AS AQA Further Maths Support Pack 1: Core Revision Guide | AS AQA 进阶数学支持包1：核心复习指南

📚 AS AQA Further Maths Support Pack 1: Core Revision Guide | AS AQA 进阶数学支持包1：核心复习指南

This Support Pack 1 revision guide covers the essential topics of the AQA International AS Further Mathematics (9665) course. Each section pairs a concise summary with exam-ready patterns, focusing on the techniques and common pitfalls that appear most frequently in assessments.

本支持包1复习指南涵盖 AQA 国际 AS 进阶数学（9665）课程的核心主题。每个小节将简明总结与考试解法模式配对，聚焦评估中最常出现的技巧与常见陷阱。

---

1. Complex Numbers – Arithmetic | 复数运算

the real part (Re z) and b is the imaginary part (Im z). The defining property is i² = −1.

复数写作 z = a + bi，其中 a 是实部（Re z），b 是虚部（Im z）。其定义性质为 i² = −1。

Addition and subtraction combine real and imaginary parts separately. Multiplication expands like a binomial, replacing every i² with −1. Division multiplies the numerator and denominator by the complex conjugate of the denominator.

加减法分别合并实部和虚部。乘法像二项式一样展开，并将每个 i² 替换为 −1。除法则将分子和分母同时乘以分母的共轭复数。

For z = a + bi, the complex conjugate is z̄ = a − bi, and the modulus is |z| = √a² + b². These two concepts appear constantly in division and in proofs about conjugate roots.

对于 z = a + bi，共轭复数为 z̄ = a − bi，模为 |z| = √a² + b²。这两者在除法运算和共轭根证明中反复出现。

z₁ + z₂ = (a + c) + (b + d)i, z₁z₂ = (ac − bd) + (ad + bc)i

z₁ ÷ z₂ = (z₁ × z̄₂) ÷ |z₂|²

When dividing, always write the result as a single complex number in the form a + bi. Never leave a fraction with i in the denominator; rationalise it using the conjugate.

做除法时，务必把结果写成 a + bi 形式的单一复数。绝不要留下分母含 i 的分数；要用共轭复数将其有理化。

---

2. Modulus and Argument | 模与辐角

Every complex number can be represented as a point in the Argand diagram, with the real part along the horizontal axis and the imaginary part along the vertical axis. The modulus is the distance from the origin; the argument is the angle measured anticlockwise from the positive real axis.

每个复数都可以表示为阿尔冈图中的点，实部沿水平轴，虚部沿垂直轴。模是到原点的距离；辐角是从正实轴逆时针量得的角度。

The modulus–argument form writes z = r(cos θ + i sin θ), where r = |z| nterval (−π, π]. To find θ, start with tan θ = b/a, then adjust according to the quadrant.

模-辐角形式写作 z = r(cos θ + i sin θ)，其中 r = |z|，θ = arg(z)。主辐角位于区间 (−π, π] 内。求 θ 时，先计算 tan θ = b/a，再根据象限进行调整。

z = a + bi = r(cos θ + i sin θ), r = √a² + b², tan θ = b/a

Common special values are worth memorising: cos(π/3) = ½, sin(π/3) = √3/2, cos(π/4) = √2/2, and so on. These make medium-difficulty questions fast and accurate.

常用的特殊值值得牢记：cos(π/3) = ½、sin(π/3) = √3/2、cos(π/4) = √2/2 等等。这些能使中等难度题目解得又快又准。

Multiplication and division become simpler in polar form: the moduli multiply or divide, while the arguments add or subtract. This pattern is the foundation for later work with De Moivre's theorem.

在极坐标形式下，乘除运算变得更加简单：模相乘或相除，辐角相加或相减。这一规律是后续棣莫弗定理学习的基础。

---

广告 / Sponsored

3. Complex Roots of Polynomial Equations | 多项式方程的复数根

satisfy α + β = −b/a and αβ = c/a. The same pattern extends to higher-degree polynomials.

对于二次方程 az² + bz + c = 0，若其根为 α 和 β，则根和满足 α + β = −b/a，根积满足 αβ = c/a。同样的规律可推广到更高次多项式。

For a cubic ax³ + bx² + cx + d = 0 with roots α, β, γ:

对于三次方程 ax³ + bx² + cx + d = 0，若其根为 α、β、γ：

Σα = −b/a, Σαβ = c/a, αβγ = −d/a

lving the equation explicitly. Use the identity (Σα)² = Σα² + 2Σαβ, so Σα² = (Σα)² − 2Σαβ.

这些关系使你能在不解方程的情况下求出 α² + β² + γ² 等表达式的值。利用恒等式 (Σα)² = Σα² + 2Σαβ，可得 Σα² = (Σα)² − 2Σαβ。

Complex roots of polynomials with real coefficients always occur in conjugate pairs: if α is

Published by TutorHao | AS Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）