📚 A-Level CIE Further Mathematics: Detailed Worked Examples | A-Level CIE 进阶数学：典型例题详解
This article presents a selection of carefully chosen worked examples covering core topics from the CIE A-Level Further Mathematics syllabus. Each example is solved step by step, illustrating key techniques and common pitfalls. The detailed bilingual explanations aim to strengthen your conceptual understanding and problem-solving skills, making this an ideal revision resource for final examination preparation.
本文精选了 CIE A-Level 进阶数学核心考点中的典型例题，逐一进行详细解析。每道例题均提供分步解答，展示关键方法和常见误区。中英双语详解有助于巩固概念理解和提升解题能力，是备战大考的理想复习资料。
1. Complex Numbers: Roots of a Complex Equation | 复数：复数方程的根
Problem: Find all roots of the equation z⁴ + 16 = 0 and sketch them on an Argand diagram.
问题：求方程 z⁴ + 16 = 0 的所有根，并在 Argand 图上绘出。
Step 1: Rewrite the equation as z⁴ = -16. Express -16 in polar form: 16(cos π + i sin π) = 16 eiπ.
步骤 1：将方程改写为 z⁴ = -16，将 -16 表示为极坐标形式：16(cos π + i sin π) = 16 eiπ。
Step 2: By De Moivre’s theorem, the fourth roots are z = 16¼ ei(π + 2kπ)/4 = 2 ei(π/4 + kπ/2) for k = 0, 1, 2, 3.
步骤 2：根据棣莫弗定理，四次根为 z = 16¼ ei(π + 2kπ)/4 = 2 ei(π/4 + kπ/2)，其中 k = 0, 1, 2, 3。
Step 3: Evaluate each root in Cartesian form:
k=0: z₀ = 2(cos(π/4) + i sin(π/4)) = √2 + i√2
k=1: z₁ = 2(cos(3π/4) + i sin(3π/4)) = -√2 + i√2
k=2: z₂ = 2(cos(5π/4) + i sin(5π/4)) = -√2 – i√2
k=3: z₃ = 2(cos(7π/4) + i sin(7π/4)) = √2 – i√2
步骤 3：计算各根的直角坐标形式：
k=0: z₀ = 2(cos(π/4) + i sin(π/4)) = √2 + i√2
k=1: z₁ = 2(cos(3π/4) + i sin(3π/4)) = -√2 + i√2
k=2: z₂ = 2(cos(5π/4) + i sin(5π/4)) = -√2 – i√2
k=3: z₃ = 2(cos(7π/4) + i sin(7π/4)) = √2 – i√2
Step 4: On the Argand diagram, the four roots form the vertices of a square centred at the origin with side length 2√2.
步骤 4：在 Argand 图上，这四个根构成一个以原点为中心、边长为 2√2 的正方形的顶点。
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