📚 Complex Numbers | 复数
This article unpacks the key ideas of complex numbers as they appear in the Edexcel IGCSE Further Pure Mathematics specification. Even if your regular GCSE course does not cover imaginary numbers, mastering these foundations will give you a head start for A‑level and beyond. We’ll explore the definition of i, arithmetic with complex numbers, conjugates, the complex plane, and solving quadratics with no real roots.
本文围绕 Edexcel IGCSE 进阶纯数（Further Pure Mathematics）考纲中复数的核心考点展开。即便普通 GCSE 课程不涉及虚数，吃透这些基础也会让你在 A‑level 和未来的学习中领先一步。我们将详解 i 的定义、复数四则运算、共轭复数、复平面以及无实根二次方程的求解。
1. What is a Complex Number? | 什么是复数？
he form a + bi, where a and b are real numbers and i is the imaginary unit defined by i² = – art is b (not bi). For example, in 3 + 4i, the real part is 3 and the imaginary part is 4.
复数是指形如 a + bi 的数，其中 a 和 b 为实数，i 为满足 i² = –1 的虚数单位。a 称为实部（real part），b 称为虚部（imaginary part），注意虚部是 b 而不是 bi。例如，3 + 4i 的实部为 3，虚部为 4。
2. Powers of i | i 的幂
The powers of i cycle every four steps: i¹ = i, i² = –1, i³ = i² × i = –i, i⁴ = (i²)² = 1. Then the pattern repeats: i⁵ = i, i⁶ = –1, and so on. To simplify iⁿ, divide the exponent n by 4 and use the remainder: iⁿ = iremainder.
i 的幂次每 4 次一循环：i¹ = i，i² = –1，i³ = i² × i = –i，i⁴ = (i²)² = 1。此后循环重复：i⁵ = i，i⁶ = –1，以此类推。化简 iⁿ 时，将指数 n 除以 4，取余数即可：iⁿ = i余数。
3. Adding and Subtracting Complex Numbers | 复数的加法与减法
a + bi) + (c + di), the sum is (a + c) + (b + d)i. Subtraction works analogously: (a + bi) – (c + di) = (a – c) + (b – d)i. Always express the final answer in the standard form a + bi.
复数的加减法分别合并实部与虚部即可。(a + bi) + (c + di) = (a + c) + (b + d)i。减法类似：(a + bi) – (c + di) = (a – c) + (b – d)i。最终结果务必写成 a + bi 的标准形式。
4. Multiplying Complex Numbers | 复数的乘法
omplex numbers by expanding brackets as you would with binomials, and then replace i² with –1. For (a + bi)(c + di): multiply to get ac + adi + bci + bdi². Since i² = –1, bdi² = –bd. Collecting real and imaginary parts gives (ac – bd) + (ad + bc)i. Always simplify to a + bi form.
复数乘法按多项式展开，再将 i² 换成 –1。例如 (a + bi)(c + di) = ac + adi + bci + bdi²。由 i² = –1 得 bdi² = –bd。合并实部与虚部得 (ac – bd) + (ad + bc)i。最后务必整理成 a + bi 格式。
5. Complex Conjugates | 共轭复数
The complex conjugate of z = a + bi is z* (or z̅) = a – bi. Conjugates are useful because their product is always a real number: (a + bi)(a – These properties are essential for division and for solving equations with complex roots.
复数 z = a + bi 的共轭（conjugate）记为 z* 或 z̅，即 a – bi。共轭复数的乘积始终为实数：(a + bi)(a – bi) = a² + b²。一个复数与其共轭的和等于实部的两倍：z + z* = 2a。这些性质在复数除法和处理共轭复根时不可或缺。
6. Dividing Complex Numbers | 复数的除法
s turns the denominator into a real number. For (a + bi) ÷ (c + di), calculate [(a + bi)(c – di)] / [(c + di)(c – di)] = [(ac + bd) + (bc – ad)i] / (c² + d²). Then separate into real and imaginary parts.
复数相除时，分子分母同乘以分母的共轭复数，使分母变为实数。例如 (a + bi) ÷ (c + di) = [(a + bi)(c – di)] / [(c + di)(c – di)] = [(ac + bd) + (bc – ad)i] / (c² + d²)。最后拆分为实部与虚部即可。
7. The Complex Plane (Argand Diagram) | 复平面（Argand 图）
mplex number a + bi can be represented as a point (a, b) on an Argand diagram, where the x‑axis is the real axis and the y‑axis is the imaginary axis. The modulus of z, written |z|, is the distance from the origin: |z| = √a² + b². The argument, arg(z), is the angle θ (in radians) measured from the positive real axis.
复数 a + bi 可在 Argand 图上表示为点 (a, b)，其中 x 轴为实轴，y 轴为虚轴。复数的模（modulus）记作 |z|，表示该点到原点的距离：|z| = √a² + b²。辐角（argument）arg(z) 是从正实轴逆时针测量的角度 θ，单位为弧度。
8. Solving Quadratic Equations with Complex Roots | 求解含复根的二次方程
When the discriminant b² – = 0 is negative, the roots are a conjugate complex pair. Use the quadratic formula: x = [–b ± √b² – 4ac] / (2a). Write √negative as i√positive. For example, with x² – 4x + 13 = 0, the discriminant is –36, so x = (4 ± 6i) / 2 = 2 ± 3i. The two roots are always a ± bi.
当二次方程 ax² + bx + c = 0 的判别式 b² – 4ac < 0 时，方程有一对共轭复根。代入求根公式 x = [–b ± √b² – 4ac] / (2a)，将 √负数 写成 i√正数。例如，x² – 4x + 13 = 0 的判别式为 –36，故 x = (4 ± 6i) / 2 = 2 ± 3i。两个根总是呈 a ± bi 的共轭形式。
9. Equating Real and Imaginary Parts | 实部与虚部分别相等
rinciple helps find unknown real numbers in identities. For instance, if (x + yi) + 2i = 5 – 3i, then equating real parts gives x = 5, and equating imaginary parts gives y + 2 = –3, so y = –5.
两个复数相等，当且仅当它们的实部相等且虚部相等。利用这一原理可求解恒等式中的未知实数。例如，由 (x + yi) + 2i = 5 – 3i，令实部相等得 x = 5，令虚部相等得 y + 2 = –3，故 y = –5。
10. Sums, Products and Roots of Polynomials | 多项式的根之和与根之积
quadratic with real coefficients and complex roots, the roots are conjugates u + vi and u – vi. Their sum is 2u (real) and product is u² + v² (real). This connects with Viète’s formulas: for x² + px + q = 0, sum of roots = – cubic with real coefficients can have one real root and a conjugate pair of complex roots.
对于实系数的二次方程，若其根为复数，则必为一对共轭复根 u + vi 与 u – vi。两根之和为 2u（实数），积为 u² + v²（实数）。这正好符合韦达定理：x² + px + q = 0 中，根之和 = –p，根之积 = q。对于实系数三次方程，可能有一个实根和一对共轭复根。
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