Working with Complex Numbers | 复数的运算

📚 Working with Complex Numbers | 复数的运算

Complex numbers extend the real number system by introducing the imaginary unit i, where i² = −1. This article covers the essential operations and representations needed for AQA A-Level Mathematics.

复数通过引入虚数单位 i（其中 i² = −1）扩展了实数系统。本文涵盖 AQA A-Level 数学所需的复数基本运算与表示方法。

1. The Imaginary Unit and Complex Number Forms | 虚数单位与复数形式

The imaginary unit is defined as i = √(−1), so i² = −1. A complex number is written as z = a + bi, where a is the real part and b is the imaginary part. Here a and b are real numbers, and b can be zero.

虚数单位定义为 i = √(−1)，因此 i² = −1。复数写作 z = a + bi，其中 a 是实部，b 是虚部。这里 a 和 b 是实数，且 b 可以为 0。

If b = 0, then z is a real number. If a = 0 and b ≠ 0, then z is called a pure imaginary number. The set of complex numbers is denoted by ℂ and contains the real numbers ℝ as a subset.

若 b = 0，则 z 是实数；若 a = 0 且 b ≠ 0，则 z 称为纯虚数。复数集用 ℂ 表示，实数集 ℝ 是它的子集。

2. Adding and Subtracting Complex Numbers | 复数的加法与减法

To add two complex numbers, add their real parts and imaginary parts separately. For example, (3 + 2i) + (1 − 5i) = (3 + 1) + (2 − 5)i = 4 − 3i.

将两个复数相加时，分别将实部与实部、虚部与虚部相加。例如，(3 + 2i) + (1 − 5i) = (3 + 1) + (2 − 5)i = 4 − 3i。

Subtraction follows the same rule: (a + bi) − (c + di) = (a − c) + (b − d)i. These operations obey the usual commutative and associative laws.

减法遵循相同规则：(a + bi) − (c + di) = (a − c) + (b − d)i。这些运算满足通常的交换律与结合律。

3. Multiplying Complex Numbers | 复数的乘法

Multiplication uses the distributive law and the fact that i² = −1. Given z₁ = a + bi and z₂ = c + di, their product is (a + bi)(c + di) = ac + adi + bci + bdi² = (ac − bd) + (ad + bc)i.

乘法使用分配律及 i² = −1。设 z₁ = a + bi，z₂ = c + di，则它们的乘积为 (a + bi)(c + di) = ac + adi + bci + bdi² = (ac − bd) + (ad + bc)i。

For example, (2 + 3i)(1 − 4i) = 2 − 8i + 3i − 12i² = 2 − 5i + 12 = 14 − 5i. Notice that i² is replaced by −1, which can change the sign of the real part.

例如，(2 + 3i)(1 − 4i) = 2 − 8i + 3i − 12i² = 2 − 5i + 12 = 14 − 5i。注意将 i² 替换为 −1，这可能会改变实部的符号。

4. Complex Conjugates | 共轭复数

The complex conjugate of z = a + bi is written as z̄ = a − bi. It is obtained by changing the sign of the imaginary part. The conjugate is extremely useful in division and in finding the modulus.

复数 z = a + bi 的共轭复数写作 z̄ = a − bi，即将虚部的符号取反。共轭复数在除法运算和求模中非常有用。

Key properties: (z̄)̄ = z, and z × z̄ = a² + b², which is always a non-negative real number. Also, the sum z + z̄ = 2a is real, while the difference z − z̄ = 2bi is purely imaginary.

重要性质：(z̄)̄ = z，且 z × z̄ = a² + b² 始终是非负实数。此外，和 z + z̄ = 2a 是实数，而差 z − z̄ = 2bi 是纯虚数。

5. Dividing Complex Numbers | 复数的除法

To divide by a complex number, multiply the numerator and denominator by the conjugate of the denominator. This makes the denominator real.

除以一个复数时，将分子和分母同时乘以分母的共轭复数，从而使分母变为实数。

For example, (3 + 4i) / (1 − 2i) = (3 + 4i)(1 + 2i) / ((1 − 2i)(1 + 2i)) = (3 + 6i + 4i + 8i²) / (1 + 4) = (−5 + 10i) / 5 = −1 + 2i.

例如，(3 + 4i) / (1 − 2i) = (3 + 4i)(1 + 2i) / ((1 − 2i)(1 + 2i)) = (3 + 6i + 4i + 8i²) / (1 + 4) = (−5 + 10i) / 5 = −1 + 2i。

In general, (a + bi) / (c + di) = [(a + bi)(c − di)] / (c² + d²) = (ac + bd) / (c² + d²) + (bc − ad) / (c² + d²) i.

一般地，(a + bi) / (c + di) = [(a + bi)(c − di)] / (c² + d²) = (ac + bd) / (c² + d²) + (bc − ad) / (c² + d²) i。

6. The Argand Diagram | 阿尔冈图

A complex number z = a + bi can be represented graphically as a point (a, b) in the complex plane, called an Argand diagram. The horizontal axis is the real axis, and the vertical axis is the imaginary axis.

复数 z = a + bi 可以在复平面上表示为点 (a, b)，这种图形称为阿尔冈图。水平轴是实轴，垂直轴是虚轴。

Addition and subtraction of complex numbers correspond to vector addition and subtraction on the Argand diagram. Multiplication by i corresponds to an anticlockwise rotation of 90° about the origin.

复数的加法和减法对应于阿尔冈图上的向量加减。乘以 i 对应于绕原点逆时针旋转 90°。

7. Modulus and Argument | 模与辐角

The modulus of z = a + bi is |z| = √(a² + b²), and it represents the distance from the origin to the point (a, b). The argument arg(z) is the angle θ measured anticlockwise from the positive real axis to the line joining the origin to z.

复数 z = a + bi 的模为 |z| = √(a² + b²)，它表示原点 (0,0) 到点 (a,b) 的距离。辐角 arg(z) 是从正实轴逆时针旋转到原点与 z 连线所成的角度 θ。

We can express a and b in terms of the modulus and argument: a = |z| cos θ and b = |z| sin θ. Hence z = |z| (cos θ + i sin θ).

我们可以用模和辐角表示 a 和 b：a = |z| cos θ，b = |z| sin θ。因此 z = |z| (cos θ + i sin θ)。

8. The Polar Form and Euler’s Formula | 极坐标形式与欧拉公式

The polar form of a complex number is z = r(cos θ + i sin θ), where r = |z| and θ = arg(z). A more compact notation is z = r cis θ, where cis θ means cos θ + i sin θ.

复数的极坐标形式为 z = r(cos θ + i sin θ)，其中 r = |z|，θ = arg(z)。更紧凑的记法是 z = r cis θ，其中 cis θ 表示 cos θ + i sin θ。

Using Euler’s identity, e^(iθ) = cos θ + i sin θ, so we can write z = r e^(iθ). This form is especially useful for multiplication, division and powers of complex numbers.

利用欧拉恒等式 e^(iθ) = cos θ + i sin θ，我们可以将 z 写成 z = r e^(iθ)。这种形式对于复数的乘法、除法和幂运算尤其方便。

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

If z₁ = r₁(cos θ₁ + i sin θ₁) and z₂ = r₂(cos θ₂ + i sin θ₂), then their product is z₁z₂ = r₁r₂ [cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)]. This means moduli multiply and arguments add.

若 z₁ = r₁(cos θ₁ + i sin θ₁)，z₂ = r₂(cos θ₂ + i sin θ₂)，则乘积为 z₁z₂ = r₁r₂ [cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)]。也就是说，模相乘，辐角相加。

For division, z₁ / z₂ = (r₁ / r₂) [cos(θ₁ − θ₂) + i sin(θ₁ − θ₂)], provided r₂ ≠ 0. This gives a quick geometrical way to multiply and divide complex numbers.

对于除法，z₁ / z₂ = (r₁ / r₂) [cos(θ₁ − θ₂) + i sin(θ₁ − θ₂)]，其中 r₂ ≠ 0。这提供了一种快速进行复数乘除的几何方法。

10. Powers and De Moivre’s Theorem | 幂运算与棣莫弗定理

De Moivre’s theorem states that for any integer n, [r(cos θ + i sin θ)]ⁿ = rⁿ [cos(nθ) + i sin(nθ)]. This is a powerful result for calculating integer powers and roots of complex numbers.

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)]。这是计算复数整数幂和根的有力工具。

For example, (1 + i)⁶ can be computed by first writing 1 + i = √2 (cos 45° + i sin 45°). Then (1 + i)⁶ = (√2)⁶ [cos(6 × 45°) + i sin(6 × 45°)] = 8 [cos 270° + i sin 270°] = −8i.

例如，计算 (1 + i)⁶ 可先将 1 + i 写成 √2 (cos 45° + i sin 45°)。然后 (1 + i)⁶ = (√2)⁶ [cos(6 × 45°) + i sin(6 × 45°)] = 8 [cos 270° + i sin 270°] = −8i。

11. Solving Quadratic Equations with Complex Roots | 用复数根解二次方程

When a quadratic equation ax² + bx + c = 0 has a negative discriminant, its solutions are complex conjugates. The quadratic formula still works: x = (−b ± √(b² − 4ac)) / 2a.

当二次方程 ax² + bx + c = 0 的判别式为负数时，它的解是一对共轭复数。求根公式仍然适用：x = (−b ± √(b² − 4ac)) / 2a。

Example: solve x² − 2x + 5 = 0. The discriminant is (−2)² − 4 × 1 × 5 = −16. Thus x = (2 ± √(−16)) / 2 = (2 ± 4i) / 2 = 1 ± 2i. The roots are 1 + 2i and 1 − 2i, which are conjugates.

例：解方程 x² − 2x + 5 = 0。判别式为 (−2)² − 4 × 1 × 5 = −16。因此 x = (2 ± √(−16)) / 2 = (2 ± 4i) / 2 = 1 ± 2i。根为 1 + 2i 和 1 − 2i，它们是一对共轭复数。

12. Practical Tips for Exam Success | 考试实用技巧

When working with complex numbers, always simplify i² to −1 first, especially in multiplication. Write the final answer in the form a + bi, and ensure the real and imaginary parts are clearly separated.

在处理复数时，始终先将 i² 化简为 −1，尤其是在乘法中。最终答案写成 a + bi 的形式，并确保实部和虚部清晰分开。

Check conjugates: roots of real polynomials occur in conjugate pairs.
In division, multiply by the conjugate of the denominator.
Use the Argand diagram to visualise modulus and argument.
Remember that |z|² = z z̄, which is often useful in proofs.

检查共轭：实系数多项式的根总是成对共轭出现。
做除法时，乘以分母的共轭复数。
利用阿尔冈图形象化理解模与辐角。
记住 |z|² = z z̄，这在证明中经常用到。

Practice converting between rectangular and polar forms, as this skill appears frequently in examination questions. Always draw a small diagram if you are unsure whether an argument is acute or obtuse.

练习在直角坐标形式和极坐标形式之间转换，这一技能在考试中经常出现。如果不确定辐角是锐角还是钝角，就画一个小图帮助判断。

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Working with Complex Numbers | 复数的运算