📚 Example 3.4.3 – Application to de Moivre's Theorem for Positive Integers | 正整数德莫瓦定理应用示例 3.4.3
In AQA A-Level Further Mathematics, de Moivre's theorem is a central tool for working with complex numbers in polar form. This article walks through Example 3.4.3, which focuses on applying the theorem when the exponent n is a positive integer. We will explore the theorem's statement, a typical worked example, and common ways it is used to derive trigonometric identities.
在 AQA A-Level 进阶数学中，德莫瓦定理是处理极式复数的核心工具。本文围绕示例 3.4.3 展开，重点讲解当指数 n 为正整数时如何应用该定理。我们将介绍定理内容、典型例题以及利用它推导三角恒等式的常见方法。
1. What Does the Syllabus Expect? | 考纲要求什么？
For AQA A-Level Further Mathematics, you are expected to state de Moivre's theorem for positive integer powers, apply it to complex numbers written in modulus-argument form, and use it to derive multiple-angle identities such as cos 3θ and sin 3θ. The positive integer case is tested regularly because it forms the basis for more advanced applications.
在 AQA A-Level 进阶数学中，你需要掌握正整数次幂的德莫瓦定理，能够对写成模-辐角形式的复数进行运算，并会用它推导倍角公式，例如 cos 3θ 和 sin 3θ。正整数情形经常出现在考题中，因为它是更高级应用的基础。
Example 3.4.3 is the kind of question that often appears in the first half of a paper: it tests fluency rather than heavy algebra. However, a careless sign or quadrant error can cost marks, so the steps below are written to help you develop a reliable routine.
示例 3.4.3 属于试卷前半部分常见的题型，主要考查熟练度而不是繁重的代数运算。然而，一个不小心的符号或象限错误都可能导致失分，因此下面给出的步骤旨在帮助你形成可靠的解题流程。
2. The Theorem for Positive Integers | 正整数情形的定理
De Moivre's theorem states that for any real number θ and any positive integer n,
德莫瓦定理指出，对于任意实数 θ 和任意正整数 n，
(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ
In words, raising the unit complex number cos θ + i sin θ to the n-th power simply multiplies the argument θ by n. The modulus of the unit complex number is 1, so 1ⁿ remains 1.
也就是说，将单位复数 cos θ + i sin θ 进行 n 次幂运算，只需把辐角 θ 乘以 n。单位复数的模为 1，因此 1ⁿ 仍然为 1。
(cos nθ + i sin nθ). This is the form used in Example 3.4.3, where r is not necessarily 1.
更一般地，若 z = r(cos θ + i sin θ)，则 zⁿ = rⁿ(cos nθ + i sin nθ)。这就是示例 3.4.3 中所使用的形式，其中 r 不一定为 1。
3. Why the Positive Integer Case Matters | 为什么正整数情形重要
rincipal arguments, but the positive integer version is unambiguous. For this reason, AQA often introduces de Moivre's theorem through n = 2 and n = 3 before expecting students to quote the general result.
正整数情形是定理的基础。推广到负整数和分数次幂时需要小心分支和主辐角，而正整数版本没有歧义。因此，AQA 通常会先通过 n = 2 和 n = 3 引入德莫瓦定理，再要求学生引用一般结论。
y direct expansion, then quote it for larger powers. This tests both understanding and efficient application.
许多考题会先让你通过直接展开证明 n = 2 或 n = 3 的情形，然后再引用定理处理更高次幂。这既考查理解，也考查高效应用。
s large enough that direct multiplication would be tedious but small enough that the trigonometric values can still be found exactly.
示例 3.4.3 重点处理 n = 5，这个次数足够大，直接乘法会很繁琐，但又足够小，三角函数值仍然可以精确求出。
4. Example 3.4.3 – Question | 示例 3.4.3 – 题目
A typical wording of Example 3.4.3 is:
示例 3.4.3 的典型题干如下：
'Given z = 2(cos π/6 + i sin π/6), use de Moivre's theorem to find z⁵ in modulus-argument form and in exact rectangular form.'
'已知 z = 2(cos π/6 + i sin π/6)，利用德莫瓦定理求 z⁵ 的模-辐角形式和精确的代数形式。'
The angle π/6 is chosen because its multiples have exact trigonometric values, making it ideal for a non-calculator paper. The modulus 2 is also simple to handle, but students must remember that it is raised to the power 5 as well.
选择 π/6 这个角是因为它的倍数具有精确的三角函数值，非常适合不允许使用计算器的试卷。模 2 也容易处理，但学生必须记住它同样要被提升到 5 次方。
5. Step 1 – Identify Modulus and Argument | 第一步 – 确定模与辐角
is already expressed in polar form, so no conversion from rectangular form is needed in this particular example.
将 z 写成标准形式 r(cos θ + i sin θ)。这里 r = 2，θ = π/6。该复数已经用极式表示，因此在这个例子中无需从代数形式进行转换。
Check that the angle is in radians, as required by AQA unless degrees are explicitly stated. If the question had given 30° instead, you should first convert it to π/6 radians to keep the working consistent.
确认角度使用弧度制，除非题目明确说明使用角度制，这也是 AQA 的要求。如果题目给出的是 30°，你应当先将其转换为 π/6 弧度，以保持计算过程一致。
6. Step 2 – Apply De Moivre's Theorem | 第二步 – 应用德莫瓦定理
Using zⁿ = rⁿ(cos nθ + i sin nθ), substitute n = 5, r = 2, θ = π/6:
利用公式 zⁿ = rⁿ(cos nθ + i sin nθ)，代入 n = 5, r = 2, θ = π/6：
z⁵ = 2⁵ [cos(5 × π/6) + i sin(5 × π/6)]
This gives z⁵ = 32 [cos(5π/6) + i sin(5π/6)].
由此得到 z⁵ = 32 [cos(5π/6) + i sin(5π/6)]。
The modulus has been raised to the power 5, and the argument has been multiplied by 5. This is the core operation of de Moivre's theorem: multiplication of arguments corresponds to exponentiation.
模被提升到 5 次方，辐角被乘以 5。这就是德莫瓦定理的核心运算：辐角的乘法对应于复数的乘方。
7. Step 3 – Convert to Exact Rectangular Form | 第三步 – 转换为精确代数形式
Now evaluate the trigonometric values exactly. Since 5π/6 is in the second quadrant,
现在精确计算三角函数值。因为 5π/6 在第二象限，
cos(5π/6) = -√3/2 and sin(5π/6) = 1/2
Substitute these values into z⁵:
将这些值代入 z⁵：
z⁵ = 32(-√3/2 + i × 1/2) = -16√3 + 16i
Therefore the exact rectangular form is z⁵ = -16√3 + 16i. The real part is negative and the imaginary part is positive, which is consistent with an argument in the second quadrant.
因此精确的代数形式为 z⁵ = -16√3 + 16i。实部为负，虚部为正，这与辐角在第二象限一致。
8. Interpretation on the Argand Diagram | 在阿尔冈图上的解释
ed the distance from the origin from 2 to 32, and rotated the point anticlockwise from π/6 to 5π/6. This gives a clear geometric picture of what exponentiation does in the complex plane.
变换 z → z⁵ 使点到原点的距离从 2 增加到 32，并将该点逆时针从 π/6 旋转到 5π/6。这清晰地展示了复数乘方在复平面上的几何意义。
Because 32 is positive and the argument is 5π/6, the result lies in quadrant II, matching the signs of the real part (negative) and imaginary part (positive). Drawing an Argand diagram can help you check your signs quickly.
因为 32 为正且辐角为 5π/6，所以结果位于第二象限，与实部为负、虚部为正的符号一致。画出阿尔冈图有助于快速检查符号。
9. Using the Expansion to Derive cos 5θ | 利用展开式推导 cos 5θ
Example 3.4.3 can also be extended: expand (cos θ + i sin θ)⁵ by the binomial theorem and equate real parts to find cos 5θ. This is a common AQA follow-up because it connects de Moivre's theorem with polynomial identities.
示例 3.4.3 还可以进一步拓展：用二项式定理展开 (cos θ + i sin θ)⁵，并令实部相等，从而求出 cos 5θ。这是 AQA 常见的后续问题，因为它将德莫瓦定理与多项式恒等式联系起来。
The expansion is:
展开式为：
θ)⁵ = cos⁵θ + 5i cos⁴θ sin θ - 10 cos³θ sin²θ - 10i cos²θ sin³θ + 5 cos θ sin⁴θ + i sin⁵θ
Taking the real part and using sin²θ = 1 - cos²θ leads to:
取实部并利用 sin²θ = 1 - cos²θ，可得：
cos 5θ = 16 cos⁵θ - 20 cos³θ + 5 cos θ
This is a standard AQA follow-up question and shows why the positive integer case is so powerful. By equating imaginary parts, you can similarly derive an expression for sin 5θ.
这是 AQA 常见的后续问题，也说明了正整数情形为何如此强大。通过令虚部相等，你还可以类似地推导出 sin 5θ 的表达式。
10. Common Mistakes to Avoid | 常见错误与规避方法
One common mistake is to raise only the cosine and sine terms to the power n but forget to raise the modulus r. Remember that both must be transformed: r becomes rⁿ and θ becomes nθ.
一个常见错误是只把余弦和正弦项乘方，却忘记对模 r 乘方。请记住模和辐角都要变换：r 变为 rⁿ，θ 变为 nθ。
le, 5π/6 is not the same as 30° × 5 = 150° written mistakenly as if it were in quadrant I.
另一个错误是角度误用度数，或在最终三角函数值中弄错象限，尤其是当 nθ 超过 π 或 2
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