📚 Addition, Subtraction and Multiplication of Complex Numbers | 复数 x + iy 的加法、减法与乘法
Complex numbers extend the real number system by introducing the imaginary unit i, defined by i² = -1. Every complex number can be written in the standard form x + iy, where x and y are real numbers. In this article, we develop the three fundamental operations — addition, subtraction and multiplication — for expressions of the form x + iy, exactly as required by the AQA A-Level Mathematics specification.
复数通过引入虚数单位 i（规定 i² = -1）扩展了实数系。每个复数都可以写成标准形式 x + iy，其中 x 和 y 是实数。本文将按照 AQA A-Level 数学考纲的要求，系统讲解形如 x + iy 的复数的三种基本运算——加法、减法和乘法。
1. The Form x + iy | 复数形式 x + iy
A complex number z is written as z = x + iy, where x is called the real part, denoted Re(z), and y is called the imaginary part, denoted Im(z). Note that both x and y are ordinary real numbers, and the ‘i’ is the imaginary unit satisfying i² = -1.
复数 z 写成 z = x + iy，其中 x 称为实部，记作 Re(z)；y 称为虚部，记作 Im(z)。注意 x 与 y 都是普通的实数，i 是满足 i² = -1 的虚数单位。
For example, in the complex number 3 + 4i, the real part is 3 and the imaginary part is 4. Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. This equality rule underpins every operation that follows.
例如，在复数 3 + 4i 中，实部为 3，虚部为 4。两个复数相等当且仅当它们的实部与虚部分别相等。这一相等法则是一切后续运算的基础。
2. Addition of Complex Numbers | 复数的加法
To add two complex numbers, simply add their real parts and add their imaginary parts separately. If z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂, then:
两个复数相加，只需分别将实部与实部相加、虚部与虚部相加。若 z₁ = x₁ + iy₁，z₂ = x₂ + iy₂，则：
z₁ + z₂ = (x₁ + x₂) + i(y₁ + y₂)
Example: (3 + 2i) + (5 – 7i) = (3 + 5) + (2 – 7)i = 8 – 5i.
例：(3 + 2i) + (5 – 7i) = (3 + 5) + (2 – 7)i = 8 – 5i。
The result of adding two complex numbers is always another complex number of the same form. Addition follows the commutative law, z₁ + z₂ = z₂ + z₁, and the associative law, (z₁ + z₂) + z₃ = z₁ + (z₂ + z₃), for all complex numbers.
两个复数相加的结果仍是同一形式的复数。加法满足交换律 z₁ + z₂ = z₂ + z₁，以及结合律 (z₁ + z₂) + z₃ = z₁ + (z₂ + z₃)。
3. Subtraction of Complex Numbers | 复数的减法
Subtraction is performed in exactly the same spirit: subtract the real parts and subtract the imaginary parts separately:
减法与加法的规则完全类似：实部减实部、虚部减虚部：
z₁ – z₂ = (x₁ – x₂) + i(y₁ – y₂)
Example: (6 + i) – (2 – 3i) = (6 – 2) + (1 – (-3))i = 4 + 4i. Take care with signs: subtracting a negative imaginary part produces a positive coefficient of i.
例：(6 + i) – (2 – 3i) = (6 – 2) + (1 – (-3))i = 4 + 4i。请特别注意符号：减去一个负的虚部会得到正的 i 系数。
Subtraction can also be viewed as adding the additive inverse: z₁ – z₂ = z₁ + (-z₂), where -z₂ = -x₂ – iy₂. This perspective is often useful when simplifying complicated expressions.
减法也可以看作加上相反数：z₁ – z₂ = z₁ + (-z₂)，其中 -z₂ = -x₂ – iy₂。这一视角在化简复杂表达式时常会派上用场。
4. Multiplication of Complex Numbers | 复数的乘法
Multiplication of two complex numbers follows the ordinary rules of algebra, with one crucial extra rule: i² = -1. Expanding (x₁ + iy₁)(x₂ + iy₂) as a double bracket gives:
两个复数相乘遵循普通代数法则，但多了一条关键规则：i² = -1。将 (x₁ + iy₁)(x₂ + iy₂) 按双重括号展开：
z₁z₂ = (x₁x₂ – y₁y₂) + i(x₁y₂ + x₂y₁)
Notice that the term y₁y₂i² becomes -y₁y₂ and is real, which is why the real part of the product is x₁x₂ – y₁y₂. The i terms combine to give the imaginary part x₁y₂ + x₂y₁.
注意 y₁y₂i² 转化为 -y₁y₂ 并入实部，因此乘积的实部是 x₁x₂ – y₁y₂；含 i 的项合并后得到虚部 x₁y₂ + x₂y₁。
Worked example: (2 + 3i)(4 – i) = 8 – 2i + 12i – 3i² = 8 + 10i + 3 = 11 + 10i.
例题：(2 + 3i)(4 – i) = 8 – 2i + 12i – 3i² = 8 + 10i + 3 = 11 + 10i。
5. Powers of i | i 的幂
When multiplying complex numbers, we frequently encounter powers of i. Since i² = -1, we can derive the cycle:
在复数乘法中我们经常遇到 i 的幂。由 i² = -1 可以推出如下的循环规律：
| Power | Value |
| i¹ | i |
| i² | -1 |
| i³ | -i |
| i⁴ | 1 |
| i⁵ | i |
The pattern repeats every four powers. To evaluate iⁿ for a positive integer n, divide n by 4 and use the remainder: remainder 0 gives 1, remainder 1 gives i, remainder 2 gives -1, remainder 3 gives -i.
该规律每 4 次幂循环一次。要计算正整数 n 的 iⁿ，只需用 n 除以 4 并利用余数：余 0 得 1，余 1 得 i，余 2 得 -1，余 3 得 -i。
Example: i²⁰²⁴ = 1 because 2024 ÷ 4 leaves remainder 0. This technique saves time in exam questions involving high powers of i.
例：i²⁰²⁴ = 1，因为 2024 ÷ 4 余 0。这一技巧在考试中遇到 i 的高次幂时能节省大量时间。
6. Algebraic Properties of Multiplication | 乘法的代数性质
Multiplication of complex numbers inherits the familiar algebraic laws of real numbers. For all complex numbers z₁, z₂ and z₃:
复数的乘法继承了实数中熟悉的代数律。对任意复数 z₁、z₂ 和 z₃：
- Commutative law: z₁z₂ = z₂z₁ | 交换律：z₁z₂ = z₂z₁
- Associative law: (z₁z₂)z₃ = z₁(z₂z₃) | 结合律：(z₁z₂)z₃ = z₁(z₂z₃)
- Distributive law: z₁(z₂ + z₃) = z₁z₂ + z₁z₃ | 分配律：z₁(z₂ + z₃) = z₁z₂ + z₁z₃
These laws mean we can expand and factorise complex expressions exactly as we do with real polynomials in x, remembering to replace any i² with -1 as soon as it appears.
这些运算律意味着我们可以像处理关于 x 的实系数多项式一样对复数表达式进行展开与因式分解，只要在出现 i² 时立即将其替换为 -1 即可。
7. Geometric Interpretation | 几何意义
In the Argand diagram, the complex number z = x + iy is represented by the point (x, y). Addition and subtraction of complex numbers then correspond exactly to vector addition and subtraction of their position vectors.
在阿甘图（复平面）中，复数 z = x + iy 用点 (x, y) 表示。复数的加法与减法恰好对应位置向量的向量加法与向量减法。
For instance, adding 2 + i to 3 + 4i moves the point (3, 4) by the vector (2, 1) to reach (5, 5). This parallelogram interpretation is often assessed in AQA examination questions together with the algebraic computation.
例如，将 2 + i 加到 3 + 4i 上，相当于把点 (3, 4) 沿向量 (2, 1) 平移到 (5, 5)。这一平行四边形解释常与代数计算同时出现在 AQA 考试题中。
8. Complex Conjugates and Product | 共轭复数及其乘积
The complex conjugate of z = x + iy is z̄ = x – iy. Geometrically, z̄ is the reflection of z in the real axis. Multiplying a complex number by its conjugate always gives a non-negative real number:
复数 z = x + iy 的共轭复数为 z̄ = x – iy。在几何上，z̄ 是 z 关于实轴的对称点。一个复数与其共轭相乘始终得到非负实数：
z z̄ = (x + iy)(x – iy) = x² + y²
This identity is extremely important: it is used to divide complex numbers, to rationalise denominators, and to find the modulus |z| = √(x² + y²). The middle terms ixy – ixy cancel, leaving only real terms.
这个恒等式极为重要：它用于复数的除法、分母有理化以及求模 |z| = √(x² + y²)。展开后中间项 ixy – ixy 相互抵消，只剩实数项。
9. Working with Products of Several Complex Numbers | 多个复数连乘
When multiplying three or more complex numbers, work step by step: multiply two at a time and simplify i² immediately. For example, (1 + i)(2 – i)(3 + 2i) can be computed as ((1 + i)(2 – i))(3 + 2i) = (3 + i)(3 + 2i) = 9 + 6i + 3i + 2i² = 7 + 9i.
当三个或更多复数相乘时，应逐步计算：每次先乘两个并立刻化简 i²。例如，(1 + i)(2 – i)(3 + 2i) 可先算 (1 + i)(2 – i) = 3 + i，再算 (3 + i)(3 + 2i) = 9 + 6i + 3i + 2i² = 7 + 9i。
An alternative strategy is to group conjugates when they appear, because their product is real. For example, (2 + 3i)(2 – 3i)(1 + i) = (4 + 9)(1 + i) = 13(1 + i) = 13 + 13i, which avoids expanding three brackets at once.
另一种策略是当表达式中出现共轭对时优先将其结合相乘，因为共轭乘积是实数。例如，(2 + 3i)(2 – 3i)(1 + i) = (4 + 9)(1 + i) = 13(1 + i) = 13 + 13i，这样就避免了同时展开三个括号的繁琐运算。
10. Exam-Style Tips and Common Mistakes | 考点提示与常见错误
Students sitting AQA A-Level Mathematics papers should master the following points to secure full marks on complex-number arithmetic questions:
参加 AQA A-Level 数学考试的同学们应当掌握以下要点，才能在复数运算题中确保不失分：
- Always write answers in the form x + iy, collecting real and
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