📚 AS AQA Further Maths 9665 Support Pack 2: Core Revision Guide | AS AQA 进阶数学 9665 支持包 2 核心复习指南
This support pack is designed to consolidate the essential topics from the AQA International AS Further Mathematics (9665) syllabus. It provides a focused revision of the core techniques you must master, with worked examples and exam-level tips embedded throughout. Work through each section methodically, and you will build both fluency and confidence for the examination.
本支持包旨在系统巩固 AQA 国际 AS 进阶数学（9665）考纲中的核心专题。它聚焦精要方法与技巧，配有典型例题和考试级提示。请按小节逐一击破，你将逐步建立熟练度与应试信心。
1. Complex Number Arithmetic | 复数运算
A complex number is written in the form z = a + bi, where a is the real part, b is the imaginary part, and i is defined by i² = −1. Complex numbers extend the real number system and allow us to solve equations such as x² + 1 = 0.
复数记为 z = a + bi 的形式，其中 a 为实部，b 为虚部，i 满足 i² = −1。复数扩展了实数系，使我们能够求解 x² + 1 = 0 这类方程。
Addition and subtraction are performed component-wise: (a + bi) + (c + di) = (a + c) + (b + d)i, and similarly for subtraction. Multiplication follows the distributive law, remembering that i² = −1: (a + bi)(c + di) = (ac − bd) + (ad + bc)i.
复数的加减法按分量对应进行：(a + bi) + (c + di) = (a + c) + (b + d)i，减法同理。乘法遵循分配律，并牢记 i² = −1：(a + bi)(c + di) = (ac − bd) + (ad + bc)i。
The complex conjugate of z = a + bi is z̄ = a − bi. Multiplying a complex number by its conjugate gives the real number z·z̄ = a² + b², which is then used to divide complex numbers. To compute z₁/z₂, multiply numerator and denominator by the conjugate of the denominator.
复数 z = a + bi 的共轭复数为 z̄ = a − bi。复数与其共轭相乘得到实数 z·z̄ = a² + b²，这一性质用于复数的除法。计算 z₁/z₂ 时，将分子分母同时乘以分母的共轭复数。
(2 + 3i) ÷ (1 − 2i) = [(2 + 3i)(1 + 2i)] ÷ [(1 − 2i)(1 + 2i)] = (−4 + 7i) ÷ 5 = −0.8 + 1.4i
2. Argand Diagrams and Modulus–Argument Form | 阿尔冈图与模–辐角形式
An Argand diagram is a plane in which the horizontal axis represents the real part and the vertical axis the imaginary part. Each complex number corresponds to a unique point or position vector in this plane.
阿尔冈图以水平轴表示实部、竖直轴表示虚部。每个复数都对应平面上的一个唯一点或位置向量。
The modulus of z = a + bi is the distance from the origin to the point (a, b), given by |z| = √(a² + b²). The argument, arg(z), is the angle the line from the origin to (a, b) makes with the positive real axis, measured in radians, with: tan θ = b/a.
复数 z = a + bi 的模是原点到点 (a, b) 的距离，记为 |z| = √(a² + b²)。辐角 arg(z) 是原点到点 (a, b) 的连线与正实轴之间的夹角，以弧度为单位，满足 tan θ = b/a。
This gives the modulus–argument form: z = r(cos θ + i sin θ), often abbreviated as z = r cis θ. This form is particularly powerful for multiplication and division: when multiplying, moduli are multiplied and arguments are added; when dividing, moduli are divided and arguments are subtracted.
由此得到模–辐角形式：z = r(cos θ + i sin θ)，常简写为 z = r cis θ。这一形式在乘法与除法中格外强大：乘法时模相乘、辐角相加；除法时模相除、辐角相减。
If z₁ = r₁ cis θ₁ and z₂ = r₂ cis θ₂, then z₁z₂ = r₁r₂ cis(θ₁ + θ₂) and z₁/z₂ = (r₁/r₂) cis(θ₁ − θ₂)
Use the principal value of the argument, always in the interval −π < arg(z) ≤ π. Be careful with the quadrant: adding π if the real part is negative ensures your calculator answer matches the principal range.
辐角取主值，始终落在区间 −π < arg(z) ≤ π 内。需注意象限判断：当实部为负时需加上 π，以确保计算器结果与主值范围一致。
3. De Moivre’s Theorem | 棣莫弗定理
De Moivre’s theorem states that for any integer n and any real angle θ: (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ). In its concise form, cis θ raised to the power n equals cis(nθ).
棣莫弗定理指出：对于任意整数 n 和任意实数角 θ，有 (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)。其简洁形式为：cis θ 的 n 次方等于 cis(nθ)。
This theorem works for any integer power, including negative and fractional exponents. For negative powers, rewrite (cos θ + i sin θ)⁻ⁿ = cos(−nθ) + i sin(−nθ) = cos(nθ) − i sin(nθ), which is simply the conjugate.
该定理适用于任意整数幂，包括负指数与分数指数。对于负幂，可改写为 (cos θ + i sin θ)⁻ⁿ = cos(−nθ) + i sin(−nθ) = cos(nθ) − i sin(nθ)，即共轭形式。
When applying De Moivre’s theorem to derive multiple-angle identities, expand (cos θ + i sin θ)ⁿ using the binomial theorem, then equate the real and imaginary parts. For example, the real part of (cis θ)³ gives cos 3θ = 4cos³θ − 3cos θ.
运用棣莫弗定理推导倍角公式时，先用二项式定理展开 (cos θ + i sin θ)ⁿ，再分别比较实部与虚部。例如，(cis θ)³ 的实部给出 cos 3θ = 4cos³θ − 3cos θ。
De Moivre’s theorem also leads to the roots of unity. The n-th roots of the complex number w are given by: zₖ = r^(1/n) · cis[(θ + 2kπ)/n], for k = 0, 1, 2, …, n − 1. These roots form a regular polygon centred at the origin in the Argand diagram.
棣莫弗定理还引向单位根问题。复数 w 的 n 次方根为：zₖ = r^(1/n) · cis[(θ + 2kπ)/n]，其中 k = 0, 1, 2, …, n − 1。这些根在阿尔冈图上构成一个以原点为中心的正多边形。
4. Matrices: Operations, Determinants and Inverses | 矩阵：运算、行列式与逆矩阵
A matrix is a rectangular array of numbers. The order of a matrix is given by its dimensions: rows × columns. Matrix addition and scalar multiplication are performed element-wise, while matrix multiplication requires the number of columns of the first matrix to equal the number of rows of the second.
矩阵是数字的矩形阵列。矩阵的阶由行数 × 列数给出。矩阵加法与数乘按元素逐一进行；矩阵乘法要求左矩阵的列数等于右矩阵的行数。
For a 2 × 2 matrix A = [a b; c d], the determinant is det(A) = ad − bc. The determinant is a measure of the area scale factor of the transformation represented by the matrix. If det(A) = 0, the matrix is singular and has no inverse.
对于 2 × 2 矩阵 A = [a b; c d]，行列式为 det(A) = ad − bc。行列式衡量矩阵所表示变换的面积缩放因子。若 det(A) = 0，矩阵为奇异矩阵，不存在逆矩阵。
The inverse of a 2 × 2 matrix is found using the standard formula: swap the diagonal entries, change the signs of the off-diagonal entries, and multiply by 1/det(A). Check your result by confirming that A·A⁻¹ = I, where I is the identity matrix.
2 × 2 矩阵的逆矩阵由标准公式求得：交换主对角线元素，改变副对角线元素的符号，再乘以 1/det(A)。验证结果只需确认 A·A⁻¹ = I，其中 I 为单位矩阵。
A⁻¹ = (1/(ad − bc)) · [d −b; −c a]
For 3 × 3 matrices, calculate the determinant by expansion along a row or column. The inverse is given by A⁻¹ = adj(A)/det(A), where adj(A) is the transpose of the matrix of cofactors. Practise both methods with the same matrix to check consistency.
对于 3 × 3 矩阵，行列式按某一行或某一列展开计算。逆矩阵由 A⁻¹ = adj(A)/det(A) 给出，其中 adj(A) 是代数余子式矩阵的转置。建议对同一矩阵用两种方法互验。
5. Matrix Transformations | 矩阵变换
A 2 × 2 matrix maps a point (x, y) to a new point using the rule [x’ ; y’] = M[x ; y]. Common transformations include rotations, reflections, enlargements and shears, each with its own standard matrix.
2 × 2 矩阵通过规则 [x’ ; y’] = M[x ; y] 将点 (x, y) 映射到新点。常见变换包括旋转、反射、放缩与剪切，每种变换都有对应的标准矩阵。
Rotation anticlockwise through angle θ about the origin: R(θ) = [cos θ −sin θ; sin θ cos θ]. Reflection in a line through the origin making angle θ with the x-axis: F(θ) = [cos 2θ sin 2θ; sin 2θ −cos 2θ].
绕原点逆时针旋转角 θ 的矩阵为：R(θ) = [cos θ −sin θ; sin θ cos θ]。关于过原点且与 x 轴夹角为 θ 的直线反射的矩阵为：F(θ) = [cos 2θ sin 2θ; sin 2θ −cos 2θ]。
The order of transformations matters. If a transformation S is followed by T, the combined matrix is T·S, since the first transformation is applied to the position vector first and the second acts on the result. Always check the order by testing on a simple point such as (1, 0).
变换的先后顺序至关重要。若先进行变换 S、再进行变换 T，则复合矩阵为 T·S，因为先作用于位置向量的是第一个变换，而第二个变换作用于其结果。用简单点如 (1, 0) 验证顺序是一个好习惯。
The determinant of a transformation matrix gives the area scale factor, and the sign reveals orientation: a negative determinant indicates a reflection has included a reversal of orientation. For similar figures after an enlargement, the linear scale factor is √|det(M)| for a shape of dimension 2.
变换矩阵的行列式给出面积缩放因子，其符号揭示取向：行列式为负表示反射导致取向反转。对二维图形，放大后的线性缩放因子为 √|det(M)|。
6. Roots of Polynomials | 多项式根的关系
For a quadratic equation ax² + bx + c = 0 with roots α and β, the elementary symmetric relationships are: α + β = −b/a and αβ = c/a. These relationships allow you to find new equations whose roots are functions of α and β without solving the original.
对于二次方程 ax² + bx + c = 0，设根为 α 与 β，基本对称关系为：α + β = −b/a，αβ = c/a。利用这些关系，无需解原方程即可构造以 α、β 的某种组合为根的新方程。
For a cubic equation ax³ + bx² + cx + d = 0 with roots α, β, γ: α + β + γ = −b/a, αβ + βγ + γα = c/a, and αβγ = −d/a. These can be verified by expanding a(x − α)(x − β)(x − γ) and equating coefficients.
对于三次方程 ax³ + bx² + cx + d = 0，设根为 α、β、γ：α + β + γ = −b/a，αβ + βγ + γα = c/a，αβγ = −d/a。这些结论可通过展开 a(x − α)(x − β)(x − γ) 并比较系数加以验证。
To construct a new polynomial with roots α², β², use the identities: α² + β² = (α + β)² − 2αβ and α²β² = (αβ)². Similarly, for reciprocals 1/α and 1/β: 1/α + 1/β = (α + β)/(αβ), and the product is 1/(αβ). Begin the new equation with xⁿ and multiply through by constants to clear fractions.
构造以 α²、β² 为根的新多项式，用恒等式：α² + β² = (α + β)² − 2αβ，α²β² = (αβ)²。对于倒数根 1/α、1/β：1/α + 1/β = (α + β)/(αβ)，乘积为 1/(αβ)。新方程以 xⁿ 开头，必要时乘以常数以消去分数。
For quartic equations ax⁴ + bx³ + cx² + dx + e = 0 with roots α, β, γ, δ, extend the pattern: sum of roots = −b/a, sum of products of pairs = c/a, sum of products of triples = −d/a, and product of all roots = e/a. Memorising these patterns reduces computational errors significantly.
对于四次方程 ax⁴ + bx³ + cx² + dx + e = 0，设根为 α、β、γ、δ，模式继续扩展：根之和 = −b/a，两两乘积之和 = c/a，三三乘积之和 = −d/a，所有根之积 = e/a。记住这些模式可显著减少计算错误。
7. Proof by Induction | 数学归纳法
Proof by induction is used to prove statements that depend on a positive integer n. The method has three essential parts: the base case, the inductive hypothesis, and the inductive step. Missing any one makes the proof invalid.
数学归纳法用于证明依赖于正整数 n 的命题。该方法包含三个必需部分：基础情形、归纳假设和归纳步骤。缺少任何一部分，证明都不完整。
For summation results, prove the statement is true for n = 1 first. Then assume it is true for n = k, and use this assumption to show it is true for n = k + 1. The key algebraic step is adding the (k + 1)-th term to both sides of the assumed equation.
对于求和结论，首先证明 n = 1 时命题成立。然后假设 n = k 时成立，并利用该假设推导 n = k + 1 时成立。关键的代数步骤是：在假设等式的两边同时加上第 (k + 1) 项。
Σᵣ₌₁ⁿ r = n(n + 1)/2 ⇒ assume: Σᵣ₌₁ᵏ r = k(k + 1)/2, then add (k + 1) to both sides and simplify
For divisibility proofs, such as showing 3²ⁿ − 1 is divisible by 8 for all n ∈ ℤ⁺, rewrite the (k + 1) expression in terms of the k expression. For instance, 3²⁽ᵏ⁺¹⁾ − 1 = 9(3²ᵏ − 1) + 8, which is divisible by 8 when 3²ᵏ − 1 is divisible by 8.
对于整除性证明，例如证明对一切 n ∈ ℤ⁺，3²ⁿ − 1 都能被 8 整除，需将 (k + 1) 的表达式改写为含 k 表达式。例如，3²⁽ᵏ⁺¹⁾ − 1 = 9(3²ᵏ − 1) + 8，当 3²ᵏ − 1 能被 8 整除时，该式显然也能被 8 整除。
Always write a clear conclusion: ‘By the principle of mathematical induction, the statement is true for all positive integers n.’ This final line demonstrates your understanding of the structure and earns full method marks.
务必写出清晰的结论：“由数学归纳法原理，该命题对一切正整数 n 成立。”这最后一句话表明你对证明结构的完整理解，能确保获得全部方法分。
8. Further Calculus: Integration Techniques | 进阶微积分：积分技巧
The AS Further Maths syllabus expects fluency in integration by parts, integration by substitution, and the use of partial fractions to integrate rational functions.
AS 进阶数学考纲要求熟练掌握分部积分法、换元积分法以及利用部分分式积分有理函数。
Integration by parts follows the formula: ∫ u dv = uv − ∫ v du. Choose u to be a function that simplifies when differentiated (such as x or x²), and dv to be a function that is simple to integrate (such as eˣ or sin x). For expressions like x²eˣ, apply the method twice.
分部积分公式为：∫ u dv = uv − ∫ v du。选择 u 为微分后更简单的函数（如 x 或 x²），选择 dv 为易于积分的函数（如 eˣ 或 sin x）。对于 x²eˣ 这类表达式，需连续运用两次分部积分。
When integrating rational functions, first perform algebraic division if the degree of the numerator is greater than or equal to the degree of the denominator. Then decompose the remainder using partial fractions into the form A/(x − a) + B/(x − b), which integrate to natural logarithms.
积分有理函数时，若分子的次数不低于分母的次数，先做多项式除法。再将余式用部分分式分解为 A/(x − a) + B/(x − b) 的形式，其积分结果为自然对数。
Substitution works best when you can spot a function and its derivative within the integrand. Use trigonometric substitutions for integrands containing √(a² − x²), setting x = a sin θ, which transforms the expression into a simple trigonometric identity.
换元法在能识别被积函数中的函数及其导数时最为有效。对于含 √(a² − x²) 的被积函数，采用三角换元 x = a sin θ，将表达式转化为简单三角恒等式。
9. Numerical Methods | 数值方法
When equations cannot be solved algebraically, numerical methods provide approximate solutions. The Newton–Raphson iterative formula approximates successively better solutions from an initial guess x₀. It requires that the derivative exists and is non-zero at the root.
当方程无法用代数方法求解时，数值方法提供近似解。牛顿–拉弗森迭代公式从初始猜测值 x₀ 出发，逐步逼近更精确的解。该公式要求根附近导数存在且非零。
xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ)
To apply the method: choose a sensible starting value (use the sign-change rule to locate an interval containing a root), compute f(xₙ) and f'(xₙ), then iterate until successive approximations agree to the required number of decimal places. Check your final answer by substituting back into the original equation.
应用该方法的步骤：选择合理的初始值（用变号规则确定含根的区间），计算 f(xₙ) 与 f'(xₙ)，然后反复迭代直到相邻两次近似值达到所需小数位数。将最终答案代回原方程验算。
Beware of cases where the method fails: if f'(x) = 0 near the root, the formula divides by zero; if the initial guess is far from the root, it may converge to a different root or diverge. Drawing a rough graph before starting is strongly recommended.
注意方法的失效情况：若根附近 f'(x) = 0，公式会出现除零；若初始值距离根太远，可能收敛到另一个根或发散。强烈建议开始前先绘制粗略图形。
10. Exam Strategy and Common Pitfalls | 考试策略与常见错误
Time management is critical. Start with the topics you know best to secure marks early, and label your answers clearly with parts (a), (b), (c) corresponding to the question. Show every intermediate step, as AQA awards method marks for correct processes even when the final answer is wrong.
时间管理至关重要。先做最有把握的题目以尽早锁定分数，并按 (a)、(b)、(c) 清楚标注各部分答案。务必写出每一步中间过程，因为 AQA 对正确过程给方法分，即使最终答案有误。
Common pitfalls in complex numbers include forgetting that i² = −1 when expanding products, misplacing the negative sign when calculating the principal argument in the third quadrant, and omitting the ±π adjustment in trigonometric identities. Always convert your calculator results to the principal range.
复数部分的常见错误包括：展开乘积时忘记 i² = −1；计算第三象限主辐角时正负号放错；在三角恒等式中遗漏 ±π 的调整。务必将计算器结果转换为主值范围。
In matrix work, err on the side of checking your determinant by recomputing it a different way (for example, expanding along a different row). Remember that matrix multiplication is not commutative: performing AB instead of BA will change the result completely. Verify inverses by multiplying the matrix by its claimed inverse.
矩阵运算中，建议用不同展开方式（如换一行展开）重新计算行列式以互相校验。牢记矩阵乘法不满足交换律：算成 AB 而非 BA 会得到完全不同的结果。验证逆矩阵时，用矩阵乘以其逆矩阵进行检验。
Finally, read the question carefully: does it ask for exact values or decimal approximations? Does it require radians or degrees? Does it want the answer in the form a + bi or in modulus–argument form? Answering the precise question asked is the simplest way to avoid losing easy marks.
最后，仔细审题：题目要求精确值还是近似小数？用弧度制还是角度制？答案要求写成 a + bi 形式还是模–辐角形式？精准回应用题设要求，是避免丢分的最简单方法。
Regular practice with this support pack alongside past exam papers is the most reliable route to success. Revisit each section after a short interval to reinforce the techniques, and build a personal checklist of the common pitfalls listed above. Stay consistent, and you will be fully prepared for the 9665 examination.
定期使用本支持包并结合历年真题进行练习，是通向成功的最可靠路径。每隔一段时间回看各节以强化方法技巧，并建立个人易错清单（参照上文列出的常见陷阱）。保持稳定节奏，你必将为 9665 考试做好充分准备。
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