Core Pure Maths 1 Revision: Complex Numbers, Matrices and Vectors | Edexcel 进阶数学核心纯数学 1 复习：复数、矩阵与向量

📚 Core Pure Maths 1 Revision: Complex Numbers, Matrices and Vectors | Edexcel 进阶数学核心纯数学 1 复习：复数、矩阵与向量

This revision guide covers the most frequently examined topics in Edexcel A Level Further Mathematics Core Pure 1: complex numbers, Argand diagrams, series, roots of polynomials, proof by induction, matrices, linear transformations, vectors and volumes of revolution. Each section gives the key result in English followed by the Chinese explanation, with clear formulas and exam tips.

本复习指南覆盖 Edexcel A Level 进阶数学核心纯数学 1 的高频考点：复数、Argand 图、级数、多项式根、归纳法证明、矩阵、线性变换、向量和旋转体体积。每一节先给出英文要点，再给出中文解释，并配有清晰公式与考试提示。

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1. Complex Numbers: Arithmetic and Conjugates | 复数：运算与共轭

A complex number is written as z = a + bi, where a is the real part and b is the imaginary part. The imaginary unit satisfies i² = -1. Addition and subtraction work component-wise: (a + bi) ± (c + di) = (a ± c) + (b ± d)i.

复数写作 z = a + bi，其中 a 是实部，b 是虚部。虚数单位满足 i² = -1。加减法按实部和虚部分别运算：(a + bi) ± (c + di) = (a ± c) + (b ± d)i。

Multiplication uses the distributive law and replaces i² by -1: (a + bi)(c + di) = (ac – bd) + (ad + bc)i. The complex conjugate of z is z* = a – bi, and zz* = a² + b² is always real.

乘法使用分配律并把 i² 换成 -1：(a + bi)(c + di) = (ac – bd) + (ad + bc)i。z 的共轭复数是 z* = a – bi，且 zz* = a² + b² 恒为实数。

To divide by a complex denominator, multiply numerator and denominator by the conjugate of the denominator: (a + bi)/(c + di) = [(a + bi)(c – di)] / [(c + di)(c – di)] = [(ac + bd) + (bc – ad)i] / (c² + d²).

除以一个复数时，将分子和分母同时乘以分母的共轭复数：(a + bi)/(c + di) = [(a + bi)(c – di)] / [(c + di)(c – di)] = [(ac + bd) + (bc – ad)i] / (c² + d²)。

For polynomial equations with real coefficients, complex roots occur in conjugate pairs. If z is a root, then z* is also a root. This is especially useful when finding remaining roots of cubics or quartics.

对于实系数多项式方程，复根成共轭对出现。如果 z 是根，那么 z* 也是根。这一性质在求三次或四次方程的剩余根时特别有用。

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2. Argand Diagrams and Modulus-Argument Form | Argand 图与模-辐角形式

An Argand diagram represents a complex number z = a + bi as the point (a, b) in the plane. The horizontal axis is the real axis and the vertical axis is the imaginary axis. The modulus |z| is the distance from the origin to the point: |z| = √(a² + b²).

Argand 图将复数 z = a + bi 表示为平面上的点 (a, b)。横轴是实轴，纵轴是虚轴。模 |z| 是从原点到该点的距离：|z| = √(a² + b²)。

The argument θ = arg z is the angle measured from the positive real axis to the line joining the origin to the point. The principal argument is usually taken in the interval -π < θ ≤ π. The modulus-argument form is z = r(cos θ + i sin θ), where r = |z| and θ = arg z.

辐角 θ = arg z 是从正实轴到原点与该点连线的夹角。主辐角通常取在 -π < θ ≤ π 区间。模-辐角形式为 z = r(cos θ + i sin θ)，其中 r = |z|，θ = arg z。

Multiplication and division become very tidy in modulus-argument form:

在模-辐角形式下，乘法和除法变得非常简洁：

z₁z₂ = r₁r₂ [cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)]

z₁/z₂ = r₁/r₂ [cos(θ₁ – θ₂) + i sin(θ₁ – θ₂)]

This shows that multiplying by a complex number rotates and scales the original complex number, which is essential for geometric interpretations.

这表明乘以一个复数会对原复数进行旋转和缩放，这对于几何解释至关重要。

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3. Loci and Regions in the Complex Plane | 复平面中的轨迹与区域

The equation |z – z₀| = r describes a circle with centre z₀ and radius r. For example, |z – (3 + 4i)| = 2 is a circle centred at 3 + 4i with radius 2.

方程 |z – z₀| = r 表示以 z₀ 为圆心、r 为半径的圆。例如，|z – (3 + 4i)| = 2 是以 3 + 4i 为圆心、半径为 2 的圆。

The equation arg(z – z₀) = θ describes a half-line starting at z₀ and making an angle θ with the positive real direction. The point z₀ itself is not included in the half-line.

方程 arg(z – z₀) = θ 表示从 z₀ 出发、与正实方向夹角为 θ 的射线。点 z₀ 本身不包含在该射线中。

The equation |z – z₁| = |z – z₂| is the perpendicular bisector of the line segment joining z₁ and z₂. Inequalities such as |z – z₀| < r or |z - z₁| > |z – z₂| define regions, and you should shade the required side and use a dashed boundary for strict inequalities.

方程 |z – z₁| = |z – z₂| 是连接 z₁ 与 z₂ 线段的垂直平分线。不等式如 |z – z₀| < r 或 |z - z₁| > |z – z₂| 定义区域，作答时应涂出所需一侧，并对严格不等式使用虚线边界。

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4. Summation of Series | 级数求和

Core Pure 1 expects fluency with the standard results for the sums of powers of integers:

核心纯数学 1 要求熟练掌握整数幂和的标准结果：

• Σᵣ₌₁ⁿ r = n(n + 1)/2
• Σᵣ₌₁ⁿ r² = n(n + 1)(2n + 1)/6
• Σᵣ₌₁ⁿ r³ = n²(n + 1)²/4

These are often combined with linearity rules. For example, Σ(ar² + br) = aΣr² + bΣr, where a and b are constants. You may also need to expand brackets such as Σr(r – 1) = Σr² – Σr before applying the standard results.

这些公式通常与线性规则结合使用。例如，Σ(ar² + br) = aΣr² + bΣr，其中 a 和 b 为常数。你可能还需要先展开括号，如 Σr(r – 1) = Σr² – Σr，再应用标准结果。

A common exam question asks for a sum in the form Σ(3r² + 2r + 1) from r = 1 to n. Split it into standard parts, substitute, and simplify to a single fraction.

常见考题要求求形如 Σ(3r² + 2r + 1) 从 r = 1 到 n 的和。将其拆分为标准部分，代入并化简为单个分式即可。

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5. Method of Differences | 差分法

The method of differences is used for sums where the terms can be written as a difference f(r + 1) – f(r). The sum telescopes: Σᵣ₌₁ⁿ [f(r + 1) – f(r)] = f(n + 1) – f(1).

差分法用于可将项写成 f(r + 1) – f(r) 形式的求和。该和会相互抵消：Σᵣ₌₁ⁿ [f(r + 1) – f(r)] = f(n + 1) – f(1)。

A classic example uses partial fractions: 1/[r(r + 1)] = 1/r – 1/(r + 1). Therefore Σᵣ₌₁ⁿ 1/[r(r + 1)] = (1 – 1/2) + (1/2 – 1/3) + … + (1/n – 1/(n + 1)) = 1 – 1/(n + 1) = n/(n + 1).

经典例子使用部分分式：1/[r(r + 1)] = 1/r – 1/(r + 1)。因此 Σᵣ₌₁ⁿ 1/[r(r + 1)] = (1 – 1/2) + (1/2 – 1/3) + … + (1/n – 1/(n + 1)) = 1 – 1/(n + 1) = n/(n + 1)。

When using the method of differences, show the cancellation pattern explicitly and write down the two surviving terms. For fractions, always test with n = 1 and n = 2 to check your closed form.

使用差分法时，要明确写出抵消模式并保留剩下的两项。对于分式，务必用 n = 1 和 n = 2 检验你的闭合形式。

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6. Roots of Polynomials and Coefficient Relations | 多项式根与系数关系

For a quadratic ax² + bx + c = 0 with roots α and β, the sum of roots is α + β = -b/a and the product is αβ = c/a. For a cubic ax³ + bx² + cx + d = 0 with roots α, β and γ:

对于二次方程 ax² + bx + c = 0，根为 α 和 β，则根之和为 α + β = -b/a，根之积为 αβ = c/a。对于三次方程 ax³ + bx² + cx + d = 0，根为 α、β 和 γ：

α + β + γ = -b/a, αβ + βγ + γα = c/a, αβγ = -d/a

For a quartic ax⁴ + bx³ + cx² + dx + e = 0, the four elementary symmetric sums are:

对于四次方程 ax⁴ + bx³ + cx² + dx + e = 0，四个初等对称和为：

α + β + γ + δ = -b/a, Σαβ = c/a, Σαβγ = -d/a, αβγδ = e/a

You will often need to evaluate expressions such as α² + β² + γ². Use the identity α² + β² + γ² = (α + β + γ)² – 2(αβ + βγ + γα). Similar expansions are common for α³ + β³ + γ³ and for reciprocal sums.

你经常需要计算如 α² + β² + γ² 的表达式。可使用恒等式 α² + β² + γ² = (α + β + γ)² – 2(αβ + βγ + γα)。类似的展开还常见于 α³ + β³ + γ³ 以及倒数之和。

Linear transformations of roots, such as finding a polynomial whose roots are 2α, 3 – α or 1/α, can be handled by substitution. Replace x in the original polynomial with the expression that maps the new root back to the original root.

根的线性变换，例如求根为 2α、3 – α 或 1/α 的多项式，可以通过

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