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Further Maths Core Pure 1: Complex Numbers, Matrices and Beyond | 进阶数学核心纯数1:复数、矩阵及更多

📚 Further Maths Core Pure 1: Complex Numbers, Matrices and Beyond | 进阶数学核心纯数1:复数、矩阵及更多

Further Maths Core Pure 1 is the foundational unit for A-level Further Mathematics. It extends GCSE and AS Pure content into complex numbers, matrices, proof, series, vectors and calculus techniques. This article summarises the most frequently examined topics and provides concise methods for success.

进阶数学核心纯数1是A-level进阶数学的基础模块。它将GCSE和AS纯数内容拓展到复数、矩阵、证明、级数、向量和微积分技巧。本文总结最高频考查的主题,并提供简洁的解题方法。


1. Complex Numbers: Polar Form and De Moivre’s Theorem | 复数:极坐标形式与棣莫弗定理

Complex numbers can be written in Cartesian form z = a + bi or in polar form z = r(cos θ + i sin θ), where r = |z| is the modulus and θ = arg(z) is the argument. The polar form is particularly useful for multiplication, division and powers.

复数可以用笛卡尔形式 z = a + bi 或极坐标形式 z = r(cos θ + i sin θ) 表示,其中 r = |z| 是模,θ = arg(z) 是幅角。极坐标形式在乘法、除法和乘方运算中特别方便。

Multiplying two complex numbers in polar form means multiplying the moduli and adding the arguments: z₁z₂ = r₁r₂[cos(θ₁+θ₂) + i sin(θ₁+θ₂)]. De Moivre’s theorem states that zⁿ = rⁿ(cos nθ + i sin nθ) for any integer n.

在极坐标形式下,两个复数相乘等于模相乘、幅角相加:z₁z₂ = r₁r₂[cos(θ₁+θ₂) + i sin(θ₁+θ₂)]。棣莫弗定理指出,对任意整数 n,有 zⁿ = rⁿ(cos nθ + i sin nθ)。

z = r(cos θ + i sin θ) ⇒ zⁿ = rⁿ(cos nθ + i sin nθ)

To find the n-th roots of a complex number, use the formula z_k = r^(1/n)[cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)] for k = 0, 1, …, n−1. There are exactly n distinct roots, equally spaced on a circle of radius r^(1/n).

求复数的 n 次方根时,使用公式 z_k = r^(1/n)[cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)],其中 k = 0, 1, …, n−1。共有 n 个不同的根,它们均匀分布在半径为 r^(1/n) 的圆上。


2. Roots of Polynomials and Sums of Powers | 多项式根与幂和

For a quadratic ax²+bx+c=0 with roots α and β, the sum and product of roots are α+β = −b/a and αβ = c/a. For a cubic ax³+bx²+cx+d=0 with roots α, β, γ, the key results are Σα = −b/a, Σαβ = c/a and αβγ = −d/a.

对于二次方程 ax²+bx+c=0 的根 α 和 β,根的和与积分别为 α+β = −b/a 和 αβ = c/a。对于三次方程 ax³+bx²+cx+d=0 的根 α, β, γ,关键结果是 Σα = −b/a,Σαβ = c/a 以及 αβγ = −d/a。

These symmetric sums can be used to find sums of powers of roots without solving the equation. For example, α²+β² = (α+β)² − 2αβ, and α³+β³ = (α+β)³ − 3αβ(α+β). For a cubic, α²+β²+γ² = (Σα)² − 2Σαβ.

这些对称和可以在不解方程的情况下求出根的幂和。例如,α²+β² = (α+β)² − 2αβ,α³+β³ = (α+β)³ − 3αβ(α+β)。对于三次方程,α²+β²+γ² = (Σα)² − 2Σαβ。

Σα = −b/a, Σαβ = c/a, αβγ = −d/a

In exams, you may be given one root and asked to find the others, or asked to form a new polynomial whose roots are functions of the original roots, such as 2α, 1/α or α+β.

考试中可能给出一个根并要求求出其他根,或者要求构造一个新多项式,其根是原根的某种函数,如 2α、1/α 或 α+β。


3. Matrices: Determinants, Inverses and Linear Transformations | 矩阵:行列式、逆矩阵与线性变换

For a 2×2 matrix A = [[a, b], [c, d]], the determinant is det A = ad − bc. If det A ≠ 0, the inverse is A⁻¹ = (1/det A)[[d, −b], [−c, a]]. A matrix with det A = 0 is singular and has no inverse.

对于 2×2 矩阵 A = [[a, b], [c, d]],行列式为 det A = ad − bc。如果 det A ≠ 0,逆矩阵为 A⁻¹ = (1/det A)[[d, −b], [−c, a]]。行列式为零的矩阵是奇异矩阵,没有逆矩阵。

A⁻¹ = (1/det A)[[d, −b], [−c, a]]

For 3×3 matrices, the determinant can be expanded by cofactors along any row or column. Common mistake is losing the sign pattern (+ − +; − + −; + − +). Linear transformations include rotations by θ represented by [[cos θ, −sin θ], [sin θ, cos θ]], reflections in lines through the origin, and enlargements. The area scale factor of a transformation A is |det A|, which means the image area equals the original area multiplied by |det A|.

对于 3×3 矩阵,可以沿任意行或列用余子式展开计算行列式。常见错误是丢失符号规律(+ − +;− + −;+ − +)。线性变换包括旋转 θ 角(矩阵为 [[cos θ, −sin θ], [sin θ, cos θ]])、关于过原点直线的反射以及放大。变换 A 的面积比例因子为 |det A|,即像的面积等于原面积乘以 |det A|。


4. Proof by Induction | 归纳法证明

Proof by induction follows a standard structure: prove the base case (usually n = 1), assume the statement is true for n = k, then prove it for n = k + 1. This shows the statement holds for all positive integers.

归纳法证明遵循标准结构:证明基础情形(通常 n = 1),假设命题对 n = k 成立,然后证明它对 n = k + 1 也成立。这样就证明了命题对所有正整数成立。

Common induction types include summation results such as Σr = n(n+1)/2, divisibility proofs, matrix powers, and inequalities. In the inductive step, always use the assumption for n = k to rewrite the expression for n = k + 1.

常见的归纳题型包括求和结果如 Σr = n(n+1)/2、整除性证明、矩阵的幂以及不等式。在归纳步骤中,务必利用 n = k 的假设来改写 n = k + 1 的表达式。

Example: prove 3^(2n) − 1 is divisible by 8. Base case n = 1 gives 9 − 1 = 8, divisible by 8. Assume 3^(2k) − 1 = 8m. Then for n = k+1, 3^(2k+2) − 1 = 9·3^(2k) − 1 = 9(3^(2k) − 1) + 8 = 9(8m) + 8 = 8(9m + 1), so it is divisible by 8.

例题:证明 3^(2n) − 1 可被 8 整除。基础情形 n = 1 得 9 − 1 = 8,可被 8 整除。假设 3^(2k) − 1 = 8m。那么当 n = k+1 时,3^(2k+2) − 1 = 9·3^(2k) − 1 = 9(3^(2k) − 1) + 8 = 9(8m) + 8 = 8(9m + 1),因此可被 8 整除。


5. Summation of Series | 级数求和

Core Pure 1 expects fluency with standard summation formulas. The most important results are Σr = n(n+1)/2, Σr² = n(n+1)(2n+1)/6, and Σr³ = [n(n+1)/2]². These are used to evaluate sums of polynomial expressions.

核心纯数1要求熟练运用标准求和公式。最重要的结果是 Σr = n(n+1)/2,Σr² = n(n+1)(2n+1)/6,以及 Σr³ = [n(n+1)/2]²。这些公式用于计算多项式表达式的和。

Σr = n(n+1)/2, Σr² = n(n+1)(2n+1)/6, Σr³ = [n(n+1)/2]²

The method of differences is used for telescoping series, such as Σ 1/[r(r+1)] from r=1 to n. Write 1/[r(r+1)] = 1/r − 1/(r+1), then most terms cancel, leaving 1 − 1/(n+1). Similar partial-fraction decomposition works for series like Σ 1/[(2r−1)(2r+1)].

差分法用于裂项相消的级数,例如从 r=1 到 n 的 Σ 1/[r(r+1)]。将 1/[r(r+1)] 写成 1/r − 1/(r+1),大部分项会相消,剩下 1 − 1/(n+1)。类似的部分分式分解也适用于 Σ 1/[(2r−1)(2r+1)] 这类级数。


6. Vectors and 3D Geometry | 向量与三维几何

A straight line in 3D can be written as r = a + λb, where a is a position vector on the line and b is a direction vector. A plane can be expressed as r·n = p, where n is the normal vector to the plane and p is a constant.

三维空间中的直线可以写成 r = a + λb,其中 a 是直线上一点的位置向量,b 是方向向量。平面可以表示为 r·n = p,其中 n 是平面的法向量,p 是常数。

The angle between two vectors a and b is found from cos θ = (a·b)/(|a||b|). Two lines are perpendicular if their direction vectors satisfy b₁·b₂ = 0, and parallel if one direction vector is a scalar multiple of the other.

两个向量 a 和 b 的夹角由 cos θ = (a·b)/(|a||b|) 求得。如果两条直线的方向向量满足 b₁·b₂ = 0,则它们垂直;如果其中一个方向向量是另一个的标量倍数,则它们平行。

The shortest distance from a point P to a line r = a + λb is |(p − a) × b| / |b|, and from a point P to a plane r·n = p is |(a − p)·n| / |n|. These formulas are frequently examined.

点 P 到直线 r = a + λb 的最短距离为 |(p − a) × b| / |b|,点 P 到平面 r·n = p 的最短距离为 |(a − p)·n| / |n|。这些公式经常考查。


7. Hyperbolic Functions | 双曲函数

Hyperbolic functions are defined using exponential functions: sinh x = (eˣ − e⁻ˣ)/2, cosh x = (eˣ + e⁻ˣ)/2, and tanh x

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