A Framework for Writing a Further Mathematics Paper (OCR) with a Model Essay on De Moivre’s Theorem | 进阶数学论文写作框架与范文：以棣莫弗定理为例

📚 A Framework for Writing a Further Mathematics Paper (OCR) with a Model Essay on De Moivre’s Theorem | 进阶数学论文写作框架与范文：以棣莫弗定理为例

In Year 12 OCR Further Mathematics, students are often required to produce an extended piece of writing, such as a mathematical investigation or a proof-based essay. This article sets out a clear framework for structuring such a paper and provides a complete model essay exploring De Moivre’s theorem and nth roots of unity. Every section of the model essay is presented in both English and Chinese so that you can see exactly how to combine rigorous mathematics with clear exposition.

在 Year 12 OCR 进阶数学中，学生通常需要完成一篇扩展性写作，如数学探究或基于证明的短文。本文提出了一份清晰的论文结构框架，并提供了探讨棣莫弗定理与 n 次单位根的完整范文。范文的每个部分均以英中双语呈现，让读者清楚看到如何将严谨的数学与清晰的论述结合起来。

1. Understanding the Purpose of a Further Maths Paper | 理解进阶数学论文的目的

An extended mathematical piece at this level is not simply a set of calculations. It should demonstrate your ability to research a topic, present a coherent argument, prove results logically, and discuss their significance. The core of your paper is a mathematical investigation that goes beyond textbook exercises, often weaving together pure topics such as complex numbers, trigonometry, and algebraic manipulation.

此级别的扩展性数学写作不只是一组计算，而应展现你研究某个专题、呈现连贯论证、逻辑性地证明结果并讨论其意义的能力。论文的核心是一项超越课本练习的数学探究，常常将复数、三角学、代数操作等纯粹数学主题交织在一起。

2. Overall Structure of the Paper | 论文的总体结构

A strong paper typically contains the following sections: title, abstract and keywords, introduction, main body (theorems, proofs, derivations, examples), critical analysis or discussion, conclusion, and a list of references. A clear structure helps the reader follow your mathematical reasoning and shows academic maturity. The model essay below illustrates each of these components in practice.

一篇优秀的论文通常包含以下部分：标题、摘要与关键词、引言、主体（定理、证明、推导、示例）、批判性分析或讨论、结论以及参考文献列表。清晰的结构有助于读者跟随你的数学推理，也体现出学术成熟度。下面的范文将逐一展示这些组成部分。

3. Abstract and Keywords – Model Essay | 摘要与关键词——范文

Abstract: This paper explores de Moivre’s theorem and its application to finding nth roots of unity as well as deriving multiple-angle trigonometric identities. Using induction, we prove the theorem for integer exponents. The geometric representation of roots of unity on the Argand diagram is examined, and several classical identities are derived by equating real and imaginary parts. The discussion highlights connections to Euler’s formula and the cyclic group of roots.

摘要：本文探讨棣莫弗定理及其在求 n 次单位根和推导多倍角三角恒等式中的应用。利用归纳法，我们证明了整数指数下的定理。考察了单位根在 Argand 图上的几何表示，并通过比较实部与虚部推导出若干经典恒等式。讨论部分强调了与欧拉公式及单位根循环群的联系。

Keywords: De Moivre’s theorem, roots of unity, Euler’s formula, proof by induction, trigonometric identities. 关键词：棣莫弗定理，单位根，欧拉公式，归纳法证明，三角恒等式。

4. Introduction – Model Essay | 引言——范文

Complex numbers extend the real number system by introducing the imaginary unit i, where i² = -1. Any complex number can be written in polar form as r(cos θ + i sin θ). Euler’s famous formula, e^(iθ) = cos θ + i sin θ, elegantly links exponential functions with trigonometric ones. In the early eighteenth century, Abraham de Moivre derived the relation (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ) for integers n. This theorem is a cornerstone of complex analysis and underpins many techniques for solving polynomial equations and manipulating trigonometric expressions.

复数通过引入虚数单位 i (i² = -1) 扩展了实数系统。任何复数都能写成极坐标形式 r(cos θ + i sin θ)。著名的欧拉公式 e^(iθ) = cos θ + i sin θ 巧妙地将指数函数与三角函数联系起来。18 世纪初，亚伯拉罕·棣莫弗推导出对于整数 n 的关系式 (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)。该定理是复分析的基石，支撑着求解多项式方程和变换三角表达式的许多方法。

The aim of this paper is to give a rigorous proof of de Moivre’s theorem, apply it to calculate the nth roots of unity, and demonstrate its power by deriving double-angle and triple-angle formulas. The geometric interpretation on the unit circle will be emphasised throughout.

本文旨在给出棣莫弗定理的严格证明，应用其计算 n 次单位根，并通过推导二倍角和三倍角公式展示其强大功能。全文将着重强调单位圆上的几何解释。

5. De Moivre’s Theorem: Statement and Proof | 棣莫弗定理：陈述与证明

Theorem (De Moivre): For any real number θ and any integer n, (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ).

定理（棣莫弗）： 对于任意实数 θ 和任意整数 n，有 (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)。

Proof for non-negative n by induction. Base case n = 0: (cos θ + i sin θ)⁰ = 1, and cos(0) + i sin(0) = 1. Base case n = 1 holds trivially. Now assume the statement is true for n = k, that is, (cos θ + i sin θ)ᵏ = cos(kθ) + i sin(kθ). Consider n = k+1:

对非负整数 n 的归纳证明。 基本情况 n=0 和 n=1 显然成立。假设对 n=k 为真，即 (cos θ + i sin θ)ᵏ = cos(kθ) + i sin(kθ)。考虑 n=k+1：

(cos θ + i sin θ)^(k+1) = (cos θ + i sin θ)ᵏ · (cos θ + i sin θ) = [cos(kθ) + i sin(kθ)] · [cos θ + i sin θ]

Expand the product: = cos(kθ)cos θ – sin(kθ)sin θ + i[sin(kθ)cos θ + cos(kθ)sin θ]. Using the addition formulas for cosine and sine, this simplifies to cos(kθ + θ) + i sin(kθ + θ) = cos((k+1)θ) + i sin((k+1)θ). Thus the statement holds for k+1. By the principle of mathematical induction, the theorem is true for all non-negative integers n.

展开乘积，利用余弦和正弦的和角公式，上式简化为 cos(kθ+θ) + i sin(kθ+θ) = cos((k+1)θ) + i sin((k+1)θ)。因此命题对 k+1 成立。由数学归纳法原理，定理对所有非负整数 n 成立。

For negative integers, let n = -m where m > 0. Then (cos θ + i sin θ)⁻ᵐ = 1 / (cos θ + i sin θ)ᵐ = 1 / [cos(mθ) + i sin(mθ)]. Multiplying the numerator and denominator by cos(mθ) – i sin(mθ) yields cos(mθ) – i sin(mθ) = cos(-mθ) + i sin(-mθ), since cosine is even and sine is odd. This completes the proof.

对于负整数，令 n=-m (m>0)，取倒数并乘以共轭，利用余弦的偶性和正弦的奇性即得。证明完毕。

6. nth Roots of Unity and Their Geometry | n 次单位根及其几何表示

A complex number z is an nth root of unity if zⁿ = 1. To solve this, write the right-hand side in polar form: 1 = cos(2πk) + i sin(2πk) for integer k. If z = cos φ + i sin φ, then by de Moivre’s theorem zⁿ = cos(nφ) + i sin(nφ). Equating the expressions gives cos(nφ) + i sin(nφ) = cos(2πk) + i sin(2πk). Hence nφ = 2πk + 2πm, but distinct solutions arise when we restrict the argument to [0, 2π), giving φ = 2πk/n for k = 0, 1, 2, …, n-1.

若复数 z 满足 zⁿ = 1，则称 z 为 n 次单位根。将右边写为极坐标形式 1 = cos(2πk) + i sin(2πk)。设 z = cos φ + i sin φ，由棣莫弗定理 zⁿ = cos(nφ) + i sin(nφ)。比较得 cos(nφ) + i sin(nφ) = cos(2πk) + i sin(2πk)，故 nφ = 2πk，取不同 k 可得 φ = 2πk/n，k = 0,1,…,n-1。

The n distinct roots are therefore: ωₖ = cos(2πk/n) + i sin(2πk/n) = e^(2πik/n), k = 0,1,…,n-1. Note that ω₀ = 1 always. On an Argand diagram, these points lie exactly on the unit circle, equally spaced by an angle of 2π/n. They form the vertices of a regular n-gon. For instance, the cube roots of unity (n=3) are 1, -½ + i(√3/2), -½ – i(√3/2), giving an equilateral triangle. The fifth roots of unity form a regular pentagon.

因此 n 个不同的根为 ωₖ = cos(2πk/n) + i

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