AS AQA Further Mathematics Pure Mathematics Topic Test | AS AQA 进阶数学纯数专题测试

📚 AS AQA Further Mathematics Pure Mathematics Topic Test | AS AQA 进阶数学纯数专题测试

Welcome to this comprehensive topic test revision guide for OxfordAQA International AS Level Further Mathematics (9665) Pure Mathematics. This article consolidates the core topics you need to master for your examinations, from complex numbers and matrices to series and induction.

欢迎阅读牛津AQA国际AS进阶数学（9665）纯数学专题测试复习指南。本文整合了你需要在考试中掌握的核心主题，涵盖复数、矩阵、级数与归纳法等内容。

1. Complex Number Arithmetic | 复数运算

Complex numbers extend the real number system to include solutions to equations such as x² + 1 = 0. The imaginary unit i satisfies i² = −1.

复数将实数系统扩展，以包含 x² + 1 = 0 这类方程的解。虚数单位 i 满足 i² = −1。

A complex number is written as z = x + yi, where x is the real part and y is the imaginary part. To add or subtract complex numbers, combine like terms. To multiply, expand brackets and replace i² with −1.

复数写作 z = x + yi，其中 x 为实部，y 为虚部。加减复数时合并同类项；相乘时展开括号并将 i² 替换为 −1。

For example, (2 + 3i)(1 − 2i) = 2 − 4i + 3i − 6i² = 2 − i + 6 = 8 − i.

例如，(2 + 3i)(1 − 2i) = 2 − 4i + 3i − 6i² = 2 − i + 6 = 8 − i。

The complex conjugate of z = x + yi is z̄ = x − yi. The product z z̄ = (x + yi)(x − yi) = x² + y² is always a non-negative real number.

复数 z = x + yi 的共轭复数为 z̄ = x − yi。乘积 z z̄ = (x + yi)(x − yi) = x² + y² 始终为非负实数。

Division is performed by multiplying the numerator and denominator by the conjugate of the denominator:

除法通过将分子与分母同时乘以分母的共轭复数来完成：

(1 + 2i) / (3 − i) = (1 + 2i)(3 + i) / (3 − i)(3 + i) = (1 + 7i) / 10 = 0.1 + 0.7i

2. Argand Diagrams and Modulus–Argument Form | 阿甘图与模-辐角形式

An Argand diagram represents complex numbers on a plane, with the horizontal axis for the real part and the vertical axis for the imaginary part.

阿甘图在平面上表示复数，横轴代表实部，纵轴代表虚部。

The modulus of z = x + yi is |z| = √(x² + y²), the distance from the origin. The argument is the angle θ measured anticlockwise from the positive real axis.

复数 z = x + yi 的模为 |z| = √(x² + y²)，即到原点的距离。辐角 θ 为从正实轴逆时针测量的角度。

For z = 3 + 4i, |z| = √(3² + 4²) = 5 and arg(z) = tan⁻¹(4/3) ≈ 0.927 radians.

对于 z = 3 + 4i，|z| = √(3² + 4²) = 5，arg(z) =

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