Addition, Subtraction and Multiplication of Complex Numbers | 复数的加法、减法与乘法

📚 Addition, Subtraction and Multiplication of Complex Numbers | 复数的加法、减法与乘法

Complex numbers are one of the most elegant extensions of the real number system. In this revision article, we explore how to add, subtract and multiply complex numbers — the three fundamental operations that form the backbone of complex arithmetic.

复数是实数系统最优雅的扩展之一。在这篇复习文章中，我们将探讨复数的加法、减法和乘法——这三种基本运算构成了复数运算的基石。

1. What Is a Complex Number? | 什么是复数？

A complex number is written in the standard form z = a + bi, where a and b are real numbers and i is the imaginary unit defined by i² = −1. The value a is called the real part, Re(z), and b is called the imaginary part, Im(z).

复数的标准形式为 z = a + bi，其中 a 和 b 是实数，i 是虚数单位，满足 i² = −1。a 称为实部，记作 Re(z)；b 称为虚部，记作 Im(z)。

Every real number is also a complex number with b = 0. Conversely, a number with a = 0 is called purely imaginary. This inclusion means complex numbers form a larger number system that contains the reals.

每个实数也是复数，只需令 b = 0。相反，若 a = 0，则该数称为纯虚数。这种包含关系意味着复数构成了一个包含实数的更大数系。

2. The Imaginary Unit i | 虚数单位 i

The fundamental defining property of the imaginary unit is:

虚数单位的基本定义性质是：

i² = −1

It follows that i = √(−1). This allows us to write square roots of negative numbers: for any positive real a, √(−a) = i√a. For example, √(−25) = 5i.

由此可得 i = √(−1)。这使得我们可以表示负数的平方根：对于任意正实数 a，√(−a) = i√a。例如，√(−25) = 5i。

When solving quadratic equations with a negative discriminant, the solutions involve complex numbers. For instance, x² + 1 = 0 has solutions x = ±i. The introduction of i ensures every quadratic equation has two roots, making the field of complex numbers algebraically closed for quadratics.

当二次方程的判别式为负时，其解涉及复数。例如，x² + 1 = 0 的解为 x = ±i。引入 i 后，每个二次方程都有两个根，这使得复数域对于二次方程而言是代数封闭的。

3. Addition of Complex Numbers | 复数的加法

To add two complex numbers, we add their real parts together and their imaginary parts together:

两个复数相加时，我们分别将实部与实部相加、虚部与虚部相加：

(a + bi) + (c + di) = (a + c) + (b + d)i

Addition is performed component-wise. No cross-terms appear because addition does not involve multiplying the two complex numbers together.

加法按对应分量进行，不会出现交叉项，因为加法不涉及两个复数之间的乘法。

Example 1: (3 + 4i) + (2 − 7i) = (3 + 2) + (4 + (−7))i = 5 − 3i.

例 1：(3 + 4i) + (2 − 7i) = (3 + 2) + (4 + (−7))i = 5 − 3i。

Example 2: (−5 + 2i) + (6 + 9i) = (−5 + 6) + (2 + 9)i = 1 + 11i.

例 2：(−5 + 2i) + (6 + 9i) = (−5 + 6) + (2 + 9)i = 1 + 11i。

Addition of complex numbers satisfies the commutative law z₁ + z₂ = z₂ + z₁ and the associative law (z₁ + z₂) + z₃ = z₁ + (z₂ + z₃), just like addition of real numbers.

复数加法满足交换律 z₁ + z₂ = z₂ + z₁ 和结合律 (z₁ + z₂) + z₃ = z₁ + (z₂ + z₃)，与实数加法完全一致。

4. Subtraction of Complex Numbers | 复数的减法

Subtraction is performed in exactly the same component-wise manner:

减法同样按分量进行：

(a + bi) − (c + di) = (a − c) + (b − d)i

Example 3: (7 + 3i) − (4 + 8i) = (7 − 4) + (3 − 8)i = 3 − 5i.

例 3：(7 + 3i) − (4 + 8i) = (7 − 4) + (3 − 8)i = 3 − 5i。

Example 4: (2 − 6i) − (−3 + 4i) = (2 − (−3)) + (−6 − 4)i = 5 − 10i. Remember to distribute the minus sign across both parts of the second complex number.

例 4：(2 − 6i) − (−3 + 4i) = (2 − (−3)) + (−6 − 4)i = 5 − 10i。注意，减号必须分配到第二个复数的实部和虚部两个部分。

A useful geometric interpretation: on the Argand diagram, adding z₁ and z₂ corresponds to translating the point z₁ by the vector representing z₂. Subtraction corresponds to translating by the negative vector. This vector view makes complex addition and subtraction natural extensions of 2D vector arithmetic.

一个有用的几何解释：在阿尔冈图上，z₁ + z₂ 对应于将 z₁ 点沿 z₂ 所代表的向量平移；减法对应于沿相反向量平移。这种向量视角使得复数的加减法成为二维向量运算的自然延伸。

5. Multiplication of Complex Numbers | 复数的乘法

Multiplication of complex numbers follows the distributive law, just like multiplying binomials in algebra, but with the crucial substitution i² = −1 applied at the end:

复数的乘法遵循分配律，就像代数中两个二项式相乘一样，但最终需要代入关键的 i² = −1：

(a + bi)(c + di) = (ac − bd) + (ad + bc)i

Derivation: (a + bi)(c + di) = ac + adi + bci + bdi² = ac + adi + bci − bd = (ac − bd) + (ad + bc)i.

推导：(a + bi)(c + di) = ac + adi + bci + bdi² = ac + adi + bci − bd = (ac − bd) + (ad + bc)i。

Example 5: (2 + 3i)(1 + 4i) = 2 + 8i + 3i + 12i² = 2 + 11i − 12 = −10 + 11i.

例 5：(2 + 3i)(1 + 4i) = 2 + 8i + 3i + 12i² = 2 + 11i − 12 = −10 + 11i。

Example 6: (5 − 2i)(−3 + i) = −15 + 5i + 6i − 2i² = −15 + 11i + 2 = −13 + 11i.

例 6：(5 − 2i)(−3 + i) = −15 + 5i + 6i − 2i² = −15 + 11i + 2 = −13 + 11i。

Notice that the real part of the product is ac − bd — the minus sign arises solely from i² = −1. This is the single most common source of sign errors; always write out the i² term explicitly before simplifying.

注意，乘积的实部是 ac − bd——这个减号完全来自 i² = −1。这是最常见的符号错误来源；化简前务必将 i² 项显式写出。

6. Multiplication by a Real Scalar | 实数标量乘法

When multiplying a complex number by a real constant k, we multiply both the real and imaginary parts by k:

当复数乘以实数常数 k 时，我们将实部和虚部分别乘以 k：

k(a + bi) = ka + kbi

Example 7: 4(3 − 2i) = 12 − 8i.

例 7：4(3 − 2i) = 12 − 8i。

This is a special case of general complex multiplication, obtained by setting c = k and d = 0. Geometrically, scalar multiplication scales the distance of the point from the origin but does not rotate it.

这是复数乘法的一个特例，只需令 c = k、d = 0 即可。在几何上，标量乘法放大了点与原点的距离，但不产生旋转。

Scalar multiplication is used extensively when simplifying expressions such as 3(2 + i) − 2(1 − 4i), which combines both distribution and addition: 3(2 + i) − 2(1 − 4i) = 6 + 3i − 2 + 8i = 4 + 11i.

标量乘法在化简如 3(2 + i) − 2(1 − 4i) 这样的表达式时被广泛使用，它同时涉及分配律和加法：3(2 + i) − 2(1 − 4i) = 6 + 3i − 2 + 8i = 4 + 11i。

7. Powers of i and Cyclic Patterns | i 的幂与循环规律

Since i² = −1, higher powers of i fall into a repeating cycle of length 4. This cycle is essential for simplifying expressions involving large powers of i.

由于 i² = −1，i 的更高次幂呈现出周期为 4 的循环规律。这一规律对于化简涉及 i 大次幂的表达式至关重要。

Power	Value
i⁰	1
i¹	i
i²	−1
i³	−i
i⁴	1

To compute iⁿ for any integer n, find the remainder when n is divided by 4:

计算任意整数 n 的 iⁿ 时，先求出 n 除以 4 的余数：

If n ≡ 0 (mod 4), then iⁿ = 1.

若 n ≡ 0 (模 4)，则 iⁿ = 1。
If n ≡ 1 (mod 4), then iⁿ = i.

若 n ≡ 1 (模 4)，则 iⁿ = i。
If n ≡ 2 (mod 4), then iⁿ = −1.

若 n ≡ 2 (模 4)，则 iⁿ = −1。
If n ≡ 3 (mod 4), then iⁿ = −i.

若 n ≡ 3 (模 4)，则 iⁿ = −i。

Example 8: Evaluate i²³. Since 23 = 5 × 4 + 3, the remainder is 3, so i²³ = i³ = −i.

例 8：计算 i²³。因为 23 = 5 × 4 + 3，余数为 3，所以 i²³ = i³ = −i。

This cyclic property also explains why multiplying by i rotates a point 90° anticlockwise on the Argand diagram, a fact explored further when studying the modulus-argument form.

这一循环性质也解释了为什么乘以 i 会使阿尔冈图上的点逆时针旋转 90°，这一事实在研究模-辐角形式时会进一步探讨。

8. Simplifying Expressions with Mixed Operations | 混合运算表达式的化简

AQA exam questions frequently combine addition, subtraction and multiplication in a single expression. The order of operations is identical to real-number arithmetic: brackets first, then multiplication, then addition and subtraction from left to right.

AQA 考试题经常将加法、减法和乘法组合在同一个表达式中。运算顺序与实数运算完全一致：先括号，再乘法，最后从左到右进行加减法。

Example 9: Simplify 2(3 + i) + (4 − i)(2 + 5i).

例 9：化简 2(3 + i) + (4 − i)(2 + 5i)。

Step 1 — expand each term: 2(3 + i) = 6 + 2i. And (4 − i)(2 + 5i) = 8 + 20i − 2i − 5i² = 8 + 18i + 5 = 13 + 18i.

第一步——展开每一项：2(3 + i) = 6 + 2i。以及 (4 − i)(2 + 5i) = 8 + 20i − 2i − 5i² = 8 + 18i + 5 = 13 + 18i。

Step 2 — add the results: (6 + 2i) + (13 + 18i) = 19 + 20i.

第二步——将结果相加：(6 + 2i) + (13 + 18i) = 19 + 20i。

Example 10: Simplify (1 + i)². Using (a + bi)² = a² + 2abi + b²i², we get 1 + 2i + i² = 1 + 2i − 1 = 2i.

例 10：化简 (1 + i)²。利用 (a + bi)² = a² + 2abi + b²i²，得 1 + 2i + i² = 1 + 2i − 1 = 2i。

Notice that (1 + i)² = 2i is purely imaginary. This surprising result is a hallmark of complex arithmetic and often appears in exam problems designed to test understanding of i² = −1.

注意，(1 + i)² = 2i 是纯虚数。这个出人意料的结果是复数运算的标志性特征，常出现在旨在考查 i² = −1 理解的考试题目中。

9. The Complex Conjugate in Multiplication | 复数共轭与乘法

Although the focus of this article is addition, subtraction and multiplication, multiplication by a complex conjugate is a directly related skill. The conjugate of z = a + bi is z̄ = a − bi. Their product is always a real number:

尽管本文的重点是加法、减法和乘法，但乘以共轭复数是一项直接相关的技能。z = a + bi 的共轭为 z̄ = a − bi。它们的乘积始终是一个实数：

z·z̄ = (a + bi)(a − bi) = a² + b²

Because the cross-terms +abi and −abi cancel, the imaginary part vanishes. For example, (3 + 4i)(3 − 4i) = 9 − 12i + 12i − 16i² = 9 + 16 = 25.

由于交叉项 +abi 和 −abi 相互抵消，虚部消失。例如，(3 + 4i)(3 − 4i) = 9 − 12i + 12i − 16i² = 9 + 16 = 25。

This result, a² + b², is exactly the square of the modulus |z|² = a² + b². The product of a complex number and its conjugate equals the square of its distance from the origin in the Argand diagram.

这个结果 a² + b² 恰好是模的平方 |z|² = a² + b²。复数与其共轭的乘积等于该复数在阿尔冈图上到原点距离的平方。

10. Equating Real and Imaginary Parts | 实部与虚部相等

A fundamental theorem states that two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. This theorem converts complex equations into simultaneous real equations.

基本定理指出：两个复数相等当且仅当它们的实部相等且虚部相等。这一定理将复数方程转化为实数的联立方程组。

Example 11: Find x and y given that (x + 3i)(2 − i) = 7 + yi.

例 11：已知 (x + 3i)(2 − i) = 7 + yi，求 x 和 y。

Step 1 — expand the left-hand side: (x + 3i)(2 − i) = 2x − xi + 6i − 3i² = 2x + 3 + (6 − x)i.

第一步——展开左边：(x + 3i)(2 − i) = 2x − xi + 6i − 3i² = 2x + 3 + (6 − x)i。

Step 2 — equate real parts: 2x + 3 = 7, so 2x = 4, giving x = 2.

第二步——实部相等：2x + 3 = 7，所以 2x = 4，得 x = 2。

Step 3 — equate imaginary parts: 6 − x = y, so y = 6 − 2 = 4.

第三步——虚部相等：6 − x = y，所以 y = 6 − 2 = 4。

This technique is extremely common in AQA papers, combining multiplication with algebraic solving. Always rearrange the equation so that both sides are in the form A + Bi before equating.

这种技巧在 AQA 试卷中非常常见，将乘法与代数求解相结合。在令实虚部相等之前，务必先将等式两边整理为 A + Bi 的形式。

11. Common Mistakes and Exam Tips | 常见错误与考试提示

Below are the most frequently made errors in AQA exams when handling complex arithmetic:

以下是在 AQA 考试中处理复数运算时最常出现的错误：

Forgetting i² = −1: Writing (2 + 3i)(4 + 5i) = 8 + 10i + 12i + 15i² and stopping at 15i² instead of replacing it with −15. Always replace i² with −1 as the final step of every multiplication.

忘记 i² = −1：写出 (2 + 3i)(4 + 5i) = 8 + 10i + 12i + 15i² 后就停在了 15i²，而没有把它替换为 −15。每次乘法运算的最后一步都应将 i² 替换为 −1。
Sign errors in subtraction: When subtracting (c + di), both c and d are subtracted, not just c. Write (a − c) + (b − d)i rather than skipping the imaginary part.

减法中的符号错误：减去 (c + di) 时，c 和 d 都要被减去，而不仅仅是 c。应写成 (a − c) + (b − d)i，而不是漏掉虚部。
Mixing real and imaginary parts: Adding 3 + 4i to 5 + 2i to give 8 + 6i is correct, but collecting 3 and 2i together is not valid. Keep real parts and imaginary parts separate at all times.

混淆实部与虚部：将 3 + 4i 与 5 + 2i 相加得 8 + 6i 是正确的，但把 3 和 2i 加在一起则是错误的。应始终保持实部和虚部分离。
Incorrect expansion of (a + bi)²: Some candidates write (a + bi)² = a² + b²i², omitting the 2abi cross-term. The correct expansion is a² + 2abi + b²i².

(a + bi)² 展开错误：有些考生写成 (a + bi)² = a² + b²i²，遗漏了 2abi 交叉项。正确的展开是 a² + 2abi + b²i²。

12. Practice Questions | 练习题目

Attempt these questions without a calculator, as AQA Pure Mathematics papers expect fluency in complex arithmetic by hand.

请尝试在不使用计算器的情况下完成以下题目，因为 AQA 纯数试卷要求熟练掌握手算复数运算。

Question 1: Simplify (3 + 2i) + (5 − 7i).

题目 1：化简 (3 + 2i) + (5 − 7i)。

Question 2: Simplify (8 − 4i) − (3 + 9i).

题目 2：化简 (8 − 4i) − (3 + 9i)。

Question 3: Simplify (2 + 3i)(4 − i).

题目 3：化简 (2 + 3i)(4 − i)。

Question 4: Simplify (1 + i)(1 − i) and comment on the result.

题目 4：化简 (1 + i)(1 − i) 并讨论结果。

Question 5: Given that (a + bi)(3 − 2i) = 11 + 4i, where a and b are real numbers, find the values of a and b.

题目 5：已知 (a + bi)(3 − 2i) = 11 + 4i，其中 a 和 b 是实数，求 a 和 b 的值。

Question 6: Evaluate i³³ + i²² + i¹¹.

题目 6：计算 i³³ + i²² + i¹¹。

Answers: 1) 8 − 5i 2) 5 − 13i 3) 11 + 10i 4) 2 (purely real) 5) a = 1, b = 2 6) i³³ = i, i²² = −1, i¹¹ = −i, hence i − 1 − i = −1.

答案：1) 8 − 5i 2) 5 − 13i 3) 11 + 10i 4) 2（纯实数） 5) a = 1, b = 2 6) i³³ = i，i²² = −1，i¹¹ = −i，因此 i − 1 − i = −1。

Mastering addition, subtraction and multiplication of complex numbers is essential for success in AQA A-Level Mathematics. These operations underpin later topics such as the modulus-argument form, De Moivre’s theorem, roots of unity and complex roots of polynomials. Practice each operation until the process feels natural — with i² = −1 as your constant companion.

掌握复数的加法、减法和乘法对于在 AQA A-Level 数学中取得成功至关重要。这些运算支撑着后续的主题，如模-辐角形式、棣莫弗定理、单位根以及多项式的复数根。反复练习每一种运算，直到整个过程变得自然流畅——让 i² = −1 成为你始终相伴的伙伴。

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Addition, Subtraction and Multiplication of Complex Numbers | 复数的加法、减法与乘法