AS AQA Further Maths FM01 Example Resources | AS AQA 进阶数学 FM01 例题资源

📚 AS AQA Further Maths FM01 Example Resources | AS AQA 进阶数学 FM01 例题资源

This article provides essential worked examples for the core pure topics examined in AQA International AS Further Maths unit FM01. Each example follows a step-by-step approach to build confidence in complex numbers, matrices, series and proof.

本文针对 AQA 国际 AS 进阶数学 FM01 单元中的核心纯数主题，提供关键例题及逐步解答，帮助考生掌握复数、矩阵、级数与证明等考点。

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1. The Imaginary Unit and Complex Arithmetic | 虚数单位与复数运算

bi, where a is the real part and b is the imaginary part. To divide two complex numbers, multiply the numerator and denominator by the conjugate of the denominator.

虚数单位定义为 i² = −1。复数写作 z = a + bi，其中 a 为实部，b 为虚部。两个复数相除时，将分子与分母同时乘以分母的共轭复数。

Example: Let z = 2 + 3i and w = 1 − i. Compute z / w.

例：设 z = 2 + 3i，w = 1 − i。计算 z / w。

z/w = (2+3i)/(1−i) × (1+i)/(1+i) = ((2+3i)(1+i))/(1−i²)

2i + 3i + 3i² = 2 + 5i − 3 = −1 + 5i. Since i² = −1, the denominator becomes 1 − (−1) = 2.

展开分子：2 + 2i + 3i + 3i² = 2 + 5i − 3 = −1 + 5i。因为 i² = −1，分母变为 1 − (−1) = 2。

z/w = (−1 + 5i)/2 = −0.5 + 2.5i

Always write the result in the form a + bi, with real and imaginary parts clearly separated.

结果务必写成 a + bi 的形式，并清晰区分实部与虚部。

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2. Modulus and Argument | 模与辐角

The modulus of z = a + bi is |z| = √a² + b² ut you must choose the correct quadrant to get the principal argument, usually −π < θ ≤ π.

复数 z = a + bi 的模为 |z| = √a² + b²。辐角 θ 满足 tan θ = b/a，但必须根据所在象限选择正确的主辐角，通常取 −π < θ ≤ π。

Example: Find the modulus and principal argument of z = −1 + √3 i.

例：求 z = −1 + √3 i 的模与主辐角。

|z| = √((−1)² + (√3)²) = √1 + 3 = 2

The point (−1, √3) lies in the second quadrant. The reference angle is tan⁻¹(√3/1) = π/3, so the principal argument is π − π/3 = 2π/3.

点 (−1, √3) 位于第二象限。参考角为 tan⁻¹(√3/1) = π/3，因此主辐角为 π − π/3 = 2π/3。

z = 2(cos(2π/3) + i sin(2π/3))

This exponential or polar form is especially useful for multiplication and powers of complex numbers.

这种指数或极坐标形式在复数的乘法与幂运算中尤其有用。

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3. Argand Diagram and Loci | 阿尔冈图与轨迹

ry part b on the vertical axis. A locus is a set of points satisfying a condition, such as |z − c| = r or |z − a| = |z − b|.

在阿尔冈图中，实部 a 绘制在水平轴上，虚部 b 绘制在垂直轴上。轨迹是满足某种条件的点的集合，例如 |z − c| = r 或 |z − a| = |z − b|。

Example: Describe the locus |z − 2| = 3.

例：描述轨迹 |z − 2| = 3。

The equation |z − 2| point 2 (i.e. (2,0)) is always 3. Therefore it is a circle with centre (2,0) and radius 3.

方程 |z − 2| = 3 表示 z 到点 2（即 (2,0)）的距离恒为 3，因此这是一个以 (2,0) 为圆心、半径为 3 的圆。

For a perpendicular bisector example, the set |z − 1| = |z + 1| gives the points equidistant from 1 and −1, which is the imaginary axis.

再看一个垂直平分线的例子：|z − 1| = |z + 1| 表示到 1 和 −1 距离相等的点，所得到的轨迹就是虚轴。

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4. Matrix Operations and Inverses | 矩阵运算与逆矩阵

For a 2 × 2 matrix A = [[a, b], [c, d]], the determinant is ad − bc. If the determinant is non-zero, the inverse is given by

对于 2 × 2 矩阵 A = [[a, b], [c, d]]，其行列式为 ad − bc。若行列式不为零，则逆矩阵由下式给出

A⁻¹ = 1/(ad − bc) [[d, −b], [−c, a]]

Example: Let A = [[2, 3], [1, 4]]. Find A⁻¹.

例：设 A = [[2, 3], [1, 4]]。求 A⁻¹。

det(A) = 2×4 − 3×1 = 8 − 3 = 5

A⁻¹ = 1/5 [[4, −3], [−1, 2]]

eck the result by multiplying A⁻¹ with A; the product should equal the identity matrix I = [[1,0],[0,1]].

将 A⁻¹ 与 A 相乘可验证结果，其乘积应等于单位矩阵 I = [[1,0],[0,1]]。

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5. Matrix Transformations | 矩阵变换

Matrices can represent linear transformations in the plane. Common standard matrices include rotations and reflections. The matrix for an anti-clockwise rotation by angle θ is

矩阵可以表示平面上的线性变换。常用的标准矩阵包括旋转与反射。逆时针旋转 θ 角的矩阵为

R(θ) = [[cos θ, −sin θ], [sin θ, cos θ]]

Example: Write the matrix for a 90° anti-clockwise rotation and apply it to the vector (1,0).

例：写出逆时针旋转 90° 的矩阵，并将其作用于向量 (1,0)。

R(90°) = [[0, −1], [1, 0]]

Applying it to (1,0):

将其作用于 (1,0)：

[[0, −1], [1, 0]] × [1;0] = [0;1]

This correctly maps the x-axis unit vector to the y-axis unit vector.

这正确地将 x 轴单位向量映射到 y 轴单位向量。

Another key example is reflection in the y-axis, whose matrix is [[−1, 0], [0, 1]].

另一个关键例子是关于 y 轴的反射，其矩阵为 [[−1, 0], [0, 1]]。

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6. Roots of Polynomials | 多项式根

tic equation az² + bz + c = 0 with complex roots α and β, the sum and product of roots are

对于二次方程 az² + bz + c = 0，若复数根为 α 和 β，则根的和与积分别为

α + β = −b/a, αβ = c/a

e roots of z² + 2z + 5 = 0 are α and β. Find their sum and product, and verify by solving.

例：方程 z² + 2z + 5 = 0 的根为 α 和 β。求它们的和与积，并通过求解验证。

α + β = −2/1 = −2, αβ = 5/1 = 5

Using the quadratic formula:

使用求根公式：

z = (−2 ± √4 − 20)/2 = (−2 ± √−16)/2 = −1 ± 2i

(−1 + 2i) + (−1 − 2i) = −2 and αβ = (−1)² + 2² = 1 + 4 = 5, confirming the relationships.

这两个根的加和为 (−1 + 2i) + (−1 − 2i) = −2，乘积为 (−1)² + 2² = 1 + 4 = 5，验证了上述关系。

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7. Summation of Series | 级数求和

The standard results for summing powers are essential for FM01:

求和幂的常用公式是 FM01 的核心内容：

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

Example: Find Σ_r=1ⁿ (2r + 3).

例：求 Σ_r=1ⁿ (2r + 3)。

Split the sum:

将和拆分：

Σ(2r + 3) = 2Σr + Σ3

Since Σ3 = 3n and Σr = n(n+1)/2, we get:

因为 Σ3 = 3n，Σr = n(n+1)/2，所以：

2 × n(n+1)/2 + 3n = n(n+1) + 3n = n² + 4n = n(n+4)

Always factorise the final expression neatly, as examiners reward simplified algebraic forms.

最终表达式应进行良好的因式分解，考官通常对化简后的形式给予分数。

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8. Proof by Induction | 数学归纳法

Proof by induction has three steps: base case, inductive assumption, and inductive step. This method is used for statements involving positive integers n.

数学归纳法包含三个步骤：基础情形、归纳假设和归纳递推。该方法用于证明涉及正整数 n 的命题。

Example: Prove that Σ_r=1ⁿ r³ = [n(n+1)/2]² for all positive integers n.

例：证明对所有正整数 n，Σ_r=1ⁿ r³ = [n(n+1)/2]²。

Base case: For n = 1, LHS = 1³ = 1, RHS = [1×2/2]² = 1, so the statement is true.

基础情形：当 n = 1 时，左边 = 1³ = 1，右边 = [1×2/2]² = 1，因此命题成立。

Assume true for n = k, so Σ_r=1^k r³ = [k(k+1)/2]². For n = k+1, add (k+1)³ to both sides:

假设当 n = k 时成立，即 Σ_r=1^k r³ = [k(k+1)/2]²。对于 n = k+1，两边加上 (k+1)³：

Σ_r=1^k+1 r³ = [k(k+1)/2]² + (k+1)³

Factorise (k+1)²/(4):

提出公因子 (k+1)²/4：

= (k+1)²/4 × [k² + 4(k+1)] = (k+1)²/4 × (k² + 4k + 4) = (k+1)²(k+2)²/4 = [(k+1)(k+2)/2]²

n = k+1. Hence by mathematical induction the statement is true for all positive integers.

这与 n = k+1 时的公式一致。因此由数学归纳法，命题对所有正整数成立。

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9. De Moivre's Theorem | 德莫弗定理

De Moivre's theorem states that for any complex number in polar form r(cos θ + i sin θ) and any integer n,

德莫弗定理指出，对于极坐标形式 r(cos θ + i sin θ) 的任意复数以及任意整数 n，

zⁿ = rⁿ(cos(nθ) + i sin(nθ))

Example: Compute (1 + i)⁶.

例：计算 (1 + i)⁶。

First write 1 + i in polar form: r = √2, θ = π/4.

首先将 1 + i 写成极坐标形式：r = √2，θ = π/4。

(1 + i)⁶ = (√2)⁶ (cos(6×π/4) + i sin(6×π/4))

Since (√2)⁶ = 2³ = 8 and 6π/4 = 3π/2, we have:

因为 (√2)⁶ = 2³ = 8，且 6π/4 = 3π/2，所以：

8(cos(3π/2) + i sin(3π/2)) = 8(0 − i) = −8i

De Moivre's theorem is especially efficient for high powers that would be tedious to expand binomially.

对于使用二项式展开会非常冗长的高次幂，德莫弗定理尤其高效。

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