The Cube Roots of Unity | 单位立方根

📚 The Cube Roots of Unity | 单位立方根

al for AQA A Level Further Maths, especially when studying roots of unity, polynomial factorisation, and complex algebra.

在复数学习中，单位立方根是方程 z³ = 1 的三个解。它们是 AQA A Level 进阶数学中学习单位根、多项式因式分解和复数代数的基础内容。

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1. What Are Cube Roots of Unity? | 什么是单位立方根？

plex number with modulus 1 and argument 0, there are exactly three distinct cube roots in the complex plane.

单位立方根是指满足 z³ = 1 的复数 z。由于 1 是模为 1、辐角为 0 的复数，因此复数平面内恰好有三个不同的立方根。

The term "unity" simply means the number 1. We say "cube roots of unity" rather than "cube roots of 1" to emphasise that we are working in the complex number system, not just with real numbers.

"单位" 一词指的就是数字 1。我们使用 "单位立方根" 而不是 "1 的立方根"，是为了强调我们在复数系中求解，而不仅仅在实数范围内。

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2. Solving z³ = 1 | 解方程 z³ = 1

3θ). Equating this to 1 = 1(cos 0 + i sin 0) gives the modulus equation r³ = 1, so r = 1.

为了解 z³ = 1，我们将 z 写成模-辐角形式。设 z = r(cos θ + i sin θ)。则 z³ = r³(cos 3θ + i sin 3θ)。令其等于 1 = 1(cos 0 + i sin 0)，得到模方程 r³ = 1，因此 r = 1。

πk/3. The distinct values occur for k = 0, 1 and 2, giving three angles: 0, 2π/3 and 4π/3.

对于辐角，我们需要 3θ = 0 + 2πk，其中 k 为整数。因此 θ = 2πk/3。当 k = 0、1、2 时得到三个不同的辐角：0、2π/3 和 4π/3。

z³ = 1 ⇒ r³(cos 3θ + i sin 3θ) = 1(cos 0 + i sin 0)

z³ = 1 ⇒ r³(cos 3θ + i sin 3θ) = 1(cos 0 + i sin 0)

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3. The Three Roots | 三个根

Substituting k = 0, 1 and 2 into z = cos θ + i sin θ gives the three cube roots of unity:

将 k = 0、1、2 代入 z = cos θ + i sin θ，得到三个单位立方根：

• k = 0: z = cos 0 + i sin 0 = 1
• k = 1: z = cos(2π/3) + i sin(2π/3) = −½ + (√3/2)i
• k = 2: z = cos(4π/3) + i sin(4π/3) = −½ − (√3/2)i

The three roots are often displayed as a table:

三个根通常用表格表示：

k	Angle θ	Polar form	Rectangular form
0	0	cos 0 + i sin 0	1
1	2π/3	cos(2π/3) + i sin(2π/3)	−½ + (√3/2)i
2	4π/3	cos(4π/3) + i sin(4π/3)	−½ − (√3/2)i

Notice that only one root is real, namely 1. The other two roots are non-real complex conjugates.

注意只有一个根是实数，即 1。另外两个根是非实数的共轭复数。

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4. Introducing ω | 引入 ω

To simplify notation, we often let ω denote the non-real cube root with positive imaginary part:

为了简化记号，我们通常用 ω 表示虚部为正的非实数立方根：

ω = −½ + (√3/2)i

ω = −½ + (√3/2)i

Then the other non-real root is ω², because (ω)² = (−½ + (√3/2)i)² = −½ − (√3/2)i. Therefore the three cube roots of unity can be written simply as 1, ω and ω².

那么另一个非实数根就是 ω²，因为 (ω)² = (−½ + (√3/2)i)² = −½ − (√3/2)i。因此三个单位立方根可以简洁地写成 1、ω 和 ω²。

This compact notation is used throughout AQA exam questions, especially when simplifying complicated complex expressions.

这种简洁记法在 AQA 考试题中广泛使用，尤其是在化简复杂复数表达式时。

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5. Key Property: 1 + ω + ω² = 0 | 关键性质：1 + ω + ω² = 0

0, the polynomial can be factorised completely as (z − 1)(z − ω)(z − ω²). Expanding the right-hand side gives z³ − (1 + ω + ω²)z² + (ω + ω² + ω³)z − ω³. But this must equal z³ − 1, so the coefficient of z² is 0.

因为 1、ω、ω² 是 z³ − 1 = 0 的根，该多项式可以完全分解为 (z − 1)(z − ω)(z − ω²)。展开右边得到 z³ − (1 + ω + ω²)z² + (ω + ω² + ω³)z − ω³。但右边必须等于 z³ − 1，所以 z² 项的系数为 0。

Therefore we obtain the fundamental identity:

因此我们得到基本恒等式：

1 + ω + ω² = 0

1 + ω + ω² = 0

− ω or 1 + ω = −ω². You should memorise it because many exam simplifications depend on it.

这一关系极为有用。它可以变形为 ω² = −1 − ω 或 1 + ω = −ω²。你应该牢记这个公式，因为许多考试化简都依赖于它。

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6. Powers of ω | ω 的幂

From the definition, ω³ = 1. This means powers of ω cycle with period 3:

根据定义，ω³ = 1。这意味着 ω 的幂以周期 3 循环：

• ω⁰ = 1
• ω¹ = ω
• ω² = ω²
• ω³ = 1
• ω⁴ = ω
• ω⁵ = ω²
• ω⁶ = 1

ⁿ, divide n by 3 and use the remainder. For example, ω¹⁰ = ω⁹ × ω = (ω³)³ × ω = 1 × ω = ω.

一般来说，要化简 ωⁿ，可以将 n 除以 3 并利用余数。例如，ω¹⁰ = ω⁹ × ω = (ω³)³ × ω = 1 × ω = ω。

s ω² as −1 − ω. This identity is often the quickest way to replace ω² in a simplification.

此外，由 1 + ω + ω² = 0，我们可以将 ω² 表示为 −1 − ω。在化简中，这个恒等式通常是替换 ω² 的最快方法。

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7. Conjugate Relationship | 共轭关系

The two non-real cube roots ω and ω² are complex conjugates. That is, ω = −½ + (√3/2)i has conjugate ω² = −½ − (√3/2)i.

两个非实数立方根 ω 和 ω² 是共轭复数。也就是说，ω = −½ + (√3/2)i 的共轭是 ω² = −½ − (√3/2)i。

Because |ω| ual to 1/ω. Hence ω² = 1/ω. This gives a useful reciprocal formula: 1/ω = ω² and 1/ω² = ω.

由于 |ω| = 1，ω 的共轭也等于 1/ω。因此 ω² = 1/ω。这给出了有用的倒数公式：1/ω = ω²，1/ω² = ω。

In exam work, whenever you see 1/ω, it is often helpful to replace it immediately with ω², and vice versa.

在考试中，当你看到 1/ω 时，通常可以立即将其替换为 ω²，反之亦然。

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8. Geometric Interpretation | 几何解释

On an Argand diagram, the cube roots of unity all lie on the unit circle centred at the origin. Their arguments are 0, 2π/3 and 4π/3, so they are separated by equal angles of 120°.

在阿尔冈图上，单位立方根都位于以原点为圆心的单位圆上。它们的辐角分别为 0、2π/3 和 4π/3，因此彼此之间的夹角相等，均为 120°。

The three points form an equilateral triangle with one vertex at (1, 0). This is a special case of the general result: the n-th roots of unity are equally spaced around the unit circle, forming a regular n-gon.

这三个点构成一个等边三角形，其中一个顶点位于 (1, 0)。这是一般结论的特例：任意 n 次单位根都均匀分布在单位圆上，构成正 n 边形。

This geometric view helps you check answers: the cube roots must be symmetric about the real axis, and the two non-real roots must have arguments 120° and 240°.

这种几何视角有助于检查答案：单位立方根必须关于实轴对称，且两个非实数根的辐角必须为 120° 和 240°。

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9. Factorisation of z³ − 1 | z³ − 1 的因式分解

The expression z³ − 1 can be factorised as a difference of cubes:

表达式 z³ − 1 可以按立方差公式分解：

z³ − 1 = (z − 1)(z² + z + 1)

z³ − 1 = (z − 1)(z² + z + 1)

The quadratic factor z² + z + 1 = 0 has roots given by the quadratic formula:

二次因式 z² + z + 1 = 0 的根由求根公式给出：

z = (−1 ± √1 − 4)/2 = (−1 ± i√3)/2

z = (−1 ± √1 − 4)/2 = (−1 ± i√3)/2

These two values are exactly ω and ω². So the factorisation (z − 1)(z² + z + 1) is the same as (z − 1)(z − ω)(z − ω²).

这两个值正是 ω 和 ω²。因此因式分解 (z − 1)(z² + z + 1) 与 (z − 1)(z − ω)(z − ω²) 相同。

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10. Worked Example | 例题

Example: Simplify (1 + ω)(1 + ω²).

例题：化简 (1 + ω)(1 + ω²)。

Using the identity 1 + ω + ω² = 0, we have 1 + ω = −ω² and 1 + ω² = −ω. Therefore:

利用恒等式 1 + ω + ω² = 0，我们有 1 + ω = −ω² 和 1 + ω² = −ω。因此：

(1 + ω)(1 + ω²) = (−ω²)(−ω) = ω³ = 1

(1 + ω)(1 + ω²) = (−ω²)(−ω) = ω³ = 1

So the expression simplifies to 1. This type of manipulation is very common in AQA questions involving cube roots of unity.

因此该表达式化简为 1。此类操作在涉及单位立方根的 AQA 考题中非常常见。

Another example: Show that ω² + ω + 1 = 0 implies ω² = −1 − ω, and hence simplify ω² + 2ω + 3.

另一个例题：证明 ω² + ω + 1 = 0 可推出 ω² = −1 − ω，并由此化简 ω² + 2ω + 3。

3 = ω + 2. This is already in simplest form unless you substitute a numerical value for ω.

将 ω² = −1 − ω 代入表达式：ω² + 2ω + 3 = (−1 − ω) + 2ω + 3 = ω + 2。除非代入 ω 的数值，否则这已经是最简形式。

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11. Common Exam Pitfalls | 常见考试易错点

Here are the mistakes students often make with cube roots of unity:

以下是学生在单位立方根问题上常犯的错误：

• Forgetting that z³ = 1 has three roots, not just z = 1.
• Using the wrong angles: the non-real roots are at 120° and 240°, not 60° and 300°.
• Writing 1 + ω + ω² = 1 by confusing it with ω³ = 1.
• Thinking that ω² is the negative of ω, when in fact ω² is the complex conjugate of ω.
• Forgetting that powers of ω cycle every 3, so ω⁴ = ω, not some new value.

中文对照：

忘记 z³ = 1 有三个根，而不仅仅是 z = 1。

角度使用错误：非实数根位于 120° 和 240°，而不是 60° 和 300°。

将 1 + ω + ω² = 1 与 ω³ = 1 混淆。

误以为 ω² 是 ω 的相反数，而实际上 ω² 是 ω 的共轭复数。

忘记 ω 的幂每 3 次循环一次，所以 ω⁴ = ω，而不是一个新值。

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12. Summary | 总结

The cube roots of unity are 1, ω and ω², where ω = −½ + (√3/2)i. They satisfy ω³ = 1 and the crucial identity 1 + ω + ω² = 0.

单位立方根是 1、ω 和 ω²，其中 ω = −½ + (√3/2)i。它们满足 ω³ = 1 以及关键恒等式 1 + ω + ω² = 0。

ing an equilateral triangle. The two non-real roots are complex conjugates, with ω² = 1/ω.

它们位于单位圆上的 0、2π/3 和 4π/3 角度处，构成等边三角形。两个非实数根互为共轭复数，且 ω² = 1/ω。

Master these properties and you will be able to simplify complex expressions, factorise cubic polynomials, and solve roots-of-unity problems confidently in AQA A Level Further Maths.

掌握这些性质后，你就能够自信地化简复数表达式、因式分解三次多项式，并在 AQA A Level 进阶数学中解决单位根问题。

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