AS Further Mathematics: Walkthrough of Typical Problems | AS 进阶数学：典型例题详解

📚 AS Further Mathematics: Walkthrough of Typical Problems | AS 进阶数学：典型例题详解

In AS Further Mathematics, students encounter more abstract and algebraically intensive topics than in the pure core. Typical exam questions revolve around complex numbers, matrices, polynomial roots, mathematical induction, and series. This article provides a structured walkthrough of eight classic problem types, each with a worked example and clear steps, to help you master the techniques required for top marks.

在 AS 进阶数学中，学生所接触的主题比纯数核心课程更为抽象且代数强度更大。典型的考题主要围绕复数、矩阵、多项式根、数学归纳法以及级数展开。本文以结构化的方式详解八种经典题型，每种题型均配有例题和清晰的解答步骤，帮助你掌握获取高分所需的技巧。

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1. Complex Number Arithmetic and Conjugates | 复数运算与共轭

en tested in basic algebraic manipulations. Remember that the conjugate of z = a + bi is z̅ is performed by multiplying numerator and denominator by the conjugate of the denominator.

复数的基础代数运算常常是考试重点。请记住，z = a + bi 的共轭为 z̅ = a - bi，进行除法时，需要将分子和分母同乘以分母的共轭。

Example: Express (z₁ + z₂) / (z₁ - z₂) in the form a + bi, where z₁ = 2 + 3i, z₂ = 1 - 4i.

例题：将 (z₁ + z₂) / (z₁ - z₂) 表示为 a + bi 的形式，其中 z₁ = 2 + 3i，z₂ = 1 - 4i。

= 3 - 21i - i - 7 = -4 - 22i. Therefore the result is (-4/50) - (22/50)i = -0.08 - 0.44i.

首先计算 z₁ + z₂ = (2+3i) + (1-4i) = 3 - i。再计算 z₁ - z₂ = (2+3i) - (1-4i) = 1 + 7i。因此该式为 (3 - i)/(1 + 7i)。分子分母同乘共轭 1 - 7i：分母 = 1² + 7² = 50；分子 = (3 - i)(1 - 7i) = 3×1 + 3×(-7i) - i×1 + i×7i = 3 - 21i - i - 7 = -4 - 22i。所以结果为 (-4/50) - (22/50)i = -0.08 - 0.44i。

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2. Modulus-Argument Form and Multiplication | 模与辐角形式及乘法

Any non-zero complex number can be written as r(cos θ + i sin θ), often abbreviated to r cis θ. The modulus r is √x² + y² x, with careful quadrant adjustment. Multiplication of two complex numbers in polar form multiplies the moduli and adds the arguments.

任何非零复数均可写成 r(cos θ + i sin θ) 的形式，常简记为 r cis θ。模 r 为 √x² + y²，辐角 θ 满足 tan θ = y/x，并需根据象限进行调整。两个复数的极坐标形式相乘时，模相乘，辐角相加。

Example: Express z = -√3 + i in modulus-argument form. Hence find the product z · 2 cis(π/6).

例题：将 z = -√3 + i 表示为模-辐角形式，并由此求 z · 2 cis(π/6)。

Here x = -√3, y = 1. Modulus |z| = √3 + 1 = 2. The reference angle α = arctan(|1/(-√3)| (5π/6). Then z · 2 cis(π/6) = (2)(2) cis(5π/6 + π/6) = 4 cis(π) = 4(cos π + i sin π) = -4.

此处 x = -√3，y = 1。模 |z| = √3 + 1 = 2。参考角 α = arctan(|1/(-√3)|) = π/6。由于 x<0, y>0，z 位于第二象限，故 θ = π - π/6 = 5π/6。所以 z = 2 cis(5π/6)。于是 z · 2 cis(π/6) = (2)(2) cis(5π/6 + π/6) = 4 cis(π) = 4(cos π + i sin π) = -4。

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3. De Moivre's Theorem for Powers and Roots | 棣莫弗定理求幂与根

De Moivre' ticularly useful for computing high powers of a complex number and for finding roots of complex equations by setting the argument over a full 2π cycle.

棣莫弗定理指出，对任意有理数 n，有 (r cis θ)ⁿ = rⁿ cis(nθ)。它在计算复数的高次幂以及通过辐角的完整 2π 周期求解复数方程的根时尤为有用。

Example: Find (1 + i)⁹ expressing the answer in rectangular form a + bi.

例题：求 (1 + i)⁹，并以直角坐标形式 a + bi 表示。

First write 1 + i in polar form: modulus = √1²+1² = √2; tan θ = 1/1 ⇒ θ = π/4. So 1 + i = √2 cis(π/4). Then (1 + i)⁹ = (√2)⁹ cis(9π/4) = (2^9/2) cis(9π/4) = 16√2 cis(π/4 + 2π) = 16√2 cis(π/4) after subtracting 2π. Now 16√2 cis(π/4) = 16√2(√2/2 + i√2/2) = 16(1 + i) = 16 + 16i.

先将 1 + i 写成极坐标形式：模 = √1²+1² = √2；tan θ = 1/1 ⇒ θ = π/4。因此 1 + i = √2 cis(π/4)。于是 (1 + i)⁹ = (√2)⁹ cis(9π/4) = (2^9/2) cis(9π/4) = 16√2 cis(π/4 + 2π) = 16√2 cis(π/4)，减去 2π 后即可。现 16√2 cis(π/4) = 16√2(√2/2 + i√2/2) = 16(1 + i) = 16 + 16i。

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4. Relationships Between Roots and Coefficients | 根与系数的关系

plex, the other is its conjugate. This principle extends to cubics and quartics, and helps determine unknown coefficients.

对于二次方程 ax² + bx + c = 0，其根为 α 和 β，则和 α+β = -b/a，积 αβ = c/a。当一根为复数时，另一根为其共轭。这一原则可推广到三次和四次方程，并帮助求解未知系数。

Example: iven that 3 + i is a root of x² - 6x + k = 0, find the real constant k and the other root.

例题：已知 3 + i 是方程 x² - 6x + k = 0 的根，求实数 k 以及另一根。

-(-6)/1 = 6. Product = (3+i)(3-i)=9+1=10. Hence k = product = 10. The other root is 3 - i.

由于系数为实数，另一根为其共轭 3 - i。根的和 = (3+i)+(3-i)=6，这与 -(-6)/1 = 6 一致。积 = (3+i)(3-i)=9+1=10。所以 k = 积 = 10，另一根为 3 - i。

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5. Matrix Multiplication and Inverse of 2×2 Matrices | 矩阵乘法与二阶逆矩阵

t det [[a, b], [c, d]] = ad - bc. The identity (AB)⁻¹ = B⁻¹A⁻¹ is useful for verification.

对于二阶矩阵，乘法不满足交换律，逆矩阵 A⁻¹ = (1/det A) [[d, -b], [-c, a]]。行列式 det [[a, b], [c, d]] = ad - bc。(AB)⁻¹ = B⁻¹A⁻¹ 常用于验证计算。

Example: , 1], [5, 3]] and B = [[3, -1], [-5, 2]]. Compute AB and find B⁻¹. Verify (AB)⁻¹ = B⁻¹A⁻¹.

例题：设 A = [[2, 1], [5, 3]]，B = [[3, -1], [-5, 2]]。计算 AB 并求 B⁻¹。验证 (AB)⁻¹ = B⁻¹A⁻¹。

ince B⁻¹=A implies A⁻¹=B) but actually A⁻¹ = B, thus B⁻¹A⁻¹ = A B = I. The identity holds.

首先， (-1)+1×2], [5×3+3×(-5), 5×(-1)+3×2]] = [[6-5, -2+2], [15-15, -5+6]] = [[1, 0], [0, 1]] = I。由于 AB = I，A 与 B 互为逆矩阵。B⁻¹ = A = [[2, 1], [5, 3]]。检验：(AB)⁻¹ = I⁻¹ = I；B⁻¹A⁻¹ = A A⁻¹ (因 B⁻¹=A 推出 A⁻¹=B)，实际上 A⁻¹ = B，故 B⁻¹A⁻¹ = A B = I。等式成立。

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6. Linear Transformations Represented by Matrices | 矩阵表示的线性变换

by the position vector. Invariant lines can be found by solving M v = λ v or M v = v + t.

矩阵可以表示旋转、反射、伸缩等几何变换。点的像通过将矩阵乘以位置向量得到。不变直线可通过求解 M v = λ v 或 M v = v + t 得出。

Example: ents a rotation of 90° anticlockwise about the origin. Find the image of the point (3, 4) and the equation of any invariant line passing through the origin.

例题：矩阵 R = [[0, -1], [1, 0]] 表示绕原点逆时针旋转 90°。求点 (3, 4) 的像以及任何经过原点的不变直线方程。

gin. Thus there is no invariant line of the form M v = λ v (i.e., an eigenline). Alternatively, looking for an invariant line as a whole set where points map to other points on the same line: solving y = mx gives? Already no solution. Hence R has no real invariant line through origin; the only invariant 'line' is the whole plane? Actually rotation by 90° has no invariant lines except trivial. The answer is none.

将 R 作用于 (3,4)：R[[3],[4]] = [[0×3 + (-1)×4], [1×3 + 0×4]] = [[-4], [3]]，像为 (-4, 3)。对于过原点的不变直线，解 R[[x],[y]] = λ [[x],[y]]：[[-y],[x]] = [[λx],[λy]]，得 -y = λx, x = λy。将 λ = x/y（y≠0）代入第一式：-y = (x/y)x → -y² = x² → x² + y² = 0，仅有原点。因此不存在形如 M v = λ v 的不变直线（即特征线）。因此，90°旋转没有过原点的实不变直线；答案为无。

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7. Proof by Induction for Summation | 求和公式的数学归纳法证明

pothesis (assume true for n=k), and inductive step (prove for n=k+1 using the hypothesis).

归纳法是证明涉及前 n 项和公式的标准方法。关键步骤为：基础情形 (n=1)，归纳假设 (假设 n=k 成立)，以及归纳步骤 (利用假设证明 n=k+1 成立)。

Example: Prove that Σ(r=1 to n) r(r+1) = n(n+1)(n+2)/3.

例题：证明 Σ(r=1 to n) r(r+1) = n(n+1)(n+2)/3。

ch is the RHS with n=k+1. Hence P(k+1) holds, so by mathematical induction, the statement is true for all natural numbers n.

设 P(n) 为所述命题。基础：n=1 → 左=1×2=2；右=1×2×3/3=2，成立。假设 P(k) 成立：Σ(r=1 to k) r(r+1) = k(k+1)(k+2)/3。对于 n=k+1：左 = Σ(r=1 to k) r(r+1) + (k+1)(k+2) = k(k+1)(k+2)/3 + (k+1)(k+2)。提取因子 (k+1)(k+2)：= (k+1)(k+2)[k/3 + 1] = (k+1)(k+2)[(k+3)/3] = (k+1)(k+2)(k+3)/3，这正是 n=k+1 时的右边。因此 P(k+1) 成立，根据数学归纳法，该命题对所有自然数 n 成立。

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8. Summation of Series Using Standard Results | 利用标准公式求级数和

r = n(n+1)/2, Σ r² = n(n+1)(2n+1)/6, Σ r³ = n²(n+1)²/4. Linearity allows combining these.

很多考题要求通过将多项式的有限和拆分为 r, r², r³ 的求和来计算。标准结果是：Σ r = n(n+1)/2，Σ r² = n(n+1)(2n+1)/6，Σ r³ = n²(n+1)²/4。线性性质允许我们对其进行组合。

Example: Find the sum of Σ(n=1 to 20) (3n² - 2n + 5).

例题：求 Σ(n=1 to 20) (3n² - 2n + 5) 的值。

. Σ n = 20×21/2 = 210. Σ 1 = 20. So sum = 3×2870 - 2×210 + 5×20 = 8610 - 420 + 100 = 8290.

拆分：= 3 Σ n² - 2 Σ n + 5 Σ 1，n 从 1 到 20。Σ n² = 20×21×41/6 = 2870。Σ n = 20×21/2 = 210。Σ 1 = 20。因此总和 = 3×2870 - 2×210 + 5×20 = 8610 - 420 + 100 = 8290。

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