Answers to Exercises - Further Pure 4 | 练习答案 - 进阶纯数学 4

📚 Answers to Exercises - Further Pure 4 | 练习答案 - 进阶纯数学 4

This article provides full worked solutions to a selection of typical AQA Further Pure 4 exercises. Each section gives one exam-style question, followed by a step-by-step answer that links the method to the key FP4 content: complex numbers, matrices, polar coordinates, hyperbolic functions, differential equations, series and vectors.

本文提供一组 AQA 进阶纯数学 4 典型练习的完整解答。每节给出一道考试风格的问题，并逐步给出答案，将解题方法与 FP4 的核心内容结合：复数、矩阵、极坐标、双曲函数、微分方程、级数和向量。

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1. De Moivre's Theorem and Roots of Unity | 棣莫弗定理与单位根

Question: Use de Moivre's theorem to express cos 3θ in terms of cos θ, and find the three cube roots of 1 in exact polar form.

问题：利用棣莫弗定理用 cos θ 表示 cos 3θ，并求 1 的三个立方根，结果用极坐标精确形式表示。

hen z³ = cos 3θ + i sin 3θ. Expanding (cos θ + i sin θ)³ using the binomial theorem gives:

设 z = cos θ + i sin θ。则 z³ = cos 3θ + i sin 3θ。利用二项式定理展开 (cos θ + i sin θ)³ 得：

cos 3θ + i sin 3θ = c³ + 3c²(i s) + 3c(i s)² + (i s)³ = c³ − 3c s² + i(3c²s − s³)

θ. Equating the real parts gives cos 3θ = c³ − 3c s² = c³ − 3c(1 − c²) = 4c³ − 3c. Hence:

其中 c = cos θ，s = sin θ。比较实部得 cos 3θ = c³ − 3c s² = c³ − 3c(1 − c²) = 4c³ − 3c。因此：

cos 3θ = 4 cos³θ − 3 cos θ

For the cube roots of 1, solve z³ = 1 = e^2πki for k = 0, 1, 2. The roots are z = e^2πki/3. When k = 0, z = 1. When k = 1, z = e^2πi/3 = cos(2π/3) + i sin(2π/3) = −1/2 + i√3/2. When k = 2, z = e^4πi/3 = −1/2 − i√3/2.

求 1 的立方根，解 z³ = 1 = e^2πki，其中 k = 0, 1, 2。根为 z = e^2πki/3。当 k = 0 时，z = 1。当 k = 1 时，z = e^2πi/3 = cos(2π/3) + i sin(2π/3) = −1/2 + i√3/2。当 k = 2 时，z = e^4πi/3 = −1/2 − i√3/2。

The three roots form an equilateral triangle on the unit circle, and their sum is 0. This is a standard result for roots of unity.

这三个根在单位圆上构成等边三角形，它们的和为 0。这是单位根的标准结论。

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2. Complex Loci and Transformations | 复数轨迹与变换

Question: Sketch the locus of z satisfying |z − (2 + i)| = 3. Find the image of this locus under the transformation w = iz + 1.

问题：画出满足 |z − (2 + i)| = 3 的 z 的轨迹。求该轨迹在变换 w = iz + 1 下的像。

The equation |z − (2 + i)| i, or (2, 1), and radius 3. This is because the modulus of a complex difference gives the distance between two points in the Argand diagram.

方程 |z − (2 + i)| = 3 表示以 2 + i，即 (2, 1) 为圆心、半径为 3 的圆。这是因为两个复数之差的模表示阿尔冈图中两点之间的距离。

1/i = −i, this gives z = −i(w − 1) = −i w + i. Substituting into the original locus gives:

要求变换 w = iz + 1 下的像，将 z 表示为 z = (w − 1)/i。因为 1/i = −i，所以 z = −i(w − 1) = −i w + i。代入原轨迹得：

|−i w + i − 2 − i| = |−i w − 2| = |i w + 2| = |i(w − 2i)| = |w − 2i|

Therefore the image satisfies |w − 2i| = 3. This is a circle centred at 2i, or (0, 2), with the same radius 3.

因此像满足 |w − 2i| = 3。这是一个以 2i，即 (0, 2) 为圆心、半径仍为 3 的圆。

ion through 90° anticlockwise about the origin, followed by a translation by 1. Circle centres and radii transform in a predictable way.

变换 w = iz + 1 表示先绕原点逆时针旋转 90°，再平移 1。圆心的变换和半径的变化是可预测的。

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3. Eigenvalues and Diagonalisation of 2×2 Matrices | 2×2 矩阵的特征值与对角化

find the eigenvalues and eigenvectors. Hence express A in the form PDP⁻¹, where D is diagonal, and find Aⁿ.

问题：对于矩阵 A = [3 1; 0 2]，求特征值和特征向量。由此将 A 表示为 PDP⁻¹ 的形式，其中 D 为对角矩阵，并求 Aⁿ。

ic equation is det(A − λI) = (3 − λ)(2 − λ) = 0, so the eigenvalues are λ₁ = 3 and λ₂ = 2.

特征方程为 det(A − λI) = (3 − λ)(2 − λ) = 0，因此特征值为 λ₁ = 3 和 λ₂ = 2。

he matrix A − 3I = [0 1; 0 −1] gives y = 0, with x free, so an eigenvector is v₁ = [1; 0].

对于 λ = 3，解 (A − 3I)v = 0。矩阵 A − 3I = [0 1; 0 −1] 得 y = 0，x 任意，因此一个特征向量为 v₁ = [1; 0]。

lve (A − 2I)v = 0. The matrix A − 2I = [1 1; 0 0] gives x + y = 0, so choose v₂ = [−1; 1].

对于 λ = 2，解 (A − 2I)v = 0。矩阵 A − 2I = [1 1; 0 0] 得 x + y = 0，因此可取 v₂ = [−1; 1]。

D = [3 0; 0 2]. The inverse of P is P⁻¹ = [1 1; 0 1]. Then A = P D P⁻¹, so Aⁿ = P Dⁿ P⁻¹.

因此 P = [1 −1; 0 1]，D = [3 0; 0 2]。P 的逆矩阵为 P⁻¹ = [1 1; 0 1]。于是 A = P D P⁻¹，所以 Aⁿ = P Dⁿ P⁻¹。

Carrying out the matrix multiplication gives:

进行矩阵乘法得：

Aⁿ = [3ⁿ 3ⁿ − 2ⁿ; 0 2ⁿ]

This formula works for any positive integer n. Diagonalisation is useful because powers of a diagonal matrix are simple to compute.

该公式对任意正整数 n 都成立。对角化很有用，因为对角矩阵的幂很容易计算。

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4. Polar Coordinates: Area Enclosed by a Curve | 极坐标：曲线所围面积

Question: The curve C has polar equation r = 2 + cos θ. Find the area enclosed by C.

问题：曲线 C 的极坐标方程为 r = 2 + cos θ。求 C 所围成的面积。

The area enclosed by a polar curve r

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