A-Level AQA Further Maths: Complex Numbers Revision | A-Level AQA 进阶数学：复数考点精讲

📚 A-Level AQA Further Maths: Complex Numbers Revision | A-Level AQA 进阶数学：复数考点精讲

y form a fundamental part of AQA A-Level Further Mathematics, enabling us to work with roots of negative numbers and to represent quantities in two dimensions. Mastering complex numbers is essential for progress in pure mathematics, engineering, and physics.

当我们尝试求解像 x² + 1 = 0 这样没有实数解的方程时，复数就自然而然地出现了。它们是 AQA A-Level 进阶数学的核心内容，让我们能够处理负数的平方根，并在二维平面上表示量。掌握复数对于纯数学、工程学和物理学的深入学习至关重要。

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1. Introduction to Complex Numbers | 复数介绍

the imaginary unit with the property i² = −1. The set of all complex numbers is denoted by ℂ. The real part is a, written as Re(z), and the imaginary part is b, written as Im(z).

复数是形如 a + bi 的数，其中 a 和 b 是实数，而 i 是虚数单位，满足 i² = −1。所有复数构成的集合记作 ℂ。实部为 a，记作 Re(z)，虚部为 b，记作 Im(z)。

ber can be thought of as a complex number with b = 0; a purely imaginary number has a = 0.

例如，3 + 4i 是一个复数，实部为 3，虚部为 4。实数可以看作虚部为零的复数；纯虚数则是实部为零的复数。

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2. The Imaginary Unit i | 虚数单位 i

obtain i³ = i² × i = −i, and i⁴ = (i²)² = 1. Powers of i cycle with period 4: i, −1, −i, 1, ...

虚数单位 i 由方程 i² = −1 定义。它不是实数，无法在通常的数轴上找到。根据这一定义，可得 i³ = i² × i = −i，i⁴ = (i²)² = 1。i 的幂以 4 为周期循环：i, −1, −i, 1, ...

i² = −1, i³ = −i, i⁴ = 1

When simplifying expressions involving i, reduce the power modulo 4 and use the cycle to replace higher powers with one of these four values.

在化简含有 i 的表达式时，将指数对 4 取模，并利用上述循环将高次幂替换为这四个值之一。

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3. Algebraic Form and Basic Operations | 代数形式与基本运算

e real parts and the imaginary parts separately: (a + bi) ± (c + di) = (a ± c) + (b ± d)i.

两个复数 a + bi 与 c + di 相加或相减时，分别合并实部和虚部：(a + bi) ± (c + di) = (a ± c) + (b ± d)i。

the fact that i² = −1: (a + bi)(c + di) = ac + adi + bci + bdi² = (ac − bd) + (ad + bc)i.

乘法运用分配律并利用 i² = −1：(a + bi)(c + di) = ac + adi + bci + bdi² = (ac − bd) + (ad + bc)i。

inator (see next section). For instance, (a + bi)/(c + di) = ((a + bi)(c − di))/(c² + d²).

复数的除法通过将分子和分母同时乘以分母的共轭复数来进行（见下一节）。例如，(a + bi)/(c + di) = ((a + bi)(c − di))/(c² + d²)。

It is essential to express the final answer in the form x + yi, with real and imaginary parts clearly separated.

最终答案必须写成 x + yi 的形式，实部和虚部分明。

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4. Complex Conjugates | 共轭复数

The complex conjugate of z = a + bi is denoted by z* or z̄ as a − bi. Geometrically, it is the reflection of z in the real axis on an Argand diagram.

复数 z = a + bi 的共轭复数记作 z* 或 z̄，定义为 a − bi。在几何上，它是 z 关于实轴在阿干特图上的反射。

Key properties: z + z̄ = 2a is real; z − z̄ = 2bi is purely imaginary; z × z̄ = a² + b², which is a non‑ number. The product of a complex number and its conjugate gives the square of its modulus.

重要性质：z + z̄ = 2a 为实数；z − z̄ = 2bi 为纯虚数；z × z̄ = a² + b² 是一个非负实数。一个复数与其共轭的乘积等于其模的平方。

Conjugates are used to simplify division and to find real denominators: to divide z₁ by z₂, multiply numerator and denominator by z̄₂.

共轭复数用于简化除法并获得实数分母：计算 z₁ ÷ z₂ 时，将分子和分母同乘以 z̄₂。

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5. Solving Quadratic Equations | 解二次方程

ith real coefficients has solutions given by the quadratic formula. If the discriminant b² − 4ac is negative, the roots are a conjugate pair of complex numbers.

任何实系数二次方程 ax² + bx + c = 0 均可通过求根公式求解。若判别式 b² − 4ac 为负，则两根是一对共轭复数。

x = (−b ± √b² − 4ac) / (2a)

When the discriminant is negative, √negative n in terms of i. For example, solving x² − 4x + 13 = 0 gives discriminant −36, so x = (4 ± √−36)/2 = (4 ± 6i)/2 = 2 ± 3i.

当判别式为负时，√负数 用 i 表示。例如，解方程 x² − 4x + 13 = 0，判别式 −36，于是 x = (4 ± √−36)/2 = (4 ± 6i)/2 = 2 ± 3i。

This illustrates that complex roots of polynomials with real coefficients always occur in conjugate pairs. This fact is widely used in algebraic problems and proof.

这说明实系数多项式的复数根总是成共轭对出现。这一事实在代数问题和证明中被广泛应用。

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6. The Argand Diagram | 阿干特图（复平面）

al axis is the imaginary axis. A complex number z = x + yi is plotted at the point (x, y).

阿干特图将复数表示为平面上的点或向量。横轴为实轴，纵轴为虚轴。复数 z = x + yi 绘制在点 (x, y) 处。

Addition of complex numbers corresponds to vector addition. Subtraction is equivalent to adding the negative. The diagram makes it easy to visualise operations such as conjugation and to solve loci problems.

复数的加法对应于向量加法。减法相当于加上相反向量。该图使得共轭等运算可视化，也有助于解决轨迹问题。

For instance, the set of points |z − (1 + i)| dius 2 in the Argand diagram. This geometric interpretation is a powerful tool in AQA exams.

例如，满足 |z − (1 + i)| = 2 的点集在阿干特图中是一个以 (1,1) 为圆心、半径为 2 的圆。这种几何解释是 AQA 考试中的重要工具。

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7. Modulus and Argument | 模与辐角

The modulus of a complex number z = x + yi, written |z|, is the distance from the origin to the point (x, y). It is given by |z| = √x² + y². The modulus is always non‑negative.

复数 z = x + yi 的模，记作 |z|，是原点到点 (x, y) 的距离。公式为 |z| = √x² + y²。模总是非负的。

The argument of z, denoted arg(z), is the angle θ that the line joining the origin to (x, y) makes with the positive real axis. It is usually given in radians, with the principal value lying in (−π, π].

复数 z 的辐角，记作 arg(z)，是连接原点与 (x, y) 的线段与正实轴之间的夹角 θ。通常以弧度表示，主值范围在 (−π, π]。

, but adjust by adding or subtracting π to place θ in the correct quadrant relative to the signs of x and y.

找准辐角的正确象限至关重要。利用 tan θ = y/x 计算，但需根据 x 和 y 的符号，通过加减 π 将 θ 调整到正确的象限。

For example, z = −1 − i has modulus √2 and argument −3π/4 (or 5π/4 depending on convention), but the principal argument is usually −3π/4.

例如，z = −1 − i 的模为 √2，辐角为 −3π/4（根据习惯也可为 5π/4），但主值通常取 −3π/4。

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8. Polar Form of Complex Numbers | 复数的极坐标形式

A complex number can be expressed in polar (modulus–argument) form: z = r(cos θ + i sin θ), where r = |z| g(z). This representation is particularly useful for multiplication, division, and powers.

复数可以用极坐标（模–辐角）形式表示：z = r(cos θ + i sin θ)，其中 r = |z|，θ = arg(z)。这种表示形式在乘法、除法和乘方运算中尤为有用。

To convert from Cartesian form x + yi to polar form, compute r = √x² + y² d θ = arctan(y/x) with quadrant adjustment. To convert back, use x = r cos θ, y = r sin θ.

要将代数形式 x + yi 化为极坐标形式，先计算 r = √x² + y²，然后求 θ = arctan(y/x) 并调整象限。反之，可通过 x = r cos θ, y = r sin θ 换回。

Example: 1 + √3 i has modulus √1+3 = 2 and argument π/3, so its polar form is 2(cos π/3 + i sin π/3).

示例：1 + √3 i 的模为 √1+3 = 2，辐角为 π/3，因此其极坐标形式为 2(cos π/3 + i sin π/3)。

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9. Multiplication and Division in Polar Form | 极坐标形式的乘法与除法

duct is z₁z₂ = r₁r₂[cos(θ₁+θ₂) + i sin(θ₁+θ₂)]. The moduli multiply and the arguments add.

若 z₁ = r₁(cos θ₁ + i sin θ₁) 且 z₂ = r₂(cos θ₂ + i sin θ₂)，则它们的乘积为 z₁z₂ = r₁r₂[cos(θ₁+θ₂) + i sin(θ₁+θ₂)]。模相乘，辐角相加。

cos(θ₁−θ₂) + i sin(θ₁−θ₂)], provided z₂ ≠ 0. The moduli divide and the arguments subtract.

除法：z₁/z₂ = (r₁/r₂)[cos(θ₁−θ₂) + i sin(θ₁−θ₂)]，其中 z₂ ≠ 0。模相除，辐角相减。

These rules show why the polar form is so powerful: multiplication by i (which has modulus 1, argument π/2) corresponds to a rotation of 90° anticlockwise on the Argand diagram.

这些规则彰显了极坐标形式的强大：乘以 i（模为 1，辐角为 π/2）相当于在阿干特图上逆时针旋转 90°。

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10. De Moivre's Theorem | 棣莫弗定理

De Moivre' n nθ). This is a cornerstone of complex number theory and greatly simplifies computing powers.

棣莫弗定理指出，对于任意整数 n，[r(cos θ + i sin θ)]ⁿ = rⁿ(cos nθ + i sin nθ)。这是复数理论的基石，极大简化了幂的计算。

The theorem can be proved by induction for positive integers and extended to negative integers using the reciprocal. It also holds for rational exponents when care is taken with multiple values.

该定理可用数学归纳法对正整数进行证明，并通过对倒数推广到负整数。对于有理指数，若注意多值性也同样成立。

To find (1 + i)¹⁰, first express 1 + i in polar form: √2 (cos π/4 + i sin π/4). Then (1 + i)¹⁰ = (√2)¹⁰ (cos 10π/4 + i sin 10π/4) = 32(cos 5π/2 + i sin 5π/2) = 32(0 + i) = 32i.

要计算 (1 + i)¹⁰，先将 1 + i 化为极坐标式：√2 (cos π/4 + i sin π/4)。则 (1 + i)¹⁰ = (√2)¹⁰ (cos 10π/4 + i sin 10π/4) = 32(cos 5π/2 + i sin 5π/2) = 32i。

De Moivre's theorem is also used to derive trigonometric identities, such as expressing cos 3θ and sin 3θ in terms of powers of cos θ and sin θ.

棣莫弗定理还用于推导三角恒等式，例如将 cos 3θ 和 sin 3θ 表示为 cos θ 和 sin θ 的幂次。

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11. Roots of Complex Numbers | 复数的根

The n‑ utions to zⁿ = w. If w = r(cos θ + i sin θ), then the n distinct roots are given by zₖ = r^1/n[cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)] for k = 0, 1, ..., n−1.

复数 w 的 n 次方根是方程 zⁿ = w 的解。若 w = r(cos θ + i sin θ)，则 n 个不同的根由下式给出：zₖ = r^1/n[cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)]，k = 0, 1, ..., n−1。

These n roots are equally spaced around a circle of radius r^1/n on the Argand diagram, forming a regular n‑gon. Their arguments differ by 2π/n.

这 n 个根在阿干特图上均匀分布在半径为 r^1/n 的圆周上，形成一个正 n 边形，其辐角依次相差 2π/n。

A classic example is the cube roots of unity: solving z³ = 1 gives 1, e^2πi/3, e^4πi/3 (or using cos/sin form). Their sum is 0, a property used in many problems.

典型的例子是单位立方根：解方程 z³ = 1 得到 1, e^2πi/3, e^4πi/3（或使用 cos/sin 形式）。它们的和为 0，这一性质常用于许多问题。

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12. Exponential Form and Euler's Formula | 指数形式与欧拉公式

Euler's formula links the polar form to the exponential function: e^iθ = cos θ + i sin θ. Thus any complex number can be written as z = re^iθ, which is the exponential form.

欧拉公式将极坐标形式与指数函数联系起来：e^iθ = cos θ + i sin θ。因此，任何复数都可写作 z = re^iθ，此即指数形式。

The exponential form makes multiplication and division even more compact: z₁z₂ = r₁r₂ e^{i(θ₁+θ₂)}, and z₁/z₂ = (r₁/r₂) e^{i(θ₁−θ₂)}. De Moivre's theorem becomes a simple consequence of the index law (e^iθ)ⁿ = e^inθ.

指数形式让乘除法更加简洁：z₁z₂ = r₁r₂ e^{i(θ₁+θ₂)}，z₁/z₂ = (r₁/r₂) e^{i(θ₁−θ₂)}。棣莫弗定理也因此变为指数律 (e^iθ)ⁿ = e^inθ 的简单推论。

s z⁴ = 81i using the exponential approach. The key is to express both sides in the form re^iθ and then compare moduli and arguments.

在 AQA 进阶数学中，你可能需要在代数式、极坐标式和指数式之间进行转换，或利用指数方法求解如 z⁴ = 81i 的方程。关键是将两边都化为 re

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