AS CAIE Further Mathematics: Case Study in Action | AS CAIE 进阶数学：案例分析实战演练

📚 AS CAIE Further Mathematics: Case Study in Action | AS CAIE 进阶数学：案例分析实战演练

e AS CAIE Further Mathematics syllabus: complex numbers, matrix transformations, De Moivre' tep, you will see how different chapters of the course connect in exam-style questions and learn to avoid common pitfalls.

本复习资料将通过一个统一的案例分析，串联 AS CAIE 进阶数学大纲中的核心主题：复数、矩阵变换、棣莫弗定理以及多项式根。整个案例围绕复数 α = 1 + i 展开。跟随案例逐步推进，你将看到课程中不同章节如何在考试题型中相互联系，并学会避开常见陷阱。

1. Case Overview | 案例概览

t multiplication by α as a linear transformation in the plane and find its matrix. We then use De Moivre' quiring that α be a root of a real-coefficient cubic. This single number thus acts as a bridge between matrices, complex numbers, and algebra.

本案例围绕复数 α = 1 + i 构建。我们首先将乘以 α 解释为平面上的线性变换，并求其矩阵。接着使用棣莫弗定理表示 α 的幂，并通过要求 α 是一个实系数三次方程的根，将 α 与多项式方程联系起来。这样一个单一数字充当了矩阵、复数与代数之间的桥梁。

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2. Step 1: Matrix Representation of T | 步骤一：T 的矩阵表示

) = x + iy + ix - y = (x - y) + (x + y)i. The image of (x, y) is therefore (x − y, x + y).

考虑复平面上的变换 T(z) = αz，其中 α = 1 + i。令 z = x + iy，计算 αz = (1 + i)(x + iy) = x + iy + ix − y = (x − y) + (x + y)i。因此点 (x, y) 的像是 (x − y, x + y)。

The matrix representing T is

表示 T 的矩阵为

1	−1
1	1

so that

使得

[x'; y'] = [1 −1; 1 1] · [x; y]

For example, the point with coordinates (2, 3) is mapped to (2 − 3, 2 + 3) = (−1, 5).

例如，坐标为 (2, 3) 的点被映射到 (2 − 3, 2 + 3) = (−1, 5)。

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3. Step 2: Geometric Interpretation | 步骤二：几何意义

The matrix M = [[1, −1], [1, 1]] can be written as √2 · [[1/√2, −1/√2], [1/√2, 1/√2]] = √2 [[cos 45°, −sin 45°], [sin 45°, cos 45°]]. Thus T is the composition of a rotation by 45° anticlockwise and an enlargement by scale factor √2.

矩阵 M = [[1, −1], [1, 1]] 可写作 √2 · [[1/√2, −1/√2], [1/√2, 1/√2]] = √2 [[cos 45°, −sin 45°], [sin 45°, cos 45°]]。因此 T 是逆时针旋转 45° 与以 √2 为比例因子的放大的复合。

(cos 45° + i sin 45°). Multiplying by α therefore rotates a complex number by 45° and stretches its modulus by √2.

这与 α 的极形式吻合：α = √2(cos 45° + i sin 45°)。因此乘以 α 将复数旋转 45° 并将其模拉伸 √2 倍。

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4. Step 3: Powers of α via De Moivre | 步骤三：利用棣莫弗定理求 α 的幂

Using the polar form, De Moivre's theorem gives αⁿ = (√2)ⁿ (cos(n·45°) + i sin(n·45°)). Writing θ = n·45°, the real and imaginary parts are (√2)ⁿ cos θ and (√2)ⁿ sin θ.

利用极形式，棣莫弗定理给出 αⁿ = (√2)ⁿ (cos(n·45°) + i sin(n·45°))。记 θ = n·45°，实部和虚部分别为 (√2)ⁿ cos θ 和 (√2)ⁿ sin θ。

For instance, α³ = (√2)³ (cos 135° + i sin 135°) = 2√2 (−1/√2 + i(1/√2 = −2 + 2i. This formula will later allow us to compute powers of the matrix M efficiently.

例如，α³ = (√2)³ (cos 135° + i sin 135°) = 2√2 (−1/√2 + i(1/√2)) = −2 + 2i。这一公式随后将使我们能高效地计算矩阵 M 的幂。

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5. Step 4: Building a Polynomial with Root α | 步骤四：构造以 α 为根的多项式

a root. The quadratic factor with these roots is (z − (1 + i))(z − (1 − i)) = z² − 2z + 2.

假设一个实系数三次多项式有根 α = 1 + i。由于系数为实数，其共轭 α* = 1 − i 也必为根。以这两个数为根的二次因式为 (z − (1 + i))(z − (1 − i)) = z² − 2z + 2。

− 5z² + 8z − 6. This polynomial has roots 3, 1 + i, 1 − i, and its coefficients are real.

我们可以通过乘以一个线性因式得到完整的三次式，例如 (z − k) 其中 k 为实数。取 k = 3，得到三次多项式 f(z) = (z − 3)(z² − 2z + 2) = z³ − 5z² + 8z − 6。该多项式的根为 3、1 + i、1 − i，且系数为实数。

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6. Step 5: Relationships Between Roots and Coefficients | 步骤五：根与系数的关系

For the cubic f(z) = z³ − 5z² + 8z − 6, Vieta' ese match −(coefficient of z²), coefficient of z, and −constant term respectively, confirming the result.

对于三次式 f(z) = z³ − 5z² + 8z − 6，韦达定理给出：根之和 = 3 + (1 + i) + (1 − i) = 5，两两根积之和 = 3(1 + i) + 3(1 − i) + (1 + i)(1 − i) = 6 + 2 = 8，根之积 = 3·(1 + i)·(1 − i) = 3·2 = 6。这些分别与 −(z² 系数)、z 系数和 −常数项相匹配，验证了结果。

In an exam, you may be asked to form an equation from given roots, or to deduce unknown coefficients using these relationships. This step links the complex number α directly to algebraic manipulations with polynomials.

在考试中，你可能被要求由给定的根构造方程，或利用这些关系推导未知系数。本步骤将复数 α 与多项式的代数操作直接联系起来。

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7. Step 6: Matrix Powers and De Moivre | 步骤六：矩阵的幂与棣莫弗定理

polar form of αⁿ, Mⁿ = [[Re(αⁿ), −Im(αⁿ)], [Im(αⁿ), Re(αⁿ)]]. Using De Moivre: Re(αⁿ) = (√2)ⁿ cos(n·45°), Im(αⁿ) = (√2)ⁿ sin(n·45°). Therefore

由于 Tⁿ(z) = αⁿz，Tⁿ 对应的矩阵为 Mⁿ。由 αⁿ 的极形式，Mⁿ = [[Re(αⁿ), −Im(αⁿ)], [Im(αⁿ), Re(αⁿ)]]。利用棣莫弗定理：Re(αⁿ) = (√2)ⁿ cos(n·45°)，Im(αⁿ) = (√2)ⁿ sin(n·45°)。因此

Mⁿ = (√2)ⁿ [[cos(n·45°), −sin(n·45°)], [sin(n·45°), cos(n·45°)]]

s compute Mⁿ quickly for any n, avoiding repeated matrix multiplication. For n = 4, M⁴ = (√2 80°]] = 4[[−1, 0], [0, −1]] = [[−4, 0], [0, −4]], which agrees with (1 + i)⁴ = (2i)² = −4.

该公式使我们能快速计算任意 n 的 Mⁿ，避免重复矩阵乘法。对于 n = 4，M⁴ = (√2⁴ [[cos 180°, −sin 180°], [sin 180°, cos 180°]] = 4[[−1, 0], [0, −1]] = [[−4, 0], [0, −4]]，与 (1 + i)⁴ = (2i)² = −4 吻合。

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8. Step 7: Integrated Exam-Style Question | 步骤七：综合考试题型

Let us now consider a typical Cambridge exam question that draws on several of these ideas.

现在我们来考虑一道典型的剑桥考试题，它综合了上述多个知识点。

Question: The complex number α = 1 + i. (a) Express α in polar form. (b) A linear transformation T: ℝ² → ℝ n by α. Write down M. (c) Find the image of the point (1, 2) under T. (d) Show that Mⁿ = (√2 oefficients and α as one root. Determine p, q, and r given that the sum of the roots is 1.

题目：复数 α = 1 + i。(a) 将 α 表示为极形式。(b) 线性变换 T: ℝ² → ℝ² 由 T([x; y]) = M[x; y] 给出，其中 M 是表示乘以 α 的矩阵。写出 M。(c) 求点 (1, 2) 在 T 下的像。(d) 证明 Mⁿ = (√2)ⁿ [[cos(nπ/4), −sin(nπ/4)], [sin(nπ/4), cos(nπ/4)]]。(e) 三次方程 z³ + pz² + qz + r = 0 具有实系数，且 α 为其一根。已知根之和为 1，求 p、q 和 r。

This single question assesses polar form, matrix representation, geometry, De Moivre's theorem, and root–coefficient relationships—exactly the skills practised in our case study.

这一道题同时考查了极形式、矩阵表示、几何、棣莫弗定理以及根与系数关系——正是我们的案例所训练的技能。

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9. Common Mistakes and How to Avoid Them | 常见错误与避免方法

Confusing rotation and enlargement: Students often forget that the matrix [[1, −1], [1, 1]] involves both a rotation and a scaling, because its determinant is 2, not 1. Always check whether the matrix is orthogonal before claiming it represents a pure rotation.

混淆旋转与放大： 学生常忘记矩阵 [[1, −1], [1, 1]] 同时包含旋转和缩放，因为其行列式为 2 而非 1。在声称矩阵表示纯旋转之前，务必检查它是否正交。

Real-coefficient condition: When a complex number is a root of a real polynomial, its conjugate must also be a root. Missing this leads to incorrect factorisation and wrong coefficients.

实系数条件： 当复数是实系数多项式的根时，其共轭也必为根。忽略这一点会得到错误的因式分解和系数。

Angle units: In De Moivre' ians as required by the question. Here 45° = π/4 radians. Mixing units is a classic error.

角度单位： 在棣莫弗定理中，确保角度单位正确——题目要求为度或弧度。此处 45° = π/4 弧度。混用单位是经典错误。

Matrix multiplication order: When composing transformations, the matrix for Tⁿ is the nth power of M, not n multiplied by M. Use powers, not scalar multiplication, unless the transformation is a pure enlargement.

矩阵乘法顺序： 组合变换时，Tⁿ 的矩阵是 M 的 n 次幂，而非 M 乘以 n。除非变换是纯放大，否则应使用幂运算，而非标量乘法。

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10. Summary and Extension | 总结与拓展

We have seen how the single complex number 1 + i appears in three different AS Further Mathematics contexts: as the multiplier in a linear transformation, as the base for powers via De Moivre, and as a root of a real-coefficient cubic. Mastering these connections prepares you for synoptic questions that test multiple topics at once.

我们已看到单一复数 1 + i 如何出现在 AS 进阶数学的三种不同情境中：作为线性变换的乘子、作为棣莫弗定理幂运算的底数、以及作为实系数三次方程的根。掌握这些联系能让你应对同时考查多个主题的综合类考题。

s of the form z → αz + β represent a combination of rotation, enlargement, and translation—an idea that leads to the study of fractals and iterative maps.

作为拓展，你可以探索矩阵 M 与复数特征值（A2 引入）的关系，或形如 z → αz + β 的变换如何表示旋转、放大和平移的组合——这一思路通向分形与迭代映射的研究。

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