📚 IGCSE OCR Maths: Complex Functions Key Points | IGCSE OCR 数学：复变函数考点精讲
Complex functions build on our understanding of complex numbers, extending algebraic operations into a world where every polynomial equation has a solution. For IGCSE OCR Mathematics, you are expected to define, manipulate and interpret complex numbers in various forms, solve quadratic equations with non‑real roots, and apply simple transformations that treat a complex number as an input to produce another complex number. This article covers all the core ideas, from the standard form a + bi to polar representation and basic loci, pairing every English explanation with its Chinese equivalent so you can master the topic confidently.
复变函数建立在我们对复数的理解之上，它将代数运算拓展到了一个每个多项式方程都有解的世界。在 IGCSE OCR 数学中，你需要会定义、运算并解释不同形式的复数，解带有非实数根的二次方程，并能应用简单的变换将一个复数作为输入产生另一个复数。本文涵盖从 a + bi 的标准形式到极坐标表示与基本轨迹的所有核心内容，每一条英文讲解都配有对应的中文说明，帮助你扎实掌握这一主题。
1. Introduction to Complex Numbers | 复数导论
A complex number is an expression of the form a + bi, where a and b are real numbers, and i is the imaginary unit defined by i² = −1. The set of complex numbers is denoted by ℂ and includes all real numbers (when b = 0) together with purely imaginary numbers (when a = 0). The existence of i allows us to solve equations such as x² + 1 = 0, which have no solution in the real number system.
复数是一个形如 a + bi 的表达式，其中 a 和 b 为实数，i 是满足 i² = −1 的虚数单位。复数集记为 ℂ，它包含了所有实数（当 b = 0 时）以及纯虚数（当 a = 0 时）。i 的存在使我们能解出像 x² + 1 = 0 这样在实数系中无解的方程。
The real part of z = a + bi is Re(z) = a, and the imaginary part is Im(z) = b — note that Im(z) is the coefficient b itself, not bi. Two complex numbers are equal if, and only if, their real parts are equal and their imaginary parts are equal.
复数 z = a + bi 的实部为 Re(z) = a，虚部为 Im(z) = b — 注意 Im(z) 是系数 b 本身，而不是 bi。两个复数相等，当且仅当它们的实部相等且虚部相等。
2. Standard Form and Equality | 标准形式与相等
Every complex number should be written in the standard form a + bi before carrying out operations or comparisons. For example, √(−4) is first rewritten as 2i, and an expression such as (3 + i) + 2i is simplified to 3 + 3i. Keeping numbers in this form ensures that the real and imaginary parts are clearly separated.
在进行运算或比较之前，每个复数都应写为标准形式 a + bi。例如，√(−4) 要先写成 2i，而像 (3 + i) + 2i 这样的式子要化简为 3 + 3i。保持这种形式能确保实部与虚部清晰地分开。
Equality of complex numbers is a powerful tool for finding unknowns. If a + bi = c + di, then necessarily a = c and b = d. This property is frequently used in solving equations where the real and imaginary parts on each side must match independently.
复数相等是求未知量的有力工具。若 a + bi = c + di，则必有 a = c 且 b = d。这一性质常用于解方程，此时等式两边的实部和虚部必须各自对应相等。
Example: If (x + y) + (x − y)i = 5 + i, then x + y = 5 and x − y = 1 → x = 3, y = 2.
例子：若 (x + y) + (x − y)i = 5 + i，则 x + y = 5 且 x − y = 1 → x = 3, y = 2。
3. Addition and Subtraction | 加法与减法
To add or subtract complex numbers, we simply combine the real parts and the imaginary parts separately. For z₁ = a + bi and z₂ = c + di, we have z₁ + z₂ = (a + c) + (b + d)i and z₁ − z₂ = (a − c) + (b − d)i. The operations follow the familiar rules of algebra, treating i as a symbol that is collected with its coefficients.
复数的加减法只需分别合并实部与虚部。对于 z₁ = a + bi 和 z₂ = c + di，有 z₁ + z₂ = (a + c) + (b + d)i，z₁ − z₂ = (a − c) + (b − d)i。这些运算遵循熟悉的代数规则，将 i 视为与它的系数一起归并的符号。
Geometrically, addition of complex numbers corresponds to vector addition in the complex plane. If we represent z₁ and z₂ as position vectors, their sum is the diagonal of the parallelogram they span. This visual interpretation helps when working with translations as complex functions later.
从几何角度看，复数加法对应复平面上的向量加法。若将 z₁ 和 z₂ 表示为位置向量，它们的和就是它们所张成的平行四边形的对角线。这种直观解释在以后将平移视为复变函数时很有帮助。
4. Multiplication and Division | 乘法与除法
Multiplication uses the distributive law together with the fact that i² = −1. Multiplying z₁ = a + bi and z₂ = c + di gives (ac − bd) + (ad + bc)i. Notice that the pattern is similar to expanding a pair of brackets, with the extra step of replacing any i² by −1.
乘法使用分配律并结合 i² = −1 的事实。将 z₁ = a + bi 与 z₂ = c + di 相乘得到 (ac − bd) + (ad + bc)i。注意此模式与展开一对括号相似，只是多了将 i² 替换为 −1 的步骤。
Division is performed by multiplying both the numerator and the denominator by the complex conjugate of the denominator. This process makes the denominator a real number. For z₁ = a + bi and z₂ = c + di, the quotient is z₁/z₂ = (a + bi)(c − di) / (c² + d²), which expands to a standard form.
除法通过将分子和分母同时乘以分母的共轭复数来实现。这一过程使分母变成实数。对于 z₁ = a + bi 和 z₂ = c + di，商为 z₁/z₂ = (a + bi)(c − di) / (c² + d²)，展开后即得标准形式。
Example: (2 + i) / (1 − i) = (2 + i)(1 + i) / (1² + 1²) = (2 + 2i + i + i²) / 2 = (1 + 3i) / 2 = 0.5 + 1.5i
例子：(2 + i) / (1 − i) = (2 + i)(1 + i) / (1² + 1²) = (2 + 2i + i + i²) / 2 = (1 + 3i) / 2 = 0.5 + 1.5i
5. Complex Conjugate | 复共轭
The complex conjugate of z = a + bi is written as z* (or sometimes z̄) and is defined as z* = a − bi. It is formed by changing the sign of the imaginary part. Conjugates appear when solving quadratic equations with real coefficients: if a + bi is a root, then its conjugate a − bi is also a root.
复数 z = a + bi 的共轭记作 z*（有时也记作 z̄），定义为 z* = a − bi。它通过改变虚部的符号得到。当解实系数二次方程时，共轭会出现：若 a + bi 是一个根，则其共轭 a − bi 也是根。
Conjugates have several useful properties: z + z* = 2a is real; z − z* = 2bi is purely imaginary; z z* = a² + b² = |z|² is real and non‑negative. These identities are the key to simplifying divisions and finding the modulus of a complex number.
共轭具有几个有用的性质：z + z* = 2a 为实数；z − z* = 2bi 为纯虚数；z z* = a² + b² = |z|² 为非负实数。这些恒等式是化简除法以及求复数模的关键。
6. Modulus and Argument | 模与辐角
The modulus of a complex number z = a + bi is its distance from the origin in the complex plane, denoted by |z| = √(a² + b²). Geometrically, this is the length of the vector representing z. The modulus is always a non‑negative real number and satisfies |z₁ z₂| = |z₁| |z₂| and |z₁ / z₂| = |z₁| / |z₂|.
复数 z = a + bi 的模是它在复平面上到原点的距离，记作 |z| = √(a² + b²)。从几何上看，这就是表示 z 的向量的长度。模总是非负实数，且满足 |z₁ z₂| = |z₁| |z₂| 和 |z₁ / z₂| = |z₁| / |z₂|。
The argument of z, written as arg(z), is the angle θ that the line joining the origin to z makes with the positive real axis. It is usually measured in radians, and its principal value lies in the interval (−π, π]. For a given a + bi, we have tan θ = b/a, but the quadrant must be considered to determine θ correctly.
复数 z 的辐角记作 arg(z)，是从原点到 z 的连线与正实轴形成的夹角 θ。它通常以弧度度量，主值落在区间 (−π, π] 内。对于给定的 a + bi，有 tan θ = b/a，但要正确确定 θ 必须考虑所在象限。
Example: For z = −1 + √3 i, |z| = √(1 + 3) = 2, and arg(z) = 2π/3 because the point lies in the second quadrant.
例子：对于 z = −1 + √3 i，|z| = √(1 + 3) = 2，arg(z) = 2π/3，因为该点位于第二象限。
7. Polar Form and de Moivre’s Theorem | 极形与棣莫弗定理
Using the modulus r = |z| and the argument θ, any non‑zero complex number can be written in polar form as z = r (cos θ + i sin θ). This representation is especially convenient for multiplication, division and powers, because multiplying two complex numbers multiplies their moduli and adds their arguments.
利用模 r = |z| 和辐角 θ，任何非零复数都可以写成极形式 z = r (cos θ + i sin θ)。这种表示法在乘法、除法和乘方时特别方便，因为两个复数相乘时，模相乘而辐角相加。
De Moivre’s theorem extends this idea to powers: (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ for any integer n. Combined with zⁿ = rⁿ (cos nθ + i sin nθ), it provides a shortcut for raising complex numbers to powers and finding roots. At IGCSE level you will mainly use it for small integer exponents.
棣莫弗定理将这一思想推广到乘方：对任意整数 n，(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ。结合 zⁿ = rⁿ (cos nθ + i sin nθ) 就提供了快速求复数幂和方根的捷径。在 IGCSE 阶段你主要用它来处理较小的整数指数。
Example: (cos π/3 + i sin π/3)³ = cos π + i sin π = −1.
例子：(cos π/3 + i sin π/3)³ = cos π + i sin π = −1。
8. Solving Quadratic Equations | 解二次方程
Many IGCSE problems involve quadratic equations with real coefficients that have no real roots. When the discriminant Δ = b² − 4ac is negative, the roots are complex conjugates. The quadratic formula x = [−b ± √(b² − 4ac)] / (2a) still works if we replace √(Δ) with i√(|Δ|).
许多 IGCSE 题目涉及实系数二次方程，它们没有实数根。当判别式 Δ = b² − 4ac 为负时，根为一对共轭复数。二次公式 x = [−b ± √(b² − 4ac)] / (2a) 依然有效，只需将 √(Δ) 替换为 i√(|Δ|)。
For example, solving x² + 4x + 5 = 0 yields Δ = 16 − 20 = −4, so the roots are x = (−4 ± 2i)/2 = −2 ± i. Always express the final roots in standard form a + bi. Knowing that complex roots occur in conjugate pairs helps verify answers and construct equations from given roots.
例如，解 x² + 4x + 5 = 0 得到 Δ = 16 − 20 = −4，因此根为 x = (−4 ± 2i)/2 = −2 ± i。最终根要表示为标准形式 a + bi。知道复数根成共轭对出现有助于验证答案，也可以由给定根构造方程。
9. Complex Functions and Transformations | 复变函数与变换
A complex function takes a complex number as input and produces another complex number. Simple linear functions, such as f(z) = z + w (where w is a fixed complex number), represent a translation of the complex plane by the vector corresponding to w. If f(z) = kz with a real positive k, the transformation is an enlargement centred at the origin; if k is a complex number of modulus 1, the transformation is a rotation about the origin.
复变函数以一个复数作为输入，并产生另一个复数。简单的线性函数，如 f(z) = z + w（其中 w 是一个固定的复数），表示将复平面上所有的点平移一个对应于 w 的向量。若 f(z) = kz 且 k 为正实数，该变换是以原点为中心的缩放；若 k 是一个模为 1 的复数，则该变换是绕原点的旋转。
More generally, the function f(z) = az + b, with a and b complex, combines rotation, scaling and translation. Understanding how these elementary functions change the position of points in the Argand diagram is essential when you are asked to find the image of a region or solve related loci problems.
更一般地，函数 f(z) = az + b（a 和 b 为复数）综合了旋转、缩放和平移。当你被要求求一个区域在变换下的像或解决相关的轨迹问题时，理解这些初等函数如何改变 Argand 图上点的位置至关重要。
10. Loci in the Complex Plane | 复平面上的轨迹
A locus is a set of points in the complex plane that satisfy a given condition. The simplest loci are circles and perpendicular bisectors described using the modulus. |z − a| = r represents a circle with centre a and radius r; |z − a| = |z − b| represents the perpendicular bisector of the segment joining a and b.
轨迹是复平面上满足给定条件的点的集合。最简单的轨迹是用模来描述的圆和垂直平分线。|z − a| = r 表示以 a 为圆心、半径为 r 的圆；|z − a| = |z − b| 表示连接 a 与 b 的线段的垂直平分线。
Arguments also define half‑lines: arg(z − a) = θ represents a ray starting from point a, making an angle θ with the positive real direction. When sketching such loci, it is helpful to shade or exclude the initial point if required and to consider whether the boundary should be solid or dashed.
辐角也可以定义半直线：arg(z − a) = θ 表示从点 a 出发、与正实轴方向成 θ 角的一条射线。在给这些轨迹作图时，按要求用阴影标出或排除起点，并考虑边界应是实线还是虚线。
Example: The locus given by |z − 3 + i| = 2 is a circle centred at 3 − i with radius 2.
例子：由 |z − 3 + i| = 2 给出的轨迹是以 3 − i 为圆心、半径为 2 的圆。
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