📚 Introduction to Complex Numbers | 复数入门
, they are introduced through the imaginary unit i, defined by the equation i² = -1, and they quickly become a powerful tool for algebra, geometry and problem solving. This revision guide covers the definition of i, the standard form z = x + iy, the arithmetic of complex numbers, the complex conjugate, division, the Argand diagram, modulus and argument, complex roots of quadratic equations, and the exam technique that turns these ideas into marks.
复数扩展了实数系，使得每个二次方程都有解。在 AQA A-Level 数学与进阶数学中，复数通过虚数单位 i 引入，其定义为 i² = -1，并很快成为代数、几何与解题中的有力工具。本篇复习指南涵盖 i 的定义、标准形式 z = x + iy、复数的四则运算、共轭复数、除法、阿冈图（Argand diagram）、模与辐角、二次方程的复根，以及把这些概念转化为分数的应试技巧。
1. Why We Need Complex Numbers | 为什么需要复数
gives x² = -1, which has no real solution, because the square of every real number is greater than or equal to zero. In the same way, x² - 2x + 5 = 0 has discriminant 4 - 20 = -16, so the quadratic formula produces √-16, an expression that is undefined in the real numbers.
在 AQA A-Level 数学中，当一个二次方程的判别式为负时，就会遇到复数。例如，x² + 1 = 0 得到 x² = -1，这在实数范围内无解，因为任何实数的平方都大于或等于零。同样地，x² - 2x + 5 = 0 的判别式为 4 - 20 = -16，因此求根公式会产生 √-16，而这一表达式在实数范围内没有定义。
By defining a new number whose square is -1, mathematicians removed this restriction. The real number system is extended to the complex number system, and the chain of number sets becomes ℕ ⊂ ℤ ⊂ ℚ ⊂ ℝ ⊂ ℂ, where ℂ denotes the set of all complex numbers. Every quadratic equation now has exactly two roots in ℂ, counted with multiplicity.
通过定义一种平方等于 -1 的新数，数学家消除了这一限制。实数系被扩展为复数系，数集的包含关系变为 ℕ ⊂ ℤ ⊂ ℚ ⊂ ℝ ⊂ ℂ，其中 ℂ 表示全体复数构成的集合。现在每个二次方程在 ℂ 中都恰好有两个根（按重数计）。
Complex numbers are far from being a mathematical curiosity. They are used in electrical engineering to model alternating current, in control theory, in signal processing, and in the geometry of transformations such as rotations and enlargements. In A-Level they also lay the groundwork for Further Mathematics topics such as de Moivre's theorem, roots of polynomials and loci problems.
复数绝非纯粹的数学趣闻。它们被用于电气工程中描述交流电，也用于控制理论、信号处理，以及旋转与放缩等变换的几何研究。在 A-Level 中，它们还为进阶数学的后续主题奠定基础，例如棣莫弗定理、多项式方程的根以及轨迹问题。
2. The Imaginary Unit i | 虚数单位 i
unit is written as i, and it is defined by the single equation i² = -1. Equivalently, i = √-1, so i is one of the two square roots of the negative number -1; the other is -i. Notice that i is not a real number, so we cannot place it anywhere on the ordinary number line.
虚数单位记作 i，它由唯一的一个方程 i² = -1 定义。等价地，i = √-1，因此 i 是负数 -1 的两个平方根之一，另一个是 -i。需要注意的是，i 不是实数，所以我们无法把它标在普通的数轴上。
i = √-1 and i² = -1
g i exactly as we treat algebraic letters, with the one extra rule that every occurrence of i² is replaced by -1. For example, i + i = 2i and 3 × i = 3i.
虽然 i 不是实数，但它可以按照通常的代数规则与实数结合。我们可以像处理代数字母一样对含 i 的表达式进行加减乘除，只需额外遵守一条规则：每次出现 i² 时，都用 -1 替换。例如，i + i = 2i，3 × i = 3i。
When simplifying surds, be careful with negative numbers under the square root sign. The familiar rule √a × √b = √ab is only valid when a and b are non-negative. In particular, √-1 × √-1 is not √1 but -1, which is exactly the statement i² = -1.
在化简根式时要特别小心根号下的负数。熟悉法则 √a × √b = √ab 仅在 a 和 b 非负时才成立。特别地，√-1 × √-1 并不等于 √1，而是等于 -1，这正是 i² = -1 的含义。
3. Complex Number Form and Terminology | 复数的形式与术语
d for a second complex number. The real part of z is Re(z) = x, and the imaginary part of z is Im(z) = y.
复数具有标准形式 z = x + iy，其中 x 和 y 都是实数。字母 z 通常表示复变量，而 w 常用来表示第二个复数。z 的实部为 Re(z) = x，z 的虚部为 Im(z) = y。
z = x + iy, Re(z) = x, Im(z) = y
s the letter i itself. For example, if z = 3 - 4i, then Re(z) = 3 and Im(z) = -4, not -4i.
此处最常见的考试错误是写成 Im(z) = iy。虚部是 i 的实数系数，因此它本身绝不含字母 i。例如，若 z = 3 - 4i，则 Re(z) = 3，Im(z) = -4，而不是 -4i。
and z is a real number, so every real number is also a complex number. Two complex numbers are equal only when both their real parts and their imaginary parts match, which is the idea tested in the next section.
特殊情况有专门的名称。若 x = 0，则 z = iy，称 z 为纯虚数；例如 5i 与 -7i 都是纯虚数。若 y = 0，则 z = x，z 是实数，因此每个实数同时也是复数。两个复数相等，当且仅当它们的实部与虚部分别相等，这正是下一节要考查的思想。
| Complex number z | Re(z) | Im(z) | Type |
| 3 - 4i | 3 | -4 | General complex |
| 6i | 0 | 6 | Purely imaginary |
| -5 | -5 | 0 | Real |
4. Equality of Complex Numbers | 复数的相等
Two complex numbers a + bi and c + di are equal exactly when their real parts are equal and their imaginary parts are equal. This single statement is one of the most useful tools in the whole topic, because it converts one complex equation into two ordinary real equations.
两个复数 a + bi 与 c + di 相等，当且仅当它们的实部相等且虚部相等。这一条是整章中最有用的工具之一，因为它把一个复数方程转化为两个普通的实数方程。
a + bi = c + di ⇔ a = c and b = d
The technique is always the same: expand and simplify both sides, collect all real terms on one side and all imaginary terms on the other, then equate coefficients. This method is used to find unknown real constants in an identity, to determine a missing complex number, and to prove that two complicated-looking expressions are the same.
方法总是相同的：把两边展开并化简，将所有实项归到一边、所有虚项归到另一边，然后比较系数。这一方法可用于在恒等式中求未知实常数、确定某个未知复数，以及证明两个看似复杂的表达式相同。
o y = 7. Always state clearly that you are comparing real and imaginary parts - examiners award method marks for that statement.
例如，设 (x + 2) + (y - 3)i = 5 + 4i，其中 x 与 y 为实数。比较实部得 x + 2 = 5，所以 x = 3；比较虚部得 y - 3 = 4，所以 y = 7。一定要明确写出你在比较实部与虚部，因为考官会为这一说明给出方法分。
5. Addition and Subtraction | 加法与减法
There is no interaction between the real and imaginary components, so the calculation is no harder than collecting like terms in algebra.
复数的加减法要分别合并实部和虚部。用符号表示，即 (a + bi) ± (c + di) = (a ± c) + (b ± d)i。实部与虚部之间没有交叉作用，因此这一运算并不比代数中合并同类项更难。
(a + bi) ± (c + di) = (a ± c) + (b ± d)i
e how the minus sign changes the sign of both parts in the second bracket, exactly as with ordinary brackets.
例如，(3 + 2i) + (1 - 5i) = (3 + 1) + (2 - 5)i = 4 - 3i。做减法时，(3 + 2i) - (1 - 5i) = (3 - 1) + (2 + 5)i = 2 + 7i；注意负号如何改变第二个括号中两项的符号，这与普通括号的运算法则完全一致。
rically, adding complex numbers corresponds to adding vectors in the plane, which is why complex numbers are so useful when studying transformations on an Argand diagram.
复数加法满足交换律与结合律，加法单位元是 0 + 0i。z = a + bi 的加法逆元是 -z = -a - bi。从几何上看，复数相加对应于平面上向量的相加，这正是复数在研究阿冈图上的变换时如此有用的原因。
6. Multiplication and Powers of i | 乘法与 i 的幂
ce bdi² = -bd, the result simplifies to (ac - bd) + (ad + bc)i. It is worth memorising this pattern, but it is safer to expand each time rather than to trust memory under exam pressure.
乘法按照通常的二项式展开，然后将每个 i² 替换为 -1。展开 (a + bi)(c + di) 得到 ac + adi + bci + bdi²，由于 bdi² = -bd，结果化简为 (ac - bd) + (ad + bc)i。这一模式值得记住，但考试压力下按部就班展开往往更稳妥。
(a + bi)(c + di) = (ac - bd) + (ad + bc)i
² = (a² - b²) + 2abi, which is needed when checking whether a quadratic has complex roots.
例如，(3 + 2i)(1 - 5i) = 3 - 15i + 2i - 10i² = 3 - 13i + 10 = 13 - 13i。一个特别常见的情形是平方：(a + bi)² = a² + 2abi + b²i² = (a² - b²) + 2abi，在检验二次方程是否有复根时会用到它。
. To evaluate iⁿ for a large n, divide n by 4 and look at the remainder: remainder 1 gives i, remainder 2 gives -1, remainder 3 gives -i, and remainder 0 gives 1.
i 的幂构成一个长度为 4 的循环：i¹ = i，i² = -1，i³ = i² × i = -i，i⁴ = (i²)² = 1。此后规律重复，因此 i⁵ = i，i⁶ = -1，依此类推。要计算较大的 iⁿ，只需用 n 除以 4 看余数：余 1 得 i，余 2 得 -1，余 3 得 -i，余 0 得 1。
i¹ = i i² = -1 i³ = -i i⁴ = 1
great deal of time and is frequently examined in both A-Level Mathematics and Further Mathematics papers.
例如，2025 = 4 × 506 + 1，因此 i²⁰²⁵ = i¹ = i。这一余数技巧可以节省大量时间，在 A-Level 数学与进阶数学试卷中都经常考到。
7. The Complex Conjugate and Division | 共轭复数与除法
The complex conjugate of z = a + bi is z̄ iplying a complex number by its own conjugate always produces a real, non-negative number.
复数 z = a + bi 的共轭复数为 z̄ = a - bi；实部保持不变，虚部符号取反。共轭是打开除法大门的钥匙，因为一个复数与自身共轭相乘，结果总是实数且非负。
z z̄ = (a + bi)(a - bi) = a² + b²
To divide by a complex number, multiply the numerator and the denominator by the conjugate of the denominator. This removes i from the bottom and leaves an answer in the standard form x + iy:
要除以一个复数，就把分子与分母同时乘以分母的共轭复数。这样就去掉了分母中的 i，并把结果化为标准形式 x + iy：
(a + bi) ÷ (c + di) = [(a + bi)(c - di)] ÷ (c² + d²)
0.5 + 2.5i. Notice how the denominator became the real number 2, which is exactly 1² + 1².
例如， [(3 + 2i)(1 + i)] ÷ [(1 - i)(1 + i)] = (3 + 3i + 2i + 2i²) ÷ 2 = (1 + 5i) ÷ 2 = 0.5 + 2.5i。注意分母变成了实数 2，而它正好等于 1² + 1²。
conjugate of a product is the product of the conjugates. Also, z is real if and only if z̄ = z, and z is purely imaginary if and only if z̄ = -z. These facts appear regularly in proof-style questions.
值得记住的共轭性质有：和的共轭等于共轭的和，积的共轭等于共轭的积。此外，z 为实数当且仅当 z̄ = z；z 为纯虚数当且仅当 z̄ = -z。这些结论在证明类题目中经常出现。
8. Argand Diagram, Modulus and Argument | 阿冈图、模与辐角
ented by the point (x, y) on an Argand diagram, with the real part measured along the horizontal axis and the imaginary part along the vertical axis. This geometric picture turns algebra into geometry: addition becomes vector addition, and multiplication becomes a rotation combined with an enlargement.
复数 z = x + iy 可以用阿冈图上的点 (x, y) 表示，实部沿水平轴度量，虚部沿垂直轴度量。这一几何图像把代数变成几何：加法变成向量加法，乘法变成旋转与放缩的组合。
The distance of the point from the origin is called the modulus, written |z|, and it is found using Pythagoras' theorem: |z| = √x² + y². The angle measured anticlockwise from the positive real axis to the line Oz is called the argument, written arg z. Its tangent is y ÷ x, but you must check the quadrant before stating the final answer.
该点到原点的距离称为模，记作 |z|，由勾股定理得 |z| = √x² + y²。从正实轴逆时针量到直线 Oz 的角称为辐角，记作 arg z。其正切为 y ÷ x，但在给出最终答案前必须判断象限。
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