Complex Numbers for Edexcel A-Level | A-Level Edexcel 数学：复数考点精讲

📚 Complex Numbers for Edexcel A-Level | A-Level Edexcel 数学：复数考点精讲

Complex numbers form a central pillar of Edexcel A-Level Further Mathematics, bridging algebra, geometry, and trigonometry. This revision guide unpacks every essential topic, from basic operations to De Moivre’s theorem and loci, ensuring you can tackle both Core Pure 1 and Core Pure 2 exam questions with confidence.

复数在 Edexcel A-Level 进阶数学中连接了代数、几何与三角学的核心内容，是必须掌握的基础知识。本考点精讲将从基本运算开始，逐步深入到德莫弗定理和轨迹问题，帮助你从容应对 Core Pure 1 和 Core Pure 2 中的各类考题。

1. Introduction to Complex Numbers | 复数的引入与基本运算

A complex number is an expression of the form z = a + bi, where a and b are real numbers, and i is the imaginary unit satisfying i² = –1. The set of complex numbers is denoted by ℂ. The real part of z is Re(z) = a, and the imaginary part is Im(z) = b.

复数是形如 z = a + bi 的表达式，其中 a 和 b 为实数，i 为虚数单位且满足 i² = –1。复数集合记作 ℂ。z 的实部为 Re(z) = a，虚部为 Im(z) = b。

Addition and subtraction are performed component-wise: (a + bi) ± (c + di) = (a ± c) + (b ± d)i. Multiplication uses the distributive law and the fact i² = –1: (a + bi)(c + di) = ac + adi + bci + bdi² = (ac – bd) + (ad + bc)i.

复数的加减法按实部和虚部分别运算：(a + bi) ± (c + di) = (a ± c) + (b ± d)i。乘法利用分配律及 i² = –1 进行展开：(a + bi)(c + di) = ac + adi + bci + bdi² = (ac – bd) + (ad + bc)i。

Equality of complex numbers means both real parts and imaginary parts must be equal. This principle is frequently used to solve equations by comparing real and imaginary components.

两个复数相等当且仅当它们的实部与虚部分别相等。这一原则常用来通过比较实部和虚部求解方程。

2. Complex Conjugate and Division | 共轭复数与除法

The complex conjugate of z = a + bi is denoted by z̄ or z* and is defined as z̄ = a – bi. Conjugation reflects the number across the real axis on an Argand diagram. Key properties include: z + z̄ = 2a (purely real), z – z̄ = 2bi (purely imaginary), and z·z̄ = a² + b² (a non-negative real number).

复数 z = a + bi 的共轭记作 z̄ 或 z*，定义为 z̄ = a – bi。共轭在 Argand 图上体现为关于实轴的对称。重要性质包括：z + z̄ = 2a 为纯实数，z – z̄ = 2bi 为纯虚数，且 z·z̄ = a² + b² 为非负实数。

To divide one complex number by another, multiply numerator and denominator by the conjugate of the denominator:

(a + bi) ÷ (c + di) = (a + bi)(c – di) / (c² + d²)

进行复数除法时，将分子分母同乘分母的共轭复数即可：

(a + bi) ÷ (c + di) = (a + bi)(c – di) / (c² + d²)

This yields a complex number in standard rectangular form. Always simplify the result to a + bi by separating real and imaginary parts.

所得结果应化简成标准的 a + bi 形式，即将实部和虚部分开表示。

3. Solving Quadratic Equations | 求解二次方程与判别式

For quadratic equations with real coefficients, the discriminant Δ = b² – 4ac determines the nature of the roots. If Δ ≥ 0, roots are real. If Δ < 0, the equation has a pair of complex conjugate roots, given by:

x = (–b ± i√(4ac – b²)) / 2a

对于实系数二次方程，判别式 Δ = b² – 4ac 决定了根的性质。当 Δ ≥ 0 时，根为实数。当 Δ < 0 时，方程有一对共轭复根，公式为：

x = (–b ± i√(4ac – b²)) / 2a

This means complex roots of polynomials with real coefficients always occur in conjugate pairs. If a complex number is a root of such a polynomial, its conjugate is automatically a root as well.

这意味着实系数多项式的复根总是以共轭对的形式出现。如果一个复数是该多项式的根，那么它的共轭也必然是另一个根。

4. Argand Diagram Representation | 复数的 Argand 图表示

Complex numbers can be visualised as points or position vectors on the complex plane, also called the Argand diagram. The horizontal axis represents the real part, and the vertical axis represents the imaginary part. The point (a, b) corresponds to z = a + bi.

复数可以在复平面上，即 Argand 图上表示为点或位置向量。横轴表示实部，纵轴表示虚部。点 (a, b) 即对应复数 z = a + bi。

Addition of complex numbers corresponds to vector addition. Subtraction gives the vector from one complex number to another. The Argand diagram is powerful for solving geometric problems and understanding transformations.

复数的加法对应向量加法，减法表示一个复数到另一个复数所成的向量。Argand 图在解决几何问题和理解变换时非常有用。

5. Modulus and Argument | 模与辐角

The modulus of a complex number z = a + bi, written |z|, is the distance from the origin to the point (a, b) on the Argand diagram: |z| = √(a² + b²). It is always non‑negative, and |z| = 0 if and only if z = 0.

复数 z = a + bi 的模记作 |z|，表示 Argand 图上原点到点 (a, b) 的距离：|z| = √(a² + b²)。模总是非负的，且 |z| = 0 当且仅当 z = 0。

The argument of z, denoted arg(z), is the angle θ (in radians) between the positive real axis and the line segment from the origin to (a, b). It is usually taken in the principal range (–π, π] or [0, 2π). For a ≠ 0, arg(z) = arctan(b/a) with quadrant adjustments.

复数 z 的辐角记作 arg(z)，表示正实轴与原点至 (a, b) 连线之间的夹角 θ（以弧度计）。通常取主值区间 (–π, π] 或 [0, 2π)。当 a ≠ 0 时，arg(z) = arctan(b/a)，并需根据象限进行调整。

Key properties include |z₁z₂| = |z₁||z₂| and arg(z₁z₂) = arg(z₁) + arg(z₂) (mod 2π). Similarly for division: |z₁ / z₂| = |z₁| / |z₂|, arg(z₁ / z₂) = arg(z₁) – arg(z₂).

重要性质包括：|z₁z₂| = |z₁||z₂|，arg(z₁z₂) = arg(z₁) + arg(z₂)（模 2π）。除法类似：|z₁ / z₂| = |z₁| / |z₂|，arg(z₁ / z₂) = arg(z₁) – arg(z₂)。

6. Modulus‑Argument Form and Euler’s Formula | 极坐标形式与欧拉公式

The modulus–argument form (or polar form) of a complex number is z = r(cosθ + i sinθ), where r = |z| and θ = arg(z). This form makes multiplication, division, and powers far more manageable.

复数的模–辐角形式（也称极坐标形式）为 z = r(cosθ + i sinθ)，其中 r = |z|，θ = arg(z)。该形式让乘法、除法和乘方运算变得极为简便。

Using Euler’s formula, e^(iθ) = cosθ + i sinθ, we have a compact exponential representation: z = re^(iθ). For multiplication:

z₁z₂ = r₁r₂ e^(i(θ₁ + θ₂))

利用欧拉公式 e^(iθ) = cosθ + i sinθ，可得简洁的指数表示式 z = re^(iθ)。对于乘法：

z₁z₂ = r₁r₂ e^(i(θ₁ + θ₂))

The conjugate in polar form is z̄ = r e^(–iθ) = r(cosθ – i sinθ). This representation highlights why z z̄ = r².

极坐标形式下的共轭为 z̄ = r e^(–iθ) = r(cosθ – i sinθ)。该表达式清楚地展示了为什么 z z̄ = r²。

7. De Moivre’s Theorem | 德莫弗定理

De Moivre’s theorem states that for any integer n:

(cosθ + i sinθ)ⁿ = cos(nθ) + i sin(nθ)

德莫弗定理指出，对于任意整数 n：

(cosθ + i sinθ)ⁿ = cos(nθ) + i sin(nθ)

In exponential form this is simply (e^(iθ))ⁿ = e^(inθ). The theorem is used to raise complex numbers to integer powers, to derive trigonometric identities (e.g., expressions for cos 3θ in terms of cos θ), and to find nth roots.

用指数形式表示即 (e^(iθ))ⁿ = e^(inθ)，十分简洁。该定理可用于计算复数的整数次幂、推导三角恒等式（例如用 cos θ 表示 cos 3θ），以及求 n 次方根。

For example, to express cos 3θ as a polynomial in cos θ, expand (cosθ + i sinθ)³ using the binomial theorem, then take the real part: cos 3θ = 4cos³θ – 3cosθ. The imaginary part gives sin 3θ = 3sinθ – 4sin³θ.

例如，要将 cos 3θ 表示为 cos θ 的多项式，可用二项式定理展开 (cosθ + i sinθ)³，再取实部即得 cos 3θ = 4cos³θ – 3cosθ，虚部给出 sin 3θ = 3sinθ – 4sin³θ。

8. Roots of Complex Numbers | 复数根的求解

To find the nth roots of a complex number w = r(cosφ + i sinφ), write the general polar form with +2kπi:

zₖ = r^(1/n) [cos((φ + 2kπ)/n) + i sin((φ + 2kπ)/n)]

要求复数 w = r(cosφ + i sinφ) 的 n 次方根，需使用带 +2kπ 的通式表达：

zₖ = r^(1/n) [cos((φ + 2kπ)/n) + i sin((φ + 2kπ)/n)]

Here k = 0, 1, 2, …, n‑1 gives n distinct roots equally spaced around a circle of radius r^(1/n) in the Argand diagram. The nth roots of unity (1) are of particular importance: they lie on the unit circle at angles 2kπ/n and sum to zero.

其中 k = 0, 1, 2, …, n‑1，共计 n 个不同的根，它们在 Argand 图上均匀分布在半径为 r^(1/n) 的圆周上。单位根（1 的 n 次方根）尤为重要：它们位于单位圆上，角度为 2kπ/n，且所有根之和为零。

If the original polynomial has real coefficients, the roots will be either real or occur in conjugate pairs, forming symmetric patterns on the Argand diagram.

若原始多项式系数为实数，则这些根要么为实数，要么以共轭对的形式出现，在 Argand 图上形成对称模式。

9. Loci in the Complex Plane | 复平面上的轨迹

Many exam questions ask for the set of points z that satisfy a given condition, interpreted as a locus on the Argand diagram. Common loci include:

Circles: |z – a| = r is a circle with centre at the point a and radius r.
Perpendicular bisector: |z – a| = |z – b| is the perpendicular bisector of the segment joining a and b.
Half‑lines: arg(z – a) = θ is a half‑line starting at a (but not including a) making angle θ with the positive real axis.
Regions: Inequalities such as |z – a| < r describe the interior of a circle, while combined inequalities define specific regions.

许多考题会要求找出满足特定条件的点集，即复平面上的轨迹。常见轨迹包括：

圆： |z – a| = r 表示以点 a 为圆心、半径为 r 的圆。
垂直平分线： |z – a| = |z – b| 表示连接 a 与 b 的线段的垂直平分线。
射线： arg(z – a) = θ 表示从 a 出发（不含 a 点）、与正实轴夹角为 θ 的射线。
区域： 不等式如 |z – a| < r 表示圆内部，组合不等式则可界定特定区域。

Solving locus problems often involves substituting z = x + iy, equating moduli or arguments, and simplifying to Cartesian equations. Intersection of loci can be found by solving simultaneous equations.

求解轨迹问题时常代入 z = x + iy，比较模或辐角并化简为直角坐标方程。轨迹的交点可通过解联立方程求得。

10. Polynomial Equations and Conjugate Root Theorem | 多项式方程与共轭根定理

If a polynomial with real coefficients has a complex root a + bi, then its conjugate a – bi is also a root. This conjugate root theorem is crucial for factorising polynomials and solving equations fully.

若实系数多项式有一个复根 a + bi，则其共轭 a – bi 也是该多项式的根。这一共轭根定理对于因式分解和完整求解方程极为重要。

Given one complex root, you can form the quadratic factor (z – (a + bi))(z – (a – bi)) = z² – 2az + (a² + b²) with real coefficients. The remaining roots can then be found by polynomial division or comparing coefficients.

已知一个复根，即可构造实系数的二次因式 (z – (a + bi))(z – (a – bi)) = z² – 2az + (a² + b²)。其余根可通过多项式除法或比较系数求得。

For cubic and quartic equations, this theorem guarantees that the total number of non‑real roots is even, which helps in determining the nature of all roots when one complex number is known.

对于三次和四次方程，该定理保证非实根的总数为偶数，因此一旦已知一个复数，便可推断所有根的性质。

11. Transformations of the Complex Plane | 复平面的变换

Functions from ℂ to ℂ map sets of points to new positions. Standard transformations in Edexcel include:

Translation: w = z + c shifts every point by the vector c.
Enlargement (scaling): w = kz, where k > 0 is real, multiplies all distances from the origin by k.
Rotation: w = e^(iα) z rotates points by angle α about the origin.
Reflection: w = z̄ reflects across the real axis.
Inversion: w = 1/z maps circles and lines to circles or lines (Möbius transformation).

从复平面到复平面的函数将点集映射到新的位置。Edexcel 常见的标准变换包括：

平移： w = z + c 将所有点平移向量 c。
伸缩： w = kz，其中 k > 0 为实数，将所有点到原点的距离放大 k 倍。
旋转： w = e^(iα) z 将点绕原点旋转角度 α。
反射： w = z̄ 表示关于实轴的反射。
反演： w = 1/z 将圆和直线映射为圆或直线（莫比乌斯变换）。

Exam questions often ask for the image of a locus under a given transformation. Substituting z in terms of w, or working backwards from the target locus, can reveal the required mapping.

考题常要求找出某轨迹在给定变换下的像。通过将 z 用 w 表示，或从目标轨迹反向推导，可揭示所需的映射关系。

12. Equating Real and Imaginary Parts | 实部与虚部的应用

A powerful technique for solving equations involving complex numbers is to substitute z = x + iy and equate real and imaginary parts separately. This leads to two simultaneous real equations.

求解复数方程时一个强有力的方法是代入 z = x + iy，然后将实部与虚部分别令其相等，得到两个联立的实数方程。

This method is used to find the square roots of a complex number (e.g., solve (x + iy)² = a + bi), giving x² – y² = a and 2xy = b. It also appears in problems requiring the determination of all possible complex numbers satisfying a modulus‑argument condition.

该方法常用于求复数的平方根（如求解 (x + iy)² = a + bi），得到 x² – y² = a 和 2xy = b。它也出现在需要确定所有满足模与辐角条件的复数的问题中。

Another common scenario is solving equations like z + |z| = 2 + 8i by setting x + iy + √(x² + y²) = 2 + 8i. Comparing real parts gives x + √(x² + y²) = 2, and imaginary parts give y = 8, yielding a system that can be solved algebraically.

另一种常见情形是解如 z + |z| = 2 + 8i 这样的方程，令 x + iy + √(x² + y²) = 2 + 8i，比较实部得 x + √(x² + y²) = 2，比较虚部得 y = 8，从而得到可代数求解的方程组。

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Complex Numbers for Edexcel A-Level | A-Level Edexcel 数学：复数考点精讲