📚 Complex Numbers for IB Edexcel Mathematics: Key Exam Points | IB Edexcel 数学：复数 考点精讲
tial for solving polynomial equations, describing rotations, and analysing alternating current circuits. In both IB and Edexcel A Level Mathematics, you need to master operations with complex numbers, geometric interpretations using Argand diagrams, polar forms, De Moivre' his guide consolidates the key concepts and examination techniques required for top marks.
复数通过引入虚数单位 i（其中 i² = -1）对实数系统进行了扩展。它们在求解多项式方程、描述旋转以及分析交流电路等方面至关重要。无论是 IB 还是 Edexcel A Level 数学，你都需要掌握复数的运算、利用 Argand 图进行几何解释、极坐标形式、棣莫弗定理以及求单位根的方法。本指南汇集了取得高分所需的核心概念与应试技巧。
1. Imaginary Numbers and the Definition of i | 虚数与 i 的定义
nary number can be written as bi where b is a real number. The square root of a negative number can be expressed using i: for a > 0, √-a = i√a. Remember that when simplifying products like √-4 × √-9, you must extract i first to avoid mistakes: √-4 × √-9 = 2i × 3i = -6, not √36 = 6.
虚数单位 i 由性质 i² = -1 定义。由此，任何虚数都可写作 bi，其中 b 为实数。负数的平方根可以用 i 表示：对于 a > 0，有 √-a = i√a。务必注意，化简诸如 √-4 × √-9 的乘积时，必须先提取 i 以避免错误：√-4 × √-9 = 2i × 3i = -6，而不是 √36 = 6。
i, where a and b are real numbers. a is called the real part, Re(z), and b is the imaginary part, Im(z) — ex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.
复数 z 的形式为 z = a + bi，其中 a 和 b 是实数。a 称为实部，记作 Re(z)；b 称为虚部，记作 Im(z) —— 注意 Im(z) 是 b 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。
or real x, y. Equate real and imaginary parts after expanding to obtain simultaneous equations.
在 IB 考试中，你可能会被要求求解像 (x + yi)(2 - i) = 3 + 4i 这样的方程，其中 x, y 为实数。展开后令实部与虚部分别相等，得到联立方程组即可。
2. Complex Conjugate and Basic Operations | 共轭复数与基本运算
The complex conjugate of z = a + bi is denoted by z̅ (or z*) and is defined as z̅ s the real axis in the Argand plane. The properties of conjugates are extremely useful: (z̅) = z, (z₁ ± z₂)̅ = z₁̅ ± z₂̅, (z₁z₂)̅ = z₁̅ · z₂̅, and (z₁ / z₂)̅ = z₁̅ / z₂̅. Also, z · z̅ = a² + b², which is always real and non‑negative.
复数 z = a + bi 的共轭记作 z̅（或 z*），定义为 z̅ = a - bi。它在复平面上表示关于实轴的反射。共轭的性质非常有用：(z̅) = z，(z₁ ± z₂)̅ = z₁̅ ± z₂̅，(z₁z₂)̅ = z₁̅ · z₂̅，以及 (z₁ / z₂)̅ = z₁̅ / z₂̅。此外，z · z̅ = a² + b²，该值总是实数且非负。
Addition and subtraction are performed component‑ tandard algebraic expansion along with i² = -1: (a + bi)(c + di) = (ac - bd) + (ad + bc)i.
加法和减法按分量进行：(a + bi) ± (c + di) = (a ± c) + (b ± d)i。乘法遵循标准代数展开并利用 i² = -1：(a + bi)(c + di) = (ac - bd) + (ad + bc)i。
i)] = [(ac + bd) + (bc - ad)i] / (c² + d²). Always express your answer in the form x + yi.
复数除法利用共轭使分母实数化： (c + di) = [(a + bi)(c - di)] / [(c + di)(c - di)] = [(ac + bd) + (bc - ad)i] / (c² + d²)。始终将答案表示为 x + yi 的形式。
3. Argand Diagram and Modulus | 复平面与模
The Argand diagram represents complex numbers as points or vectors on a plane, with the x‑axis as the real axis and the y‑ = a + bi corresponds to the point (a, b) or the position vector from the origin to (a, b).
Argand 图将复数表示为平面上的点或向量，其中 x 轴为实轴，y 轴为虚轴。复数 z = a + bi 对应于坐标为 (a, b) 的点或从原点到 (a, b) 的位置向量。
The modulus of z, written |z|, is the distance from the origin to (a, b), given by |z| = √a² + b². It is always a non‑negative real number. Geometrically, |z₁ - z₂| represents the distance between the points z₁ and z₂. This is frequently tested when describing circles, lines, or loci.
复数 z 的模记作 |z|，是原点到 (a, b) 的距离，公式为 |z| = √a² + b²。它始终是非负实数。从几何角度看，|z₁ - z₂| 表示两点 z₁ 与 z₂ 之间的距离。这在描述圆、直线或轨迹时经常考查。
Key properties of modulus include: |z₁z₂| = |z₁| · |z₂|, |z₁ / z₂| = |z₁| / |z₂| (z₂ ≠ 0), and the triangle inequality |z₁ + z₂| ≤ |z₁| + |z₂|. The relation |z|² = z · z̅ links modulus to conjugate.
模的关键性质包括：|z₁z₂| = |z₁| · |z₂|，|z₁ / z₂| = |z₁| / |z₂|（z₂ ≠ 0），以及三角不等式 |z₁ + z₂| ≤ |z₁| + |z₂|。关系式 |z|² = z · z̅ 将模与共轭联系起来。
4. Argument of a Complex Number | 复数的辐角
The argument of a non‑ arg(z), is the angle θ measured from the positive real axis to the vector representing z. The principal argument is usually taken in the interval (-π, π] (IB) or [0, 2π) depending on convention. You must state the range you are using clearly in exam questions.
非零复数 z = a + bi 的辐角记作 arg(z)，是从正实轴到表示 z 的向量所成的角度 θ。主辐角通常取在 (-π, π]（IB）或 [0, 2π)，取决于约定。考试中你必须清楚地说明所使用的范围。
θ = π + arctan(b/a) (adjusted to the principal range). Never just blindly apply the inverse tangent.
要求 arg(z)，可利用 tan θ = b/a，但务必考虑点 (a, b) 所在的象限。a 与 b 的符号决定了正确的角度。例如，若 a < 0 且 b > 0，则 θ = π + arctan(b/a)（再调整到主值范围）。切勿盲目套用反正切函数。
) - arg(z₂). These properties lead directly to the geometric interpretation of multiplication as a rotation and scaling.
乘积的辐角满足 arg(z₁z₂) = arg(z₁) + arg(z₂)（模 2π 同余）。类似地，arg(z₁ / z₂) = arg(z₁) - arg(z₂)。这些性质直接引出了乘法的几何解释：旋转与伸缩。
5. Polar Form of Complex Numbers | 复数的极坐标形式
A complex number can be expressed in polar form as z = r (cos θ + i sin θ), where r = |z| and θ = arg(z). This is also called the modulus‑ sin θ. Both IB and Edexcel accept the cis notation, but you must define it if you use it.
复数可以用极坐标形式表示为 z = r (cos θ + i sin θ)，其中 r = |z|，θ = arg(z)。这也称为模‑辐角形式，常缩写为 z = r cis θ，其中 cis θ 代表 cos θ + i sin θ。IB 和 Edexcel 都接受 cis 记法，但使用时需先定义。
n, simply evaluate: a = r cos θ, b = r sin θ. To go from Cartesian to polar, calculate r = √a² + b² and θ from the quadrant‑adjusted arctan(b/a).
直角坐标形式 (a + bi) 与极坐标形式的相互转换是核心技能。从极坐标转换为直角坐标只需计算：a = r cos θ，b = r sin θ。从直角坐标转换为极坐标，先计算 r = √a² + b²，再通过象限调整的 arctan(b/a) 得到 θ。
Examiners often test this alongside complex equations. For instance, express z = -1 - √3 i in polar form. Here r = √1 + 3 = 2, and tan θ = √3 θ = -2π/3 (if principal range is (-π, π]) or 4π/3. Thus z = 2 (cos(-2π/3) + i sin(-2π/3)).
考官常将此与复数方程结合考查。例如，将 z = -1 - √3 i 表示为极坐标形式。这里 r = √1 + 3 = 2，tan θ = √3，但由于 a < 0, b < 0，θ = -2π/3（若主值范围为 (-π, π]）或 4π/3。因此 z = 2 (cos(-2π/3) + i sin(-2π/3))。
6. Multiplication and Division in Polar Form | 极坐标形式的乘法与除法
en z₁z₂ = r₁r₂ [cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)]. The moduli multiply and the arguments add.
极坐标形式下两个复数的乘法非常简洁：若 z₁ = r₁ (cos θ₁ + i sin θ₁) 且 z₂ = r₂ (cos θ₂ + i sin θ₂)，则 z₁z₂ = r₁r₂ [cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)]。模相乘，辐角相加。
₁ - θ₂) + i sin(θ₁ - θ₂)], provided z₂ ≠ 0. The moduli divide and the arguments subtract. These rules are directly derived from the compound angle formulae and are much faster than expanding Cartesian products.
对于除法，z₁ / z₂ = (r₁ / r₂) [cos(θ₁ - θ₂) + i sin(θ₁ - θ₂)]，其中 z₂ ≠ 0。模相除，辐角相减。这些规则直接由复角公式推导而来，比展开直角坐标形式的乘积快得多。
A typical exam question: Given z = √2 (π/4)) and w = 3 (cos (π/3) + i sin (π/3)), find zw and z/w in polar form. Answer: zw = 3√2 (cos (7π/12) + i sin (7π/12)); z/w = (√2/3) (cos (-π/12) + i sin (-π/12)) = (√2/3) (cos (π/12) - i sin (π/12)). Always adjust the argument into the principal range if required.
典型的考试题：已知 z = √2 (cos (π/4) + i sin (π/4)) 和 w = 3 (cos (π/3) + i sin (π/3))，求 zw 和 z/w 的极坐标形式。答案：zw = 3√2 (cos (7π/12) + i sin (7π/12))；z/w = (√2/3) (cos (-π/12) + i sin (-π/12)) = (√2/3) (cos (π/12) - i sin (π/12))。若要求主辐角，需调整到规定范围内。
7. De Moivre's Theorem | 棣莫弗定理
De Moivre' ). This theorem is a powerful tool for raising complex numbers to integer powers and for deriving trigonometric identities.
棣莫弗定理指出，对任意整数 n，有 (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)。该定理是将复数求整数次幂以及推导三角恒等式的强大工具。
After applying De Moivre, you may convert back to Cartesian form if the problem demands it.
要计算 (a + bi)ⁿ，首先转换为极坐标形式：a + bi = r (cos θ + i sin θ)。然后 (a + bi)ⁿ = rⁿ (cos(nθ) + i sin(nθ))。应用棣莫弗定理后，若题目要求，可再转回直角坐标形式。
nding the left side using the binomial theorem gives cos³θ + 3i cos²θ sin θ - 3 cos θ sin²θ - i sin³θ. Equating real parts yields cos 3θ = 4 cos³θ - 3 cos θ.
试题经常要求利用棣莫弗定理将 cos 3θ 用 cos θ 表示，或证明恒等式。例如，(cos θ + i sin θ)³ = cos 3θ + i sin 3θ。用二项式定理展开左边得 cos³θ + 3i cos²θ sin θ - 3 cos θ sin²θ - i sin³θ。比较实部即得 cos 3θ = 4 cos³θ - 3 cos θ。
De Moivre also holds for negative integers, giving a method to compute z⁻ⁿ. The modulus becomes r⁻ⁿ and the argument -nθ.
棣莫弗定理对负整数也成立，提供了计算 z⁻ⁿ 的方法：模变为 r⁻ⁿ，辐角变为 -nθ。
8. Roots of Complex Numbers | 复数的根
oots, known as the nth roots of w. If w = r (cos θ + i sin θ), the solutions are given by:
zₖ = √[n]{r} [ cos((θ + 2πk)/n) + i sin((θ + 2πk)/n) ], for k = 0, 1, 2, ..., n-1.
这里 √[n]{r} 表示 r 的 n 次正实数根。注意需遍历 k = 0 到 n-1。这些根在复平面上均匀地分布在以原点为中心、半径为 √[n]{r} 的圆上，相邻两根的辐角差为 2π/n。
求复数 w 的 n 次方根是考试中的热门题型。方程 zⁿ = w 恰好有 n 个不同的根，称为 w 的 n 次方根。若 w = r (cos θ + i sin θ)，则解由下式给出：zₖ = √[n]{r} [ cos((θ + 2πk)/n) + i sin((θ + 2πk)/n) ]，其中 k = 0, 1, 2, ..., n-1。这些根在复平面上均匀分布在以原点为圆心、半径为 √[n]{r} 的圆上，相邻两根的辐角差为 2π/n。
ty are zₖ = cos(2πk/n) + i sin(2πk/n). They sum to zero, a fact frequently used in proofs.
对于单位根（1 的 n 次方根）这种特殊情况，w = 1 = cos 0 + i sin 0，因此 n 次单位根为 zₖ = cos(2πk/n) + i sin(2πk/n)。它们的和为零，这一事实经常用于证明。
When drawing roots on an Argand diagram, mark the centre, radius, and all n points clearly. Questions may ask for the roots in exact Cartesian form, so you must evaluate cos and sin for standard angles.
在 Argand 图上绘制根时，要清晰标出圆心、半径以及所有 n 个点的位置。题目可能要求以精确的直角坐标形式写出根，因此你必须会计算标准角的正弦和余弦值。
9. Loci in the Complex Plane | 复平面上的轨迹
This topic links complex numbers to coordinate geometry. Common loci include:
- |z - z₀| = r: a circle with centre z₀ and radius r.
- |z - z₁| = |z - z₂|: the perpendicular bisector of the line segment joining z₁ and z₂.
- arg(z - z₀) = θ: a half‑ ing from z₀ (excluding the point itself), making angle θ with the positive real direction.
- |z - z₁| + |z - z₂| = constant: an ellipse with foci z₁ and z₂.
本主题将复数与坐标几何联系起来。常见的轨迹包括：|z - z₀| = r 表示以 z₀ 为圆心、半径为 r 的圆；|z - z₁| = |z - z₂| 表示连接 z₁ 与 z₂ 的线段的垂直平分线；arg(z - z₀) = θ 表示从 z₀ 出发（不含该点）且与正实轴夹角为 θ 的射线；|z - z₁| + |z - z₂| = 常数则表示以 z₁ 和 z₂ 为焦点的椭圆。
When sketching loci, find Cartesian equations by substituting z = x + yi. For example, |z - i| = |z + 1| y², which simplifies to y = -x. Identify and label key features such as centre and radius.
绘制轨迹时，可通过代入 z = x + yi 求得直角坐标方程。例如，|z - i| = |z + 1| 可化为 x² + (y-1)² = (x+1)² + y²，化简得 y = -x。要识别并标注关键特征，如圆心和半径。
Intersections of loci are found by solving their Cartesian equations simultaneously. In IB exams, you may be asked to find the minimum or maximum value of |z - a| given a locus constraint, which uses geometry to minimise distance from a point to a line or circle.
轨迹的交点可通过联立求解其直角坐标方程得到。在 IB 考试中，你可能需要求在给定轨迹约束下 |z - a| 的最小值或最大值，这就要利用几何方法求点到直线或圆的最短距离。
10. Exponential Form and Euler's Formula | 指数形式与欧拉公式
Euler's formula, eiθ = cos θ + i sin θ, allows complex numbers to be written in exponential form: z = r eiθ. This representation is extremely useful in calculus, differential equations, and electrical engineering, and it is part of the IB Higher Level and Edexcel Further Mathematics syllabuses.
欧拉公式 eiθ = cos θ + i sin θ 使得复数可以用指数形式表示：z = r eiθ。这种表示法在微积分、微分方程和电气工程中极为有用，也是 IB 高水平课程和 Edexcel 进阶数学课程内容。
Exponential form inherits all the advantages of polar form. Multiplication: r₁eiθ₁ × r₂eiθ₂ = r₁r₂ e^(i(θ₁+θ₂)). Division: r₁eiθ₁ / r₂eiθ₂ = (r₁/r₂) e^(i(θ₁-θ₂)). De Moivre becomes (eiθ)ⁿ = einθ, which is a trivial consequence of exponent rules.
指数形式继承了极坐标形式的所有优点。乘法：r₁eiθ₁ × r₂eiθ₂ = r₁r₂ e^(i(θ₁+θ₂))。除法：r₁eiθ₁ / r₂eiθ₂ = (r₁/r₂) e^(i(θ₁-θ₂))。棣莫弗定理变为 (eiθ)ⁿ = einθ，这仅是幂运算规则的直接结果。
Complex conjugate in exponential form: if z = r eiθ, then z̅ = r e-iθ e argument across the real axis. The real and imaginary parts can be extracted: cos θ = (eiθ + e-iθ)/2, sin θ = (eiθ - e-iθ)/(2i).
指数形式下的共轭复数：若 z = r eiθ，则 z̅ = r e-iθ。这凸显了共轭将辐角关于实轴反射的事实。实部和虚部也可由此提取：cos θ = (eiθ + e-iθ)/2，sin θ = (eiθ - e-iθ)/(2i)。
In exams, using exponential form can simplify problems involving roots and powers where arguments are not standard angles. For instance, expressing 1+i in exponential form as √2 eiπ/4 allows quick calculation of (1+i)¹⁰.
在考试中，当辐角不是标准角时，使用指数形式可简化涉及方根和幂次的问题。例如，将 1+i 表示为指数形式 √2 eiπ/4 后，便可快速计算 (1+i)¹⁰。
11. Solving Equations Involving Complex Numbers | 求解含复数的方程
ary parts. For quadratic equations with real coefficients, if the discriminant is negative, the roots are complex conjugates. For example, x² - 4x + 13 = 0 gives x = 2 ± 3i.
含有复数系数的方程的处理方法与实数方程类似，但通常需要借助共轭来分离实部和虚部。对于实系数的二次方程，若判别式为负，则其根为共轭复数。例如，x² - 4x + 13 = 0 的解为 x = 2 ± 3i。
ients are complex, standard quadratic formula still works: for az² + bz + c = 0, z = [-b ± √b² - 4ac] / (2a), but you must compute the square root of a complex number properly — usually by writing the discriminant in polar form and then finding its square roots.
当系数为复数时，标准求根公式仍然适用：对于 az² + bz + c = 0，z = [-b ± √b² - 4ac] / (2a)，但你必须正确计算复数的平方根 —— 通常需将判别式写成极坐标形式，再求其平方根。
h another relation. Alternatively, treat complex numbers as unknowns and use matrix inversion if comfortable.
通过令实部和虚部分别相等，可以求解诸如 (1+i)z + (2-i)w = 3+4i 以及另一关系的联立方程组。或者，将复数视为未知数，若熟悉矩阵求逆也可使用此法。
Always check that your final answers are simplified and presented in the required form, such as a + ib, and that any argument lies within the principal range specified.
务必检查最终答案是否已化简并按要求的形式（如 a + ib）给出，以及辐角是否位于指定的主值范围内。
12. Key Formulae and Exam Tips | 核心公式与应试技巧
Here is a summary of the essential formulae you must memorise and apply accurately:
| Concept | Formula |
| Modulus | |a + bi| = √a² + b² |
| Argument | θ = arg(z), tan θ = b/a with quadrant check |
| Conjugate | z̅ = a - bi, z·z̅ = |z|² |
| Polar form | z = r(cos θ + i sin θ) = r cis θ |
| Exponential form | z = r eiθ |
| Multiplication | r₁r₂ cis(θ₁+θ₂) |
| Division | (r₁/r₂) cis(θ₁-θ₂) |
| De Moivre | (r cis θ)ⁿ = rⁿ cis(nθ) |
| nth roots | zₖ = √[n]{r} cis((θ+2πk)/n), k = 0,...,n-1 |
以下是你必须准确记忆并应用的核心公式汇总：模 |a + bi| = √a² + b²；辐角 θ = arg(z)，tan θ = b/a 并做象限判断；共轭 z̅ = a - bi, z·z̅ = |z|²；极坐标形式 z = r(cos θ + i sin θ) = r cis θ；指数形式 z = r eiθ；乘法 r₁r₂ cis(θ₁+θ₂)；除法 (r₁/r₂) cis(θ₁-θ₂)；棣莫弗定理 (r cis θ)ⁿ = rⁿ cis(nθ)；n 次方根 zₖ = √[n]{r} cis((θ+2πk)/n), k = 0,...,n-1。
Exam success tip: Always draw a quick Argand sketch to visualise the problem, especially for loci and roots. Check that the argument is expressed in the correct principal range and simplify answers fully. When proving trigonometric identities using De Moivre, equate real and imaginary parts carefully, and remember to use the binomial theorem where needed.
应考成功技巧：始终快速画出 Argand 示意图以直观理解问题，尤其对于轨迹和根的问题。检查辐角是否表达了正确的主值范围，并彻底化简答案。在利用棣莫弗定理证明三角恒等式时，仔细令实部与虚部分别相等，并在需要时记住使用二项式定理。
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