📚 AS Eduqas Further Maths: Case Study Walkthroughs | AS Eduqas 进阶数学：案例分析实战演练
In AS Eduqas Further Mathematics, students encounter a range of advanced pure topics including complex numbers, matrices, hyperbolic functions, series, and vectors. To excel, it’s crucial to apply theoretical knowledge to problem-solving. This article presents detailed case study walkthroughs, breaking down step-by-step solutions to typical exam-style questions. Each case study is explained in both English and Chinese to reinforce understanding.
在AS Eduqas进阶数学中，学生需要掌握一系列高级纯数主题，包括复数、矩阵、双曲函数、级数和向量。要取得优异成绩，将理论知识应用于解题至关重要。本文提供详细的案例分析演练，逐步解析典型考题的解题过程。每个案例均用中英双语解释以加深理解。
1. Complex Roots: Solving z⁴ = -16 | 复数根：求解 z⁴ = -16
Find all roots of the equation z⁴ = -16, giving answers in the form reiθ where r > 0 and -π < θ ≤ π.
求方程 z⁴ = -16 的所有根，结果表示为 reiθ 形式，其中 r > 0 且 -π < θ ≤ π。
Express -16 in polar form: modulus is 16, argument is π because -16 lies on the negative real axis. Hence -16 = 16eiπ.
将 -16 表示为极坐标形式：模为 16，辐角为 π，因为 -16 位于负实轴上。因此 -16 = 16eiπ。
Let z = reiθ. Then z⁴ = r⁴ei4θ = 16ei(π + 2kπ) for integer k. Equate moduli: r⁴ = 16 → r = 2. Equate arguments: 4θ = π + 2kπ → θ = (π + 2kπ)/4.
设 z = reiθ，则 z⁴ = r⁴ei4θ = 16ei(π + 2kπ)（k 为整数）。比较模长：r⁴ = 16 → r = 2。比较辐角：4θ = π + 2kπ → θ = (π + 2kπ)/4。
Now substitute k = 0, 1, 2, 3 to generate four distinct roots, then adjust angles into the range (-π, π].
代入 k = 0, 1, 2, 3 得到四个不同的根，然后将角度调整到区间 (-π, π]。
k=0: θ = π/4 → z₀ = 2eiπ/4
k=1: θ = 3π/4 → z₁ = 2ei3π/4
k=2: θ = 5π/4, which exceeds π, so subtract 2π: -3π/4 → z₂ = 2e-i3π/4
k=3: θ = 7π/4, equivalent to -π/4 → z₃ = 2e-iπ/4
k=0：θ = π/4 → z₀ = 2eiπ/4
k=1：θ = 3π/4 → z₁ = 2ei3π/4
k=2：θ = 5π/4，超出范围，减去2π得 -3π/4 → z₂ = 2e-i3π/4
k=3：θ = 7π/4，等价于 -π/4 → z₃ = 2e-iπ/4
The four roots are therefore 2eiπ/4, 2ei3π/4, 2e-i3π/4, 2e-iπ/4. In Cartesian form they are √2 + i√2, -√2 + i√2, -√2 – i√2, √2 – i√2.
因此四个根为 2eiπ/4, 2ei3π/4, 2e-i3π/4, 2e-iπ/4。用笛卡尔坐标表示为 √2 + i√2, -√2 + i√2, -√2 – i√2, √2 – i√2。
2. Matrix Characteristic Equation and Inverse | 矩阵的特征方程与逆矩阵
Given matrix A =
| 2 | 1 |
| -1 | 3 |
, find A² and determine constants p and q such that A² + pA + qI = 0. Hence obtain A⁻¹.
已知矩阵 A =
| 2 | 1 |
| -1 | 3 |
，求 A²，并确定常数 p 和 q 使得 A² + pA + qI = 0。由此求出 A⁻¹。
First compute A² = A × A:
首先计算 A² = A × A：
A² =
| 2 | 1 |
| -1 | 3 |
×
| 2 | 1 |
| -1 | 3 |
=
| 2×2 + 1×(-1) | 2×1 + 1×3 |
| (-1)×2 + 3×(-1) | (-1)×1 + 3×3 |
=
| 3 | 5 |
| -5 | 8 |
Now set up the equation A² + pA + qI = 0. Substituting A², pA and qI gives:
现在建立方程 A² + pA + qI = 0。代入 A²、pA 和 qI 得：
| 3 | 5 |
| -5 | 8 |
+
| 2p | p |
| -p | 3p |
+
| q | 0 |
| 0 | q |
=
| 0 | 0 |
| 0 | 0 |
Equate corresponding entries: (1,1): 3 + 2p + q = 0; (1,2): 5 + p = 0 → p = -5; (2,1): -5 – p = 0 → confirms p = -5; (2,2): 8 + 3p + q = 0. Substitute p = -5 into first equation: 3 – 10 + q = 0 → q = 7. The second equation also holds: 8 -15 + 7 = 0.
比较对应元素：(1,1)：3 + 2p + q = 0；(1,2)：5 + p = 0 → p = -5；(2,1)：-5 – p = 0 → 确认 p = -5；(2,2)：8 + 3p + q = 0。将 p = -5 代入第一个方程：3 – 10 + q = 0 → q = 7。第二个方程也成立：8 -15 + 7 = 0。
Thus p = -5, q = 7. The matrix satisfies A² – 5A + 7I = 0. Rearranging: A² – 5A = -7I → A(A – 5I) = -7I → A [-(1/7)(A – 5I)] = I. Therefore A⁻¹ = -(1/7)(A – 5I) = (1/7)(5I – A).
因此 p = -5，q = 7。矩阵满足 A² – 5A + 7I = 0。变形得：A² – 5A = -7I → A(A – 5I) = -7I → A [-(1/7)(A – 5I)] = I。所以 A⁻¹ = -(1/7)(A – 5I) = (1/7)(5I – A)。
Compute 5I – A =
| 5-2 | 0-1 |
| 0-(-1) | 5-3 |
=
| 3 | -1 |
| 1 | 2 |
. Hence A⁻¹ = (1/7)
| 3 | -1 |
| 1 | 2 |
.
计算 5I – A =
| 5-2 | 0-1 |
| 0-(-1) | 5-3 |
=
| 3 | -1 |
| 1 | 2 |
。因此 A⁻¹ = (1/7)
| 3 | -1 |
| 1 | 2 |
。
3. Hyperbolic Identity and Equation | 双曲恒等式与方程
Prove that cosh²x – sinh²x = 1. Hence solve the equation 2cosh²x + 3sinh x = 5.
证明 cosh²x – sinh²x = 1。由此求解方程 2cosh²x + 3sinh x = 5。
Recall definitions: cosh x = (eˣ + e⁻ˣ)/2, sinh x = (eˣ – e⁻ˣ)/2. Then cosh²x – sinh²x = [(eˣ+e⁻ˣ)/2]² – [(eˣ-e⁻ˣ)/2]² = (e²ˣ+2+e⁻²ˣ)/4 – (e²ˣ-2+e⁻²ˣ)/4 = (4)/4 = 1.
回忆定义：cosh x = (eˣ + e⁻ˣ)/2, sinh x = (eˣ – e⁻ˣ)/2。那么 cosh²x – sinh²x = [(eˣ+e⁻ˣ)/2]² – [(eˣ-e⁻ˣ)/2]² = (e²ˣ+2+e⁻²ˣ)/4 – (e²ˣ-2+e⁻²ˣ)/4 = 4/4 = 1。
Now use the identity to rewrite cosh²x = 1 + sinh²x. Substitute into the equation: 2(1 + sinh²x) + 3sinh x = 5 → 2sinh²x + 3sinh x + 2 – 5 = 0 → 2sinh²x + 3sinh x – 3 = 0.
现在使用恒等式将 cosh²x 改写为 1 + sinh²x。代入方程：2(1 + sinh²x) + 3sinh x = 5 → 2sinh²x + 3sinh x + 2 – 5 = 0 → 2sinh²x + 3sinh x – 3 = 0。
Treat this as a quadratic in u = sinh x: 2u² + 3u – 3 = 0. The discriminant: Δ = 3² – 4×2×(-3) = 9 + 24 = 33. Thus u = [-3 ± √33] / 4.
将其视为关于 u = sinh x 的二次方程：2u² + 3u – 3 = 0。判别式：Δ = 3² – 4×2×(-3) = 9 + 24 = 33。因此 u = [-3 ± √33] / 4。
Sinh x takes all real values; check which solutions are valid. √33 ≈ 5.744, so u₁ = (-3+5.744)/4 = 0.686, u₂ = (-3-5.744)/4 = -2.186. Both are real, so we accept both.
Sinh x 取所有实数；检查哪些解有效。√33 ≈ 5.744，所以 u₁ = (-3+5.744)/4 = 0.686，u₂ = (-3-5.744)/4 = -2.186。两者均为实数，故均接受。
Hence sinh x = 0.686 or sinh x = -2.186. Solve using x = arsinh(u) = ln(u + √(u²+1)). For u=0.686: x₁ = ln(0.686 + √(0.686²+1)) = ln(0.686 + √1.4706) = ln(0.686+1.2127) = ln(1.8987) ≈ 0.641. For u=-2.186: x₂ = ln(-2.186 + √(5.778+1)) = ln(-2.186 + √6.778) = ln(-2.186+2.603) = ln(0.417) ≈ -0.874. These are the real solutions.
因此 sinh x = 0.686 或 sinh x = -2.186。使用 x = arsinh(u) = ln(u + √(u²+1)) 求解。对于 u=0.686：x₁ = ln(0.686 + √(0.686²+1)) = ln(0.686 + √1.4706) = ln(0.686+1.2127) = ln(1.8987) ≈ 0.641。对于 u=-2.186：x₂ = ln(-2.186 + √(5.778+1)) = ln(-2.186 + √6.778) = ln(-2.186+2.603) = ln(0.417) ≈ -0.874。这些是实数解。
4. Summation of Series Using Standard Results | 利用标准结果对级数求和
Evaluate Σ(r=1 to n) (r² + 4r – 5).
计算 Σ（r=1 到 n）(r² + 4r – 5)。
Use standard formulas: Σr = n(n+1)/2, Σr² = n(n+1)(2n+1)/6, and Σ1 = n. Split the sum:
使用标准公式
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