📚 AS AQA Further Maths: Formula & Theorem Quick Reference Handbook | AS AQA 进阶数学：公式定理速查手册
This quick reference handbook compiles essential formulas and theorems for the AQA AS Further Mathematics (7366) specification. It covers Further Pure 1 topics including complex numbers, matrices, roots of polynomials, series, proof by induction, coordinate geometry, and inequalities. Keep it handy for revision.
本速查手册汇编了 AQA AS 进阶数学（7366）所需的核心公式与定理，涵盖 Further Pure 1 的主题，包括复数、矩阵、多项式根、级数、归纳证明、坐标几何与不等式，是考前复习的得力助手。
1. Complex Numbers – Algebra | 复数 – 代数
Complex numbers are expressed as z = x + iy, where x and y are real numbers and i² = –1.
复数表示为 z = x + iy，其中 x 和 y 是实数，且 i² = –1。
To add or subtract, combine real parts and imaginary parts separately.
进行加减时，分别合并实部和虚部。
(a + ib) + (c + id) = (a + c) + i(b + d)
Multiplication uses the distributive law and the fact that i² = –1.
乘法利用分配律以及 i² = –1。
(a + ib)(c + id) = (ac – bd) + i(ad + bc)
Division is performed by multiplying numerator and denominator by the complex conjugate of the denominator.
除法通过分子分母同乘分母的共轭复数来实现。
2. Complex Conjugate and Division | 共轭复数与除法
If z = x + iy, its complex conjugate is denoted by z* = x – iy.
若 z = x + iy，其共轭复数记为 z* = x – iy。
The product z z* = x² + y² is always a non-negative real number.
乘积 z z* = x² + y² 总是一个非负实数。
To divide z₁ = a + ib by z₂ = c + id, multiply top and bottom by z₂*:
要计算 z₁ ÷ z₂，上下同乘 z₂ 的共轭复数：
(a + ib)/(c + id) = [(a + ib)(c – id)] / (c² + d²)
The conjugate of a sum, product or quotient follows straightforward rules: (z₁ + z₂)* = z₁* + z₂*, (z₁z₂)* = z₁* z₂*, (z₁/z₂)* = z₁*/z₂*.
共轭运算满足和、积、商的规则：(z₁ + z₂)* = z₁* + z₂*, (z₁z₂)* = z₁* z₂*, (z₁/z₂)* = z₁*/z₂*。
3. Modulus, Argument and Polar Form | 模、辐角与极形式
The modulus of z = x + iy is |z| = √(x² + y²), representing the distance from the origin in the Argand diagram.
复数 z = x + iy 的模为 |z| = √(x² + y²)，表示阿干特图上到原点的距离。
The argument arg(z) is the angle θ (in radians) measured from the positive real axis to the line joining the origin to (x, y).
辐角 arg(z) 是从正实轴到连接原点与 (x, y) 的有向线段所成的角 θ（以弧度表示）。
Polar (modulus-argument) form: z = r(cos θ + i sin θ) where r = |z| and θ = arg(z).
极（模-辐角）形式：z = r(cos θ + i sin θ)，其中 r = |z|，θ = arg(z)。
Multiplication in polar form: if z₁ = r₁(cos θ₁ + i sin θ₁) and z₂ = r₂(cos θ₂ + i sin θ₂), then z₁z₂ = r₁r₂[cos(θ₁+θ₂) + i sin(θ₁+θ₂)].
极形式乘法：若 z₁ = r₁(cos θ₁ + i sin θ₁) 且 z₂ = r₂(cos θ₂ + i sin θ₂)，则 z₁z₂ = r₁r₂[cos(θ₁+θ₂) + i sin(θ₁+θ₂)]。
4. De Moivre’s Theorem | 棣莫弗定理
For any integer n, De Moivre’s theorem states: (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ).
对于任意整数 n，棣莫弗定理：(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)。
This is particularly useful for raising complex numbers in polar form to integer powers:
该定理特别适合将极形式复数进行整数次幂运算：
[r(cos θ + i sin θ)]ⁿ = rⁿ[cos(nθ) + i sin(nθ)]
The theorem can also be used to derive trigonometric identities for multiple angles or to find nth roots of a complex number (covered further at A2 level).
该定理也可用于推导多倍角三角恒等式，或求复数的 n 次方根（A2 阶段深入）。
5. Matrices – Operations | 矩阵运算
A 2×2 matrix is written as:
一个 2×2 矩阵写作：
| a | b |
| c | d |
Addition and subtraction are element-wise: each entry is added or subtracted with the corresponding entry.
加减法逐元素进行：每个元素与对应位置的元素相加或相减。
Scalar multiplication multiplies every entry by the scalar.
标量乘法将每个元素乘以该标量。
Matrix multiplication is defined by row-into-column rule. For A =
| a | b |
| c | d |
and B =
| p | q |
| r | s |
, the product AB is:
矩阵乘法按行乘列规则定义。对于上述矩阵，乘积 AB 为：
| ap+br | aq+bs |
| cp+dr | cq+ds |
Matrix multiplication is not commutative: AB ≠ BA in general. The identity matrix I =
| 1 | 0 |
| 0 | 1 |
satisfies AI = IA = A.
矩阵乘法不满足交换律：通常 AB ≠ BA。单位矩阵 I 满足 AI = IA = A。
6. Determinant and Inverse of 2×2 Matrix | 二行列式与逆矩阵
For a 2×2 matrix M =
| a | b |
| c | d |
, the determinant is det(M) = ad – bc.
对于矩阵 M，行列式为 det(M) = ad – bc。
If det(M) ≠ 0, the inverse M⁻¹ exists and is given by:
若 det(M) ≠ 0，逆矩阵 M⁻¹ 存在且由下式给出：
M⁻¹ = (1 / (ad – bc))
| d | –b |
| –c | a |
A matrix with zero determinant is called singular and has no inverse.
行列式为零的矩阵称为奇异矩阵，不存在逆矩阵。
7. Roots of Polynomials | 多项式根
For the quadratic equation ax² + bx + c = 0 with roots α and β, the sum and product are:
对于二次方程 ax² + bx + c = 0，设根为 α 和 β，和与积为：
α + β = –b/a, αβ = c/a
For the cubic equation ax³ + bx² + cx + d = 0 with roots α, β, γ:
对于三次方程 ax³ + bx² + cx + d = 0，设根为 α, β, γ：
∑α = α + β + γ = –b/a
∑αβ = αβ + βγ + γα = c/a
αβγ = –d/a
These symmetric sums can be used to find coefficients from given roots, or to rewrite expressions without solving the equation explicitly.
这些对称和可用于从已知根求系数，或在不解方程的情况下重写表达式。
8. Summation of Series | 级数求和
Standard summations of natural numbers, squares and cubes are essential:
自然数、平方与立方的标准求和公式非常重要：
∑r=1n r = n(n+1)/2
∑r=1n r² = n(n+1)(2n+1)/6
∑r=1n r³ = n²(n+1)²/4
More complicated series can be tackled by splitting them into simpler sums, using linearity: ∑(ar + b) = a∑r + ∑b.
更复杂的级数可拆分为较简单的求和，利用线性性质：∑(ar + b) = a∑r + ∑b。
The method of differences is used when terms cancel telescopically; a common form is ∑ [f(r) – f(r+1)] = f(1) – f(n+1).
当项能望远镜式消去时使用差分法，常见形式为 ∑ [f(r) – f(r+1)] = f(1) – f(n+1)。
9. Proof by Induction | 数学归纳法
Proof by induction is a powerful technique for proving statements that involve a positive integer n. It consists of four key stages:
数学归纳法是证明与正整数 n 有关的命题的强大技巧，包含四个关键步骤：
-
Basis step: verify the statement for n = 1 (or the smallest suitable integer).
奠基步骤：验证 n = 1（或最小合适整数）时命题成立。
-
Assumption step: assume true for n = k.
假设步骤：假设 n =
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