Complex Numbers Revision for AQA Mathematics | AQA 数学：复数考点精讲

📚 Complex Numbers Revision for AQA Mathematics | AQA 数学：复数考点精讲

Complex numbers extend the real number system by incorporating the imaginary unit i, where i² = -1. This revision guide covers the key AQA Further Mathematics syllabus: from algebraic operations and conjugates to Argand diagrams, polar form, and De Moivre’s theorem. Master these concepts to solve polynomial equations and tackle complex geometry with confidence.

复数通过引入虚数单位 i（满足 i² = -1）扩展了实数系统。本复习指南涵盖 AQA 进阶数学大纲的核心内容：从代数运算、共轭到阿尔冈图、极坐标形式和棣莫弗定理。掌握这些概念，自信地解决多项式方程并处理复几何问题。

1. The Imaginary Unit i | 虚数单位 i

The imaginary unit i is defined by the equation i² = -1. It is not a real number; no real number squared gives a negative result. From this definition, higher powers of i follow a cyclic pattern: i¹ = i, i² = -1, i³ = -i, i⁴ = 1, and then the cycle repeats every four powers.

虚数单位 i 由等式 i² = -1 定义。它不是实数；没有任何实数的平方为负数。由此定义，i 的高次幂遵循循环规律：i¹ = i, i² = -1, i³ = -i, i⁴ = 1，之后每四次幂循环一次。

When simplifying expressions with i, always reduce powers modulo 4. For example, i⁷ = i³ = -i, because 7 mod 4 = 3. This cyclic behaviour is fundamental when working with products and sums involving imaginary terms.

化简含 i 的表达式时，始终将幂次对 4 取模。例如 i⁷ = i³ = -i，因为 7 除以 4 余 3。在处理涉及虚数项的乘积与和时，这种循环性质是基础。

2. Complex Numbers in the Form a + bi | a + bi 形式的复数

A complex number z is expressed as z = a + bi, where a and b are real numbers. Here, a is called the real part, Re(z), and b is the imaginary part, Im(z). The imaginary part is the coefficient of i, not the whole term bi.

复数 z 表示为 z = a + bi，其中 a 和 b 是实数。这里 a 称为实部，记作 Re(z)；b 称为虚部，记作 Im(z)。虚部是 i 的系数，而不是整个 bi 项。

Two complex numbers are equal if and only if their real and imaginary parts are both equal: a + bi = c + di ⇔ a = c and b = d. This property is used to solve equations involving z by equating real and imaginary parts separately.

两个复数相等，当且仅当它们的实部和虚部分别相等：a + bi = c + di ⇔ a = c 且 b = d。这一性质常用于通过分别比较实部与虚部来求解含 z 的方程。

3. Addition and Subtraction | 加法与减法

To add or subtract complex numbers, combine the real parts and combine the imaginary parts separately. If z₁ = a + bi and z₂ = c + di, then:

复数相加或相减时，只需分别合并实部与虚部。若 z₁ = a + bi, z₂ = c + di，则：

z₁ + z₂ = (a + c) + (b + d)i

z₁ – z₂ = (a – c) + (b – d)i

This mirrors vector addition in the plane, which will be explored later in the Argand diagram. Subtracting a complex number effectively adds its additive inverse: -z = -a – bi.

这类似于平面内向量的加法，将在后续的阿尔冈图中探讨。减去一个复数相当于加上它的加法逆元：-z = -a – bi。

Be careful when negatives appear. For instance, (5 – 2i) – (3 + i) = (5 – 3) + (-2 – 1)i = 2 – 3i. Always distribute the minus sign across both real and imaginary parts of the second number.

当出现负数时要小心。例如 (5 – 2i) – (3 + i) = (5 – 3) + (-2 – 1)i = 2 – 3i。务必将负号分配给第二个数的实部和虚部。

4. Multiplication | 乘法

Multiplication of complex numbers uses the ordinary distributive law and the fact that i² = -1. Expand (a + bi)(c + di) as if they were binomials, then simplify:

复数乘法利用普通的分配律以及 i² = -1 的事实。像展开二项式那样展开 (a + bi)(c + di)，然后化简：

(a + bi)(c + di) = ac + adi + bci + bdi² = (ac – bd) + (ad + bc)i

This formula can be remembered, but it is safer to perform the expansion step by step to avoid sign errors, especially when some coefficients are negative.

这个公式可以记住，但更稳妥的做法是逐步展开以避免符号错误，尤其是当某些系数为负时。

A special product arises when multiplying a complex number by its own conjugate (see next section), yielding a real number. For instance, (x + yi)(x – yi) = x² + y², since the cross terms cancel.

当复数与其自身的共轭复数相乘时，会得到一个特殊的实数乘积（见下一节）。例如 (x + yi)(x – yi) = x² + y²，因为交叉项互相抵消。

5. Complex Conjugate | 共轭复数

The complex conjugate of z = a + bi is denoted as z̄ (or z*) and is defined as z̄ = a – bi. Geometrically, the conjugate reflects the point across the real axis in the Argand diagram.

复数 z = a + bi 的共轭复数记作 z̄（或 z*），定义为 z̄ = a – bi。从几何上看，共轭复数将点关于实轴反射到阿尔冈图上。

Conjugates have useful algebraic properties:

Sum: z + z̄ = 2a = 2 Re(z) is real.
Product: z z̄ = a² + b² = |z|², always a non-negative real number.

共轭复数有一些有用的代数性质：

和： z + z̄ = 2a = 2 Re(z)，是一个实数。
积： z z̄ = a² + b² = |z|²，总是一个非负实数。

These properties are essential when dividing complex numbers and when proving results about moduli. Also note that the conjugate of a conjugate returns the original number: (z̄)̄ = z.

这些性质在复数除法以及证明有关模长的结论时至关重要。另外注意，共轭的共轭等于原复数：(z̄)̄ = z。

6. Division | 除法

To divide two complex numbers, multiply the numerator and denominator by the complex conjugate of the denominator. This turns the denominator into a real number.

两个复数相除时，将分子和分母同时乘以分母的共轭复数。这会把分母转化为实数。

z₁ / z₂ = (a + bi) / (c + di) = [(a + bi)(c – di)] / [(c + di)(c – di)]

This results in a new complex number expressed as p + qi. The real denominator is c² + d².

结果得到一个新的复数，表示为 p + qi。实分母为 c² + d²。

For example, (3 + i) / (2 – i) = (3 + i)(2 + i) / (4 + 1) = (6 + 3i + 2i + i²) / 5 = (6 + 5i – 1) / 5 = (5 + 5i)/5 = 1 + i.

例如，(3 + i) / (2 – i) = (3 + i)(2 + i) / (4 + 1) = (6 + 3i + 2i + i²) / 5 = (6 + 5i – 1) / 5 = (5 + 5i)/5 = 1 + i。

Always simplify the final answer into the form a + bi, ensuring a and b are fully reduced fractions if necessary.

始终将最终答案化简为 a + bi 的形式，必要时确保 a 和 b 是完全约分的分数。

7. Argand Diagram | 阿尔冈图

An Argand diagram represents complex numbers as points or vectors on a plane, with the real part on the x-axis and the imaginary part on the y-axis. The point representing z = a + bi is (a, b).

阿尔冈图将复数表示为平面上的点或向量，其中实部在 x 轴上，虚部在 y 轴上。表示 z = a + bi 的点坐标为 (a, b)。

Addition of complex numbers corresponds to vector addition on the Argand diagram. Subtraction yields the vector from the endpoint of the subtracted number to the endpoint of the minuend.

复数的加法对应阿尔冈图上的向量加法。减法得到从减数终点指向被减数终点的向量。

If z₁ and z₂ are plotted, then z₁ + z₂ is the diagonal of the parallelogram formed by the two position vectors. This geometric interpretation helps visualise relationships like |z₁ – z₂| as the distance between two points.

若绘出 z₁ 和 z₂，则 z₁ + z₂ 就是由这两个位置向量构成的平行四边形的对角线。这种几何解释有助于将 |z₁ – z₂| 理解为两点间的距离。

8. Modulus and Argument | 模与辐角

The modulus of a complex number z = a + bi is denoted |z| and defined as the distance from the origin to the point (a, b):

复数 z = a + bi 的模记作 |z|，定义为从原点到点 (a, b) 的距离：

|z| = √(a² + b²)

The argument of z, written arg(z), is the angle θ measured from the positive real axis to the line segment joining the origin to (a, b). It is usually given in radians, with principal value in (-π, π].

z 的辐角记作 arg(z)，是从正实轴到连接原点与 (a, b) 的线段所成的角 θ。通常以弧度给出，主值位于 (-π, π]。

Thus, tan θ = b/a, but care must be taken to use the correct quadrant. For example, if a < 0 and b > 0, the angle is in the second quadrant, so θ = π – arctan(|b/a|).

因此，tan θ = b/a，但需要注意使用正确的象限。例如，若 a < 0 且 b > 0，角度在第二象限，则 θ = π – arctan(|b/a|)。

The modulus corresponds to the magnitude of a vector, and its properties include |z₁ z₂| = |z₁| |z₂| and |z₁ / z₂| = |z₁| / |z₂|. The argument follows arg(z₁ z₂) = arg(z₁) + arg(z₂) modulo 2π.

模对应向量的大小，其性质包括 |z₁ z₂| = |z₁| |z₂| 和 |z₁ / z₂| = |z₁| / |z₂|。辐角满足 arg(z₁ z₂) = arg(z₁) + arg(z₂)，模 2π 考虑主值。

9. Polar Form | 极坐标形式

The polar (or modulus-argument) form of a complex number expresses z in terms of its modulus r = |z| and argument θ = arg(z):

复数的极坐标形式（或称模-辐角形式）用模长 r = |z| 和辐角 θ = arg(z) 来表示 z：

z = r (cos θ + i sin θ)

This form is extremely useful for multiplication, division, and raising to powers, because trigonometric identities reveal simple rules in polar form.

这种形式对乘法、除法和求幂极为有用，因为三角恒等式在极坐标形式下展现出简洁的规则。

Two complex numbers in polar form multiply by multiplying their moduli and adding their arguments: r₁(cos θ₁ + i sin θ₁) × r₂(cos θ₂ + i sin θ₂) = r₁r₂ [cos(θ₁+θ₂) + i sin(θ₁+θ₂)].

两个极坐标形式的复数相乘：模长相乘，辐角相加：r₁(cos θ₁ + i sin θ₁) × r₂(cos θ₂ + i sin θ₂) = r₁r₂ [cos(θ₁+θ₂) + i sin(θ₁+θ₂)]。

Similarly, division gives r₁/r₂ [cos(θ₁-θ₂) + i sin(θ₁-θ₂)]. This highlights why polar form is often preferred over Cartesian a+bi when dealing with products.

类似地，除法得到 r₁/r₂ [cos(θ₁-θ₂) + i sin(θ₁-θ₂)]。这也说明了在处理乘积时为什么极坐标形式常优于笛卡尔 a+bi 形式。

10. Euler’s Formula and Exponential Form | 欧拉公式与指数形式

Euler’s formula states that e^(iθ) = cos θ + i sin θ. This connects complex exponentials with trigonometric functions and allows the compact exponential form of a complex number:

欧拉公式表明 e^(iθ) = cos θ + i sin θ。它将复指数与三角函数联系起来，并使复数可以用紧凑的指数形式表示：

z = r e^(iθ)

where r = |z| and θ = arg(z). This notation makes multiplication obvious: r₁ e^(iθ₁) × r₂ e^(iθ₂) = r₁r₂ e^(i(θ₁+θ₂)).

其中 r = |z|, θ = arg(z)。这一记法使得乘法显而易见：r₁ e^(iθ₁) × r₂ e^(iθ₂) = r₁r₂ e^(i(θ₁+θ₂))。

The exponential form is particularly elegant when applying De Moivre’s theorem or solving equations. It also leads to the famous identity e^(iπ) + 1 = 0.

在应用棣莫弗定理或解方程时，指数形式尤为优雅。它还引出了著名的恒等式 e^(iπ) + 1 = 0。

To convert between forms, compute r = √(a²+b²) and θ appropriately. For instance, -1 + i has modulus √2 and argument 3π/4, so -1 + i = √2 e^(i 3π/4).

在形式之间转换时，计算 r = √(a²+b²) 并恰当确定 θ。例如，-1 + i 的模为 √2，辐角为 3π/4，故 -1 + i = √2 e^(i 3π/4)。

11. De Moivre’s Theorem | 棣莫弗定理

De Moivre’s theorem extends the polar multiplication rule to powers. For any integer n:

棣莫弗定理将极坐标乘法规则推广到幂次。对于任意整数 n：

[r (cos θ + i sin θ)]ⁿ = rⁿ (cos nθ + i sin nθ)

When r = 1, this reduces to (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ). This identity is extremely useful for deriving trigonometric multiple-angle formulas, such as expressing cos 3θ in terms of cos θ.

当 r = 1 时，简化为 (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)。这个恒等式对推导三角函数的倍角公式极为有用，例如用 cos θ 表示 cos 3θ。

De Moivre’s theorem can also be used to find nth roots of a complex number. To solve zⁿ = w, express w in polar form, then the n distinct roots are given by:

棣莫弗定理还可用于求复数的 n 次方根。要解 zⁿ = w，先将 w 写成极坐标形式，则 n 个不同的根为：

zₖ = r^(1/n) [cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)], for k = 0, 1, …, n-1

These roots are equally spaced around a circle of radius r^(1/n) in the Argand diagram, forming a regular n-gon.

这些根在阿尔冈图上均匀分布在半径为 r^(1/n) 的圆周上，构成一个正 n 边形。

12. Solving Polynomial Equations | 解多项式方程

Complex numbers complete the number system so that every polynomial equation of degree n has exactly n roots (counting multiplicity) in the complex plane. This is the Fundamental Theorem of Algebra.

复数完备了数系，使得每个 n 次多项式方程在复平面内恰好有 n 个根（计重数）。这就是代数基本定理。

If a polynomial with real coefficients has a complex root a + bi, then its conjugate a – bi is also a root. This property helps in factorising real polynomials into irreducible quadratic factors.

如果一个实系数多项式有一个复数根 a + bi，那么它的共轭 a – bi 也是根。这一性质有助于将实多项式因式分解为不可约的二次因式。

For a quadratic equation ax² + bx + c = 0 with real coefficients, if the discriminant Δ = b² – 4ac < 0, the roots are complex conjugates given by:

对于实系数的二次方程 ax² + bx + c = 0，若判别式 Δ = b² – 4ac < 0，则根为共轭复数：

x = [-b ± i √(4ac – b²)] / (2a)

When working with cubic or quartic equations, one real root can often be found by trial, and then the remaining quadratic factor yields a conjugate complex pair. Always present roots in exact form using surds and i.

对于三次或四次方程，常可通过试根找到一个实根，然后剩下的二次因式会产生一对共轭复数。始终用根式和 i 给出根的精确形式。

For example, solve z³ – 1 = 0: one root is z = 1; the other two come from z² + z + 1 = 0, giving z = (-1 ± i√3)/2. These are the complex cube roots of unity.

例如，解 z³ – 1 = 0：一个根是 z = 1；另两个由 z² + z + 1 = 0 给出，即 z = (-1 ± i√3)/2。这些就是 1 的复立方根。

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Complex Numbers Revision for AQA Mathematics | AQA 数学：复数 考点精讲