)i。两种选择本质上只是交换 ω 和 ω²。

In exponential form, ω = e^(2πi/3) and ω² = e^(4πi/3). This comes from De Moivre’s theorem applied to z³ = cos(2π) + i sin(2π).

在指数形式下，ω = e^(2πi/3)，ω² = e^(4πi/3)。这是将棣莫弗定理应用于 z³ = cos(2π) + i sin(2π) 得到的。

Note that ω³ = 1. This property is the cornerstone of all calculations involving ω.

注意 ω³ = 1。这个性质是所有涉及 ω 运算的基石。

3. Key algebraic properties | 关键代数性质

The most important properties of ω are:

ω 最重要的性质是：

ω³ = 1
1 + ω + ω² = 0
ω² = ω̄ (the complex conjugate of ω)
ω⁻¹ = ω², since ω · ω² = ω³ = 1

The second property follows from the fact that the sum of all roots of z³ – 1 = 0 is zero (by Vieta’s formulas).

第二个性质可由韦达定理得出：方程 z³ – 1 = 0 的所有根之和为零。

These properties allow us to replace ω² with -1 – ω, and to reduce any power of ω modulo 3.

这些性质允许我们将 ω² 替换为 -1 – ω，并将 ω 的任意幂次按模 3 化简。

4. Geometric interpretation | 几何解释

The three cube roots of unity are equally spaced on the unit circle in the Argand plane, separated by angles of 120°.

在复平面中，单位立方根均匀分布在单位圆上，彼此相隔 120°。

1 = e^(0i), ω = e^(2πi/3), ω² = e^(4πi/3)

These points form the vertices of an equilateral triangle inscribed in the unit circle. The centroid of this triangle is the origin, which reflects the fact that 1 + ω + ω² = 0.

这三个点构成单位圆内接等边三角形的三个顶点。三角形的重心是原点，这正反映了 1 + ω + ω² = 0。

Multiplying any complex number by ω represents a rotation by 120° anticlockwise. Repeated multiplication cycles through the cube roots.

任意复数乘以 ω 相当于逆时针旋转 120°。重复相乘会在三个单位立方根之间循环。

5. Factorisation of z³ – 1 | z³ – 1 的因式分解

Using the roots, we can factorise z³ – 1 over the complex numbers as:

利用这些根，我们可以在复数范围内将 z³ – 1 因式分解为：

z³ – 1 = (z – 1)(z – ω)(z – ω²)

Over the reals, the quadratic factor z² + z + 1 is irreducible, so the real factorisation is:

在实数范围内，二次因子 z² + z + 1 不可约，因此实数因式分解为：

z³ – 1 = (z – 1)(z² + z + 1)

Similarly, z³ + 1 = (z + 1)(z² – z + 1), with roots -1, e^(πi/3), e^(5πi/3). However, the cube roots of unity specifically refer to z³ = 1.

类似地，z³ + 1 = (z + 1)(z² – z + 1)，其根为 -1、e^(πi/3)、e^(5πi/3)。但“单位立方根”特指 z³ = 1 的根。

6. Reducing powers of ω | 化简 ω 的幂次

Since ω³ = 1, any integer power of ω can be reduced using the remainder upon division by 3.

因为 ω³ = 1，所以 ω 的任意整数幂都可以用除以 3 的余数来化简。

For example:

例如：

ω⁴ = ω³ · ω = ω
ω⁵ = ω³ · ω² = ω²
ω⁷ = ω⁶ · ω = ω
ω⁻² = ω / ω³ = ω

In general, ωⁿ = ω^(n mod 3), where the remainder is 0, 1, or 2. A remainder of 0 gives ω⁰ = 1.

一般而言，ωⁿ = ω^(n mod 3)，其中余数为 0、1 或 2。余数为 0 时得到 ω⁰ = 1。

This reduction is essential in sequences, series, and arithmetic progressions involving complex numbers.

这种化简在处理涉及复数的数列、级数和等差问题中至关重要。

7. Sum and product identities | 求和与乘积恒等式

From the properties of ω, we can derive several useful identities:

根据 ω 的性质，我们可以推导出几个有用的恒等式：

Identity	Result
1 + ω + ω²	0
1 · ω · ω²	ω³ = 1
(1 + ω)(1 + ω²)	1
(1 – ω)(1 – ω²)	3

For example, (1 + ω)(1 + ω²) = 1 + ω + ω² + ω³ = 0 + 1 = 1.

例如，(1 + ω)(1 + ω²) = 1 + ω + ω² + ω³ = 0 + 1 = 1。

These identities often appear in exam questions requiring simplification of algebraic expressions.

这些恒等式常出现在考试题目中，用于化简代数表达式。

8. Applications to polynomial equations | 在多项式方程中的应用

The cube roots of unity are useful for solving and factorising polynomials that contain z² + z + 1 as a factor.

单位立方根对于求解和分解含有 z² + z + 1 因子的多项式非常有用。

Consider the polynomial P(z) = z⁴ + z² + 1. We can factorise it as:

考虑多项式 P(z) = z⁴ + z² + 1。我们可以将其因式分解为：

z⁴ + z² + 1 = (z² + z + 1)(z² – z + 1)

Its roots are solutions of z² + z + 1 = 0 and z² – z + 1 = 0, giving ω, ω², e^(πi/3), e^(5πi/3).

它的根是 z² + z + 1 = 0 和 z² – z + 1 = 0 的解，即 ω、ω²、e^(πi/3)、e^(5πi/3)。

Another common example: show that (x + y)(x + ωy)(x + ω²y) = x³ + y³. Expanding the factors gives the result directly, a formula sometimes tested in further algebra questions.

另一个常见例子：证明 (x + y)(x + ωy)(x + ω²y) = x³ + y³。展开因子即可直接得到结果，这是一个有时会在代数进阶题中考查的公式。

9. Relation to trigonometric identities | 与三角恒等式的关系

Since ω = cos(2π/3) + i sin(2π/3), we can use it to derive trigonometric identities for angles that are multiples of 2π/3.

由于 ω = cos(2π/3) + i sin(2π/3)，我们可以利用它来推导与 2π/3 的倍数有关的三角恒等式。

For instance, ω + ω² = -1 yields:

例如，ω + ω² = -1 给出：

2 cos(2π/3) = -1, so cos(2π/3) = -1/2

Similarly, using ω – ω² = i√3 gives sin(2π/3) = √3/2.

类似地，ω – ω² = i√3 得到 sin(2π/3) = √3/2。

More advanced applications involve evaluating sums like cos(2πk/3) + cos(4πk/3) using the real parts of ωᵏ + ω²ᵏ.

更高级的应用包括利用 ωᵏ + ω²ᵏ 的实部来求 cos(2πk/3) + cos(4πk/3) 这类和的值。

10. Relationship between ω and quadratic roots | ω 与二次方程根的关系

The quadratic z² + z + 1 = 0 has discriminant Δ = 1 – 4 = -3, so its roots are exactly (-1 ± i√3)/2.

二次方程 z² + z + 1 = 0 的判别式为 Δ = 1 – 4 = -3，所以它的根正是 (-1 ± i√3)/2。

This means that ω is a primitive cube root of unity. It is called “primitive” because ω ≠ 1 and ω³ = 1, and no smaller positive integer power of ω equals 1.

这意味着 ω 是本原单位立方根。称为“本原”是因为 ω ≠ 1 且 ω³ = 1，并且不存在比 3 更小的正整数幂使 ω 等于 1。

The other non-real root, ω², is also primitive. 1 itself is a non-primitive root because 1¹ = 1.

另一个非实数根 ω² 也是本原的。1 本身是非本原根，因为 1¹ = 1。

11. Extension to higher roots of unity | 推广至更高次单位根

The ideas above generalise to the nth roots of unity, defined as solutions of zⁿ = 1. They are given by e^(2πik/n) for k = 0, 1, …, n-1.

以上思想可以推广到 n 次单位根，即 zⁿ = 1 的解。它们是 e^(2πik/n)，其中 k = 0, 1, …, n-1。

In general, the sum of all nth roots of unity is zero for n > 1, and their product is (-1)ⁿ⁺¹.

一般来说，当 n > 1 时，所有 n 次单位根之和为零，它们的乘积为 (-1)ⁿ⁺¹。

The cube roots form the simplest non-trivial example and are often used as a stepping stone to understanding cyclotomic polynomials.

立方根是最简单的非平凡例子，常被用作理解分圆多项式的跳板。

12. Summary and exam tips | 总结与考试提示

In summary, the cube roots of unity are the three complex numbers 1, ω, and ω², with ω = -1/2 + (√3/2)i. The key facts are:

总结：单位立方根是三个复数 1、ω 和 ω²，其中 ω = -1/2 + (√3/2)i。关键知识点是：

ω³ = 1 and ω² + ω + 1 = 0
ω² = ω̄ and 1/ω = ω²
Any power of ω reduces to 1, ω, or ω²
The roots form an equilateral triangle in the Argand plane

When solving AQA exam questions, always check whether you can replace ω² with -1 – ω to simplify a quadratic expression. Also remember that ω and ω² are interchangeable in most problems.

在解答 AQA 考试题时，始终检查是否可以用 -1 – ω 替换 ω² 来化简二次表达式。同时记住，在大多数问题中 ω 和 ω² 是对称的，可以互换。

Practice with past papers: typical questions ask you to evaluate expressions like (1 + ω)ⁿ or to show that a given polynomial is divisible by z² + z + 1. Master these, and the cube roots of unity become a reliable source of marks.

多练习历年真题：典型题目要求计算 (1 + ω)ⁿ 或证明某多项式能被 z² + z + 1 整除。掌握这些，单位立方根就会成为可靠的得分点。

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The Cube Roots of Unity | 单位立方根