Complex Numbers | 复数考点精讲

📚 Complex Numbers | 复数考点精讲

Complex numbers extend the real number system and are fundamental for solving equations that have no real solutions, such as x² + 1 = 0. In the Edexcel A-Level Mathematics syllabus, you will learn to manipulate complex numbers in algebraic and polar forms, apply De Moivre’s theorem, find roots, and use the Argand diagram for geometric representation.

复数在实数系的基础上进行了扩展，是解决像 x² + 1 = 0 这样没有实数解的方程的基础。在 Edexcel A-Level 数学大纲中，你将学习如何对复数进行代数形式和极坐标形式的运算，应用棣莫弗定理，求根，并利用 Argand 图进行几何表示。

1. The Imaginary Unit and Standard Form | 虚数单位与标准形式

A complex number is written as z = a + bi, where a and b are real numbers. The symbol i denotes the imaginary unit, defined by the property i² = –1. The real part of z is Re(z) = a, and the imaginary part is Im(z) = b (note that the imaginary part is the coefficient b, not bi).

复数写作 z = a + bi，其中 a 和 b 是实数。符号 i 表示虚数单位，定义为 i² = –1。z 的实部记作 Re(z) = a，虚部为 Im(z) = b（注意虚部是系数 b，而不是 bi）。

If b = 0, the number is purely real; if a = 0, the number is purely imaginary, e.g. 3i, –5i. The set of all complex numbers is denoted by ℂ.

如果 b = 0，该数就是纯实数；如果 a = 0，则为纯虚数，例如 3i、–5i。全体复数的集合用 ℂ 表示。

When solving quadratic equations, a negative discriminant leads to complex roots. For instance, x² + 4x + 13 = 0 gives roots x = –2 ± 3i.

在解二次方程时，若判别式为负，则得到复根。例如方程 x² + 4x + 13 = 0 的根为 x = –2 ± 3i。

2. Complex Arithmetic: Addition and Subtraction | 复数运算：加法与减法

To add or subtract complex numbers, simply combine their real parts and their imaginary parts separately: (a + bi) ± (c + di) = (a ± c) + (b ± d)i. The operations follow the usual rules of algebra, treating i as a variable while remembering i² = –1.

复数的加减法只需分别合并实部和虚部：(a + bi) ± (c + di) = (a ± c) + (b ± d)i。运算遵循代数的常规规则，将 i 视为变量，但需谨记 i² = –1。

For example, (3 + 5i) + (2 – 4i) = 5 + i, and (6 – i) – (4 + 3i) = 2 – 4i. These operations are straightforward and form the basis of complex algebra.

例如，(3 + 5i) + (2 – 4i) = 5 + i，(6 – i) – (4 + 3i) = 2 – 4i。这些运算直接明了，是复数代数的基础。

3. Multiplication | 乘法运算

Multiplication of complex numbers uses the distributive law (FOIL) together with the substitution i² = –1: (a + bi)(c + di) = ac + adi + bci + bdi² = (ac – bd) + (ad + bc)i.

复数乘法利用分配律（FOIL）并代入 i² = –1 来进行：(a + bi)(c + di) = ac + adi + bci + bdi² = (ac – bd) + (ad + bc)i。

For instance, (2 + 3i)(4 – i) = 8 – 2i + 12i – 3i² = 8 + 10i – 3(–1) = 11 + 10i. Multiplying by a real number k simply scales both parts: k(a + bi) = ka + kbi.

例如，(2 + 3i)(4 – i) = 8 – 2i + 12i – 3i² = 8 + 10i – 3(–1) = 11 + 10i。乘以实数 k 只是对实部和虚部进行缩放：k(a + bi) = ka + kbi。

In particular, multiplying by i leads to a rotation: i(a + bi) = –b + ai. This rotation property is a cornerstone of geometric interpretation.

特别地，乘以 i 相当于一次旋转：i(a + bi) = –b + ai。这种旋转性质是几何解释的基础。

4. Complex Conjugates and Division | 共轭复数与除法

The complex conjugate of z = a + bi is written as z̄ = a – bi. It is obtained by changing the sign of the imaginary part. Conjugates have important properties: zz̄ = a² + b², which is always a non-negative real number. Also, (z̄) = z, and the sum z + z̄ = 2a is twice the real part.

复数 z = a + bi 的共轭记作 z̄ = a – bi，通过改变虚部的符号得到。共轭具有重要性质：zz̄ = a² + b²，总是一个非负实数。此外，(z̄) = z，而 z + z̄ = 2a 是实部的两倍。

To divide two complex numbers, multiply the numerator and denominator by the conjugate of the denominator: (a + bi)/(c + di) = (a + bi)(c – di) / (c² + d²). This yields a complex number in standard form.

进行两个复数的除法时，分子分母同时乘以分母的共轭：(a + bi)/(c + di) = (a + bi)(c – di) / (c² + d²)，结果化为标准形式。

For example, (3 + 2i)/(1 – i) = (3 + 2i)(1 + i)/(1² + 1²) = (3 + 3i + 2i + 2i²)/2 = (1 + 5i)/2 = 0.5 + 2.5i. Division is crucial when solving equations involving complex coefficients.

例如，(3 + 2i)/(1 – i) = (3 + 2i)(1 + i)/(1² + 1²) = (3 + 3i + 2i + 2i²)/2 = (1 + 5i)/2 = 0.5 + 2.5i。在处理含有复数系数的方程时，除法至关重要。

5. The Argand Diagram | 阿尔冈图（复平面）

The Argand diagram represents complex numbers as points or vectors in a plane. The horizontal axis corresponds to the real part, and the vertical axis to the imaginary part. Thus, z = a + bi is plotted at (a, b).

阿尔冈图是将复数表示为平面上的点或向量。横轴对应实部，纵轴对应虚部。因此，z = a + bi 就绘制于点 (a, b) 处。

This geometric view makes operations like addition and subtraction visualisable as vector additions. The conjugate z̄ is the reflection of z across the real axis. The distance from the origin gives the modulus, and the angle with the positive real axis gives the argument.

这一几何视角使加减法等运算可视化为向量加法。共轭 z̄ 就是 z 关于实轴的反射。点到原点的距离就是模，与正实轴的夹角就是辐角。

6. Modulus and Argument | 模与辐角

The modulus of z = a + bi, denoted |z|, is the distance from the origin to the point (a, b): |z| = √(a² + b²). It is always non-negative and equals √(zz̄). The modulus represents the ‘size’ of the complex number.

复数 z = a + bi 的模，记作 |z|，是原点到点 (a, b) 的距离：|z| = √(a² + b²)。模总是非负的，且等于 √(zz̄)。它表示复数的“大小”。

The argument of z, arg(z), is the angle θ measured from the positive real axis to the vector representing z, usually in radians. The principal argument is the value in the interval (–π, π]. It can be found using tanθ = b/a, with care taken for the quadrant in which z lies.

z 的辐角 arg(z) 是从正实轴到表示 z 的向量的夹角 θ，通常以弧度为单位。主辐角取值在区间 (–π, π] 内。它可以利用 tanθ = b/a 求得，但需注意 z 所在的象限。

For example, for z = –1 + i, modulus is √(1+1) = √2, and the argument is 3π/4 (since it lies in the second quadrant).

例如，对于 z = –1 + i，模为 √(1+1) = √2，辐角为 3π/4（因为位于第二象限）。

7. Polar Form of Complex Numbers | 复数的极形式

Using modulus r = |z| and argument θ = arg(z), any non-zero complex number can be expressed in polar form: z = r(cosθ + i sinθ). This is often abbreviated as r cis θ. Polar form is particularly useful for multiplication, division and exponentiation.

利用模 r = |z| 和辐角 θ = arg(z)，任何非零复数都可以用极形式表示：z = r(cosθ + i sinθ)。这常简记为 r cis θ。极形式在进行乘、除和幂运算时尤为有用。

To convert from Cartesian to polar form, compute r = √(a² + b²) and find θ using arctan(b/a) adjusted by quadrant. To convert back, a = r cosθ and b = r sinθ.

要从直角坐标形式转换为极形式，计算 r = √(a² + b²) 并通过 arctan(b/a) 及象限调整求得 θ。转换回来时，a = r cosθ，b = r sinθ。

8. Multiplication and Division in Polar Form | 极形式下的乘除

When multiplying two complex numbers in polar form, multiply their moduli and add their arguments: r₁(cosθ₁ + i sinθ₁) × r₂(cosθ₂ + i sinθ₂) = r₁r₂[cos(θ₁+θ₂) + i sin(θ₁+θ₂)].

两个极形式复数相乘时，将其模相乘、辐角相加：r₁(cosθ₁ + i sinθ₁) × r₂(cosθ₂ + i sinθ₂) = r₁r₂[cos(θ₁+θ₂) + i sin(θ₁+θ₂)]。

For division, divide the moduli and subtract the arguments: r₁/r₂ [cos(θ₁–θ₂) + i sin(θ₁–θ₂)], provided r₂ ≠ 0.

除法则是把模相除、辐角相减：r₁/r₂ [cos(θ₁–θ₂) + i sin(θ₁–θ₂)]，前提是 r₂ ≠ 0。

This makes geometric sense: multiplication by a complex number scales and rotates the other number. For instance, multiplying by i (which has modulus 1 and argument π/2) rotates a point by 90° anticlockwise.

这具有几何意义：乘以一个复数相当于对另一个复数进行缩放和旋转。例如，乘以 i（模为 1、辐角为 π/2）会将一个点逆时针旋转 90°。

9. De Moivre’s Theorem | 棣莫弗定理

De Moivre’s theorem is an essential tool for raising complex numbers to integer powers. If z = r(cosθ + i sinθ), then for any integer n, zⁿ = rⁿ(cos nθ + i sin nθ).

棣莫弗定理是将复数进行整数次幂运算的重要工具。如果 z = r(cosθ + i sinθ)，那么对于任意整数 n，有 zⁿ = rⁿ(cos nθ + i sin nθ)。

The theorem can be proved by induction for positive n, and then extended to negative integers using the reciprocal property. It is particularly useful in deriving trigonometric identities and solving equations of the form zⁿ = w.

该定理可通过归纳法对正整数证明，并利用倒数性质推广到负整数。它在推导三角恒等式以及求解形如 zⁿ = w 的方程时特别有用。

For example, to compute (1 + i)⁵, write 1 + i = √2(cos π/4 + i sin π/4). Then (1 + i)⁵ = (√2)⁵ [cos(5π/4) + i sin(5π/4)] = 4√2 (–√2/2 – i√2/2) = –4 – 4i.

例如，计算 (1 + i)⁵，先将 1 + i 写成 √2(cos π/4 + i sin π/4)。则 (1 + i)⁵ = (√2)⁵ [cos(5π/4) + i sin(5π/4)] = 4√2 (–√2/2 – i√2/2) = –4 – 4i。

10. Roots of Complex Numbers | 复数的根

To find the nth roots of a complex number w, write w in polar form w = r(cosθ + i sinθ). The nth roots are given by: zₖ = r^(1/n) [cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)], for k = 0, 1, 2, …, n–1.

求复数 w 的 n 次方根时，先将 w 写为极形式 w = r(cosθ + i sinθ)。n 次方根为：zₖ = r^(1/n) [cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)]，k = 0, 1, 2, …, n–1。

These n roots lie equally spaced on a circle of radius r^(1/n) in the Argand diagram, separated by an angle of 2π/n. They form the vertices of a regular n-gon centred at the origin.

这 n 个根均匀分布在复平面内半径为 r^(1/n) 的圆周上，彼此间隔 2π/n 的角度，构成以原点为中心的正 n 边形的顶点。

For instance, the cube roots of 8 are found by writing 8 = 8(cos0 + i sin0). The roots are 2, 2(cos 2π/3 + i sin 2π/3) = –1 + i√3, and 2(cos 4π/3 + i sin 4π/3) = –1 – i√3.

例如，8 的立方根可将 8 写作 8(cos0 + i sin0) 求得，根为 2，2(cos 2π/3 + i sin 2π/3) = –1 + i√3，以及 2(cos 4π/3 + i sin 4π/3) = –1 – i√3。

11. Solving Polynomial Equations with Complex Roots | 解带复根的多项式方程

Complex numbers ensure that every polynomial equation of degree n has exactly n roots in ℂ (counting multiplicity), known as the Fundamental Theorem of Algebra. For real-coefficient polynomials, non-real complex roots always occur in conjugate pairs.

复数保证了任何 n 次多项式方程在复数域 ℂ 内恰好有 n 个根（计重数），这称为代数基本定理。对于实系数多项式，非实复数根总是成对共轭出现。

Thus, if a + bi is a root, so is a – bi. This fact helps to factorise polynomials and find remaining roots when one complex root is known. For cubics, if one real root is found, the quadratic factor can be solved to give two complex conjugate roots.

因此，若 a + bi 是一个根，则 a – bi 也是根。利用这一事实可以分解多项式，并在已知一个复数根时求出其余根。对于三次方程，若找到一个实根，二次因式可解出两个共轭复根。

Example: Given that 2 + i is a root of f(z) = z³ – 5z² + 9z – 5, the conjugate 2 – i is also a root. The product (z – (2+i))(z – (2–i)) = z² – 4z + 5 divides f(z), giving the remaining factor (z – 1). So roots are 1, 2 ± i.

例如：已知 2 + i 是 f(z) = z³ – 5z² + 9z – 5 的一个根，则其共轭 2 – i 也是根。乘积 (z – (2+i))(z – (2–i)) = z² – 4z + 5 可整除 f(z)，得到剩余因式 (z – 1)。故根为 1, 2 ± i。

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

In Edexcel A-Level, students are often asked to sketch loci – sets of points satisfying certain conditions involving modulus and argument. Common loci include circles, half-lines, and perpendicular bisectors.

在 Edexcel A-Level 考试中，常要求学生绘制满足特定模和辐角条件的点集轨迹。常见轨迹包括圆、射线和垂直平分线。

A condition |z – a| = r describes a circle centred at the complex number a with radius r. |z – a| = |z – b| gives the perpendicular bisector of the segment joining a and b. arg(z – a) = θ represents a half-line starting from a at angle θ to the positive real axis.

条件 |z – a| = r 描述的是以复数 a 为圆心、r 为半径的圆；|z – a| = |z – b| 表示连接 a 和 b 线段的垂直平分线；arg(z – a) = θ 表示从 a 出发、与正实轴成 θ 角的射线。

Sketching these loci and finding intersections involves converting the conditions to Cartesian form or using geometric reasoning directly on the Argand diagram.

绘制这些轨迹并求交点时，可将条件转化为直角坐标形式，或直接在阿尔冈图上进行几何推理。

For example, the intersection of |z – 2i| = 2 and arg(z) = π/4 gives a point that can be found by solving both constraints simultaneously.

例如，轨迹 |z – 2i| = 2 与 arg(z) = π/4 的交点可同时解这两个约束条件求得。

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Complex Numbers | 复数考点精讲