FP2 Masterclass: De Moivre, Series & Differential Equations | FP2 通关指南：De Moivre 定理、级数与微分方程

📚 FP2 Masterclass: De Moivre, Series & Differential Equations | FP2 通关指南：De Moivre 定理、级数与微分方程

Further Pure 2 (FP2) is the second pure mathematics unit in the AQA International A-level Further Mathematics specification. It builds directly on the methods introduced in FP1 and deepens your command of complex numbers, series, differential equations and polar coordinates. Every FP2 exam rewards clear algebraic structure, fluent use of standard results and disciplined checking of conditions. This guide condenses all core topics into a structured revision route, with each concept paired in English and Chinese so you can study with full clarity.

进阶纯数 2（FP2）是 AQA 国际 A-level 进阶数学大纲中的第二门纯数单元。它直接建立在 FP1 所学内容之上，进一步深化你对复数、级数、微分方程和极坐标的掌握。FP2 考试注重清晰的代数结构、熟练运用标准结论，以及严格检验适用条件。本指南将所有核心考点浓缩为一条系统的复习路径，每个概念均配有中英双语讲解，帮助你透彻理解。

1. De Moivre’s Theorem | De Moivre 定理

De Moivre’s theorem is the most powerful tool in FP2. It states that for any integer n and any real angle θ, the equality below holds exactly:

De Moivre 定理是 FP2 中最强大的工具。它指出：对任意整数 n 和任意实数角 θ，下式精确成立：

(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)

This single identity turns repeated multiplication of complex numbers into simple angle scaling. For example, to evaluate (1 + i)¹⁰, first write 1 + i = √2 (cos π/4 + i sin π/4). Then apply the theorem: (1 + i)¹⁰ = (√2)¹⁰ [cos(10 × π/4) + i sin(10 × π/4)] = 32 [cos(5π/2) + i sin(5π/2)] = 32i.

这一恒等式将复数的连乘转化为简单的角度缩放。例如，计算 (1 + i)¹⁰，先将 1 + i 写成 √2 (cos π/4 + i sin π/4)，再套用定理：(1 + i)¹⁰ = (√2)¹⁰ [cos(10 × π/4) + i sin(10 × π/4)] = 32 [cos(5π/2) + i sin(5π/2)] = 32i。

Always convert a complex number to modulus-argument form before raising it to a power. Remember that the modulus r must be raised to the same power, while the argument θ is simply multiplied by n.

在求复数幂之前，务必先将其化为模-辐角形式。记住：模 r 要取相应的幂，而辐角 θ 只需乘以 n。

2. Roots of Complex Numbers & Roots of Unity | 复数方根与单位根

Given a non-zero complex number z = r(cos θ + i sin θ), the equation wⁿ = z has exactly n distinct complex solutions. These nth roots are given by the formula below, for k = 0, 1, 2, …, n − 1:

给定非零复数 z = r(cos θ + i sin θ)，方程 wⁿ = z 恰好有 n 个不同的复数解。这些 n 次方根由下式给出，其中 k = 0, 1, 2, …, n − 1：

wₖ = ⁿ√r [cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)]

Geometrically, the n roots lie equally spaced on a circle of radius ⁿ√r centred at the origin, separated by angle 2π/n. For example, the fourth roots of unity (solutions of z⁴ = 1) are 1, i, −1, −i. Their sum is 0 and their product is −1; in general the sum of all nth roots of unity is always 0.

从几何上看，这 n 个根均匀分布在以原点为圆心、半径为 ⁿ√r 的圆上，相邻夹角为 2π/n。例如，1 的四次单位根（即 z⁴ = 1 的解）为 1、i、−1、−i，其和为 0，其积为 −1；一般地，全体 n 次单位根之和恒为 0。

When solving zⁿ = w, always add 2kπ to the argument before dividing by n. This is the step that produces all n distinct roots; forgetting it will lose solutions and marks.

解 zⁿ = w 时，切记先将辐角加上 2kπ 再除以 n。正是这一步产生全部 n 个不同的根；遗漏它会丢解、丢分。

3. Multiple-Angle Formulae & Integration | 倍角公式与积分

De Moivre’s theorem also generates the multiple-angle identities. Expanding (cos θ + i sin θ)ⁿ using the binomial theorem,

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