📚 AS Further Mathematics: Last-Minute Revision Notes | AS 进阶数学：考前冲刺笔记
This revision guide summarises the key topics in AS Further Mathematics, including complex numbers, matrices, series and proof techniques. Use these notes to consolidate your understanding and boost your exam performance.
本考前冲刺笔记涵盖了 AS 进阶数学的核心专题，包括复数、矩阵、级数与证明方法。利用这些笔记巩固理解、提升考试表现。
1. Complex Numbers | 复数
A complex number is expressed as z = a + bi, where a = Re(z), b = Im(z). The imaginary unit i has the property i2 = –1.
复数表示为 z = a + bi，其中 a 是实部，b 是虚部。虚数单位 i 满足 i2 = –1。
i2 = –1
The complex conjugate of z is z* = a – bi. The modulus is |z| = √(a2 + b2). Division is carried out by multiplying numerator and denominator by the conjugate of the denominator.
共轭复数 z* = a – bi。模为 |z| = √(a2 + b2)。复数的除法通过分子分母同乘分母的共轭来实现。
On the Argand diagram, a + bi corresponds to the point (a, b) or a vector. The argument θ satisfies tan θ = b/a.
在阿甘特图上，a + bi 对应于点 (a, b) 或向量。幅角 θ 满足 tan θ = b/a。
Polar form: z = r(cos θ + i sin θ) where r = |z|. Using Euler’s identity eiθ = cos θ + i sin θ, we can write:
极坐标形式：z = r(cos θ + i sin θ)，其中 r = |z|。由欧拉恒等式 eiθ = cos θ + i sin θ，可得：
z = r eiθ
De Moivre’s theorem: (cos θ + i sin θ)n = cos nθ + i sin nθ. This is essential for raising complex numbers to powers and extracting roots.
德莫弗定理：(cos θ + i sin θ)n = cos nθ + i sin nθ。这是求复数的幂与根的核心工具。
To find n-th roots, express z in polar form with an additional 2kπ: z = r[cos(θ + 2kπ) + i sin(θ + 2kπ)], then apply De Moivre’s theorem for k = 0, 1, …, n–1.
求 n 次方根时，将 z 写成带 2kπ 的极坐标形式：z = r[cos(θ + 2kπ) + i sin(θ + 2kπ)]，然后对 k = 0,1,…,n–1 应用德莫弗定理。
2. Matrices & Transformations | 矩阵与变换
A 2×2 matrix is written as M = [a, b; c, d]. The determinant is det(M) = ad – bc. If det(M) ≠ 0, the inverse matrix exists:
2×2 矩阵记作 M = [a, b; c, d]。行列式为 det(M) = ad – bc。若行列式不为零，则存在逆矩阵：
M–1 = 1/(ad–bc) [d, –b; –c, a]
Matrix multiplication is row-by-column. For AB, the element in row i, column j is the dot product of row i of A with column j of B. Matrix multiplication is not commutative.
矩阵乘法按行乘列进行。AB 中第 i 行第 j 列的元素是 A 的第 i 行与 B 的第 j 列的点积。矩阵乘法不满足交换律。
Common transformation matrices: rotation by θ anticlockwise about the origin:
常见变换矩阵：绕原点逆时针旋转 θ：
| cos θ | –sin θ |
| sin θ | cos θ |
Reflection in the x-axis: [1, 0; 0, –1]. Enlargement scale factor k: [k, 0; 0, k]. Using matrix equations, systems of linear equations can be solved: Ax = b ⇒ x = A–1b.
关于 x 轴的反射：[1, 0; 0, –1]。比例因子为 k 的放大：[k, 0; 0, k]。利用矩阵方程可解线性方程组：Ax = b ⇒ x = A–1b。
3. Roots of Polynomial Equations | 多项式方程的根
For a quadratic ax2 + bx + c = 0 with roots α and β, the sum α + β = –b/a and product αβ = c/a.
对于二次方程 ax2 + bx + c = 0，两根为 α 和 β，则和 α + β = –b/a，积 αβ = c/a。
For a cubic ax3 + bx2 + cx + d = 0 with roots α, β, γ:
对于三次方程 ax3 + bx2 + cx + d = 0，根为 α, β, γ：
∑α = –b/a, ∑αβ = c/a, αβγ = –d/a
Similar relationships extend to quartics. Using these, you can find expressions like α2 + β2 = (α+β)2 – 2αβ or symmetric sums without solving for the roots explicitly.
类似关系可推广至四次方程。利用这些关系可以求 α2 + β2 = (α+β)2 – 2αβ 等对称表达式，而不必直接求解根。
4. Summation of Series | 级数求和
Standard results for sums of natural numbers and their powers are crucial:
自然数及其幂的求和标准结果非常重要：
- ∑r=1n r = n(n+1)/2
- ∑r=1n r2 = n(n+1)(2n+1)/6
- ∑r=1n r3 = [n(n+1)/2]2
These can be combined to sum polynomial series. For example, to find ∑(r2 – 2r), separate into ∑r2 – 2∑r and apply the formulas.
这些公式可以组合来求多项式级数的和。例如，计算 ∑(r2 – 2r) 时，分离为 ∑r2 – 2∑r 然后套用公式。
For series given in terms of f(r+1) – f(r), the method of differences can collapse the sum to a few terms.
对于形如 f(r+1) – f(r) 的级数，可用差分法将和式折叠为少数几项。
5. Proof by Induction | 数学归纳法
Mathematical induction is a powerful technique for proving statements involving natural numbers n. The three steps are:
数学归纳法是证明涉及自然数 n 的命题的有效方法。三个步骤为：
- Base case: verify the statement is true for n = 1 (or the smallest relevant integer).
- Inductive hypothesis: assume the statement holds for n = k.
- Inductive step: prove that the statement then holds for n = k+1.
- 基础情形：验证 n = 1（或最小适用整数）时命题成立。
- 归纳假设：假设 n = k 时命题成立。
- 归纳步骤：证明由此可推出 n = k+1 时命题成立。
Common applications include proving summation formulas, divisibility results (e.g., 4n – 1 is divisible by 3), and statements about matrix powers.
常见应用包括证明求和公式、整除性结论（如 4n – 1 能被 3 整除）以及关于矩阵幂的命题。
6. Vectors in 3D | 三维向量
A three-dimensional vector can be written as v = xi + yj + zk or as a column vector. The magnitude is |v| = √(x2 + y2 + z2).
三维向量可写为 v = xi + yj + zk 或列向量。模长为 |v| = √(x2 + y2 + z2)。
The scalar (dot) product of two vectors a and b is a · b = a1b1 + a2b2 + a3b3 = |a||b| cos θ, where θ is the angle between them. This is used to find angles and perpendicularity.
两向量的点积（数量积）为 a · b = a1b1 + a2b2 + a3b3Published by TutorHao | Mathematics Revision Series | aleveler.com
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