IB Maths: Complex Numbers Key Points | IB 数学：复数考点精讲

📚 IB Maths: Complex Numbers Key Points | IB 数学：复数考点精讲

Complex numbers extend the real number system to include solutions to equations like x² + 1 = 0, which have no real answers. In the IB Mathematics Analysis and Approaches (AA) Higher Level course, complex numbers appear as a key topic that connects algebra, trigonometry, and geometry. Mastering this area requires confidence with algebraic manipulation, polar representation, De Moivre’s theorem, and the geometric interpretation of complex loci.

复数将实数系扩展，使得像 x² + 1 = 0 这样无实数解的方程也有了答案。在 IB 数学分析与方法 (AA) 高水平课程中，复数是一个连接代数、三角和几何的重要考点。掌握好这部分内容需要熟练进行代数运算，理解极坐标表示、棣莫弗定理以及复平面上轨迹的几何意义。

1. The Imaginary Unit and Its Powers | 虚数单位及其幂次

The imaginary unit i is defined by the property i² = -1. From this definition, higher powers of i follow a cyclic pattern: i¹ = i, i² = -1, i³ = -i, i⁴ = 1, and then the cycle repeats every four powers. This allows simplification of expressions like i²⁰²³ by dividing the exponent by 4 and using the remainder.

虚数单位 i 由性质 i² = −1 定义。从这一定义出发，更高次幂的 i 呈现出循环模式：i¹ = i, i² = −1, i³ = −i, i⁴ = 1，此后每四个幂次循环一次。运用这一规律，可以将 i²⁰²³ 这类表达式化简，只需将指数除以4，利用余数即可。

In IB problems, you are often expected to simplify sums, products, or quotients containing powers of i. An efficient strategy is to reduce each iⁿ term by writing n = 4k + r where r ∈ {0, 1, 2, 3}, then replace iⁿ with iʳ.

在 IB 考题中，常常需要化简含 i 的幂次的求和、乘积或商。一个高效的策略是将每个 iⁿ 写成 n = 4k + r 的形式，其中 r ∈ {0, 1, 2, 3}，再用 iʳ 替换 iⁿ。

i⁴ᵏ = 1, i⁴ᵏ⁺¹ = i, i⁴ᵏ⁺² = −1, i⁴ᵏ⁺³ = −i

2. Algebraic Form and Basic Operations | 代数形式与基本运算

A complex number is written in algebraic (Cartesian) form as z = a + bi, where a and b are real numbers, called the real part Re(z) and the imaginary part Im(z) respectively. Two complex numbers are equal if and only if their real and imaginary parts are both equal.

复数可以写成代数（笛卡尔）形式 z = a + bi，其中 a 和 b 为实数，分别称为实部 Re(z) 和虚部 Im(z)。两个复数相等，当且仅当它们的实部和虚部分别相等。

Addition and subtraction are performed component‑wise: (a + bi) ± (c + di) = (a ± c) + (b ± d)i. Multiplication uses the distributive law and the fact that i² = −1: (a + bi)(c + di) = ac + adi + bci + bdi² = (ac − bd) + (ad + bc)i. These operations are frequently combined with the algebraic form in IB exam questions.

加法和减法按分量进行：(a + bi) ± (c + di) = (a ± c) + (b ± d)i。乘法利用分配律以及 i² = −1：(a + bi)(c + di) = ac + adi + bci + bdi² = (ac − bd) + (ad + bc)i。在 IB 考试中，这些运算常与代数形式一起出现。

3. Complex Conjugate and Its Properties | 共轭复数及其性质

The complex conjugate of z = a + bi is denoted by z̄ = a − bi. Geometrically, conjugation reflects z across the real axis. Key algebraic properties include: (z̄)̄ = z, z + z̄ = 2a = 2Re(z), z − z̄ = 2bi = 2i Im(z), and z · z̄ = a² + b² = |z|², which is always a non‑negative real number.

复数 z = a + bi 的共轭记作 z̄ = a − bi。从几何上看，共轭相当于将 z 关于实轴进行反射。重要的代数性质包括：(z̄)̄ = z，z + z̄ = 2a = 2Re(z)，z − z̄ = 2bi = 2i Im(z)，以及 z · z̄ = a² + b² = |z|²，后者总是非负实数。

The conjugate is particularly useful when dividing complex numbers. To express (a + bi)/(c + di) in algebraic form, multiply numerator and denominator by the conjugate of the denominator: (a + bi)/(c + di) = [(a + bi)(c − di)] / [(c + di)(c − di)] = [(ac + bd) + (bc − ad)i] / (c² + d²).

共轭在复数除法中尤为重要。要将 (a + bi)/(c + di) 写成代数形式，只需将分子和分母同乘以分母的共轭：(a + bi)/(c + di) = [(a + bi)(c − di)] / [(c + di)(c − di)] = [(ac + bd) + (bc − ad)i] / (c² + d²)。

For polynomials with real coefficients, complex roots occur in conjugate pairs. If α = p + qi is a root, then ᾱ = p − qi is also a root. This fact is often used to find unknown coefficients or to factorise polynomials over the real numbers.

对于实系数多项式，复数根总是以共轭对的形式出现。如果 α = p + qi 是一个根，那么 ᾱ = p − qi 也必定是根。这一性质常用于求解未知系数，或在实数范围内对多项式进行因式分解。

4. Modulus and Argument | 模与辐角

For a complex number z = a + bi, its modulus (or absolute value) is the distance from the origin to the point (a, b) in the complex plane: |z| = √(a² + b²). The argument arg(z) is the directed angle measured from the positive real axis to the line segment joining the origin to z, usually taken in the interval (−π, π] or [0, 2π).

对于复数 z = a + bi，它的模（或绝对值）是从原点到复平面上点 (a, b) 的距离：|z| = √(a² + b²)。辐角 arg(z) 是从正实轴旋转到连接原点与 z 的线段所成的有向角，通常取值于 (−π, π] 或 [0, 2π)。

It is essential to locate the quadrant of z when finding the argument. The principal argument Arg(z) is often obtained using tan θ = b/a, corrected by adding or subtracting π depending on the signs of a and b. For example, if z = −1 − √3 i, then a = −1, b = −√3, giving θ = arctan(√3/1) = π/3, but the complex number lies in the third quadrant, so Arg(z) = −π + π/3 = −2π/3 (if principal range is (−π, π]).

求辐角时，必须确定 z 所在的象限。主辐角 Arg(z) 通常利用 tan θ = b/a 求得，并根据 a 和 b 的符号加减 π 进行修正。例如，若 z = −1 − √3 i，则 a = −1，b = −√3，可得 θ = arctan(√3/1) = π/3，但该复数位于第三象限，因此 Arg(z) = −π + π/3 = −2π/3（若主值区间为 (−π, π]）。

5. Polar Form and Euler’s Formula | 极坐标形式与欧拉公式

Using the modulus r = |z| and argument θ = arg(z), any non‑zero complex number can be expressed in polar (trigonometric) form: z = r(cos θ + i sin θ). Euler’s formula establishes a profound connection between trigonometric and exponential functions: e^(iθ) = cos θ + i sin θ. This gives the compact exponential form z = r e^(iθ).

利用模 r = |z| 和辐角 θ = arg(z)，任何一个非零复数都可以表示为极坐标（三角）形式：z = r(cos θ + i sin θ)。欧拉公式在三角函数与指数函数之间建立了深刻联系：e^(iθ) = cos θ + i sin θ。由此得到更为简洁的指数形式 z = r e^(iθ)。

e^(iθ) = cos θ + i sin θ

In IB HL papers, you will be expected to convert freely between algebraic, polar, and exponential forms. Exponential form is particularly efficient for multiplication, division, and exponentiation. It also underpins many proofs involving trigonometric identities.

在 IB 高水平试题中，你需要能够在代数形式、极坐标形式和指数形式之间自由转换。指数形式在进行乘法、除法和乘方运算时尤其高效，也是很多涉及三角恒等式证明的理论基础。

6. Multiplication and Division in Polar Form | 极坐标形式下的乘法和除法

When complex numbers are written in polar form as z₁ = r₁ e^(iθ₁) and z₂ = r₂ e^(iθ₂), multiplication becomes straightforward: the moduli multiply and the arguments add. That is, z₁z₂ = r₁r₂ e^(i(θ₁+θ₂)). Division similarly yields: z₁/z₂ = (r₁/r₂) e^(i(θ₁−θ₂)).

当复数用极坐标形式表示为 z₁ = r₁ e^(iθ₁) 和 z₂ = r₂ e^(iθ₂) 时，乘法变得非常简单：模长相乘，辐角相加。即 z₁z₂ = r₁r₂ e^(i(θ₁+θ₂))。除法同样得出：z₁/z₂ = (r₁/r₂) e^(i(θ₁−θ₂))。

These rules can also be expressed in trigonometric notation: z₁z₂ = r₁r₂ [cos(θ₁+θ₂) + i sin(θ₁+θ₂)]. The geometric interpretation is that multiplying by a complex number rotates and scales the other: multiplication by e^(iθ) purely rotates a point by angle θ around the origin, while multiplication by a real number simply scales it.

这些规则也可以用三角记法表达：z₁z₂ = r₁r₂ [cos(θ₁+θ₂) + i sin(θ₁+θ₂)]。几何解释是，乘以一个复数相当于对另一个复数进行旋转和伸缩：乘以 e^(iθ) 纯粹将点绕原点旋转角度 θ，而乘以实数仅进行伸缩。

7. De Moivre’s Theorem | 棣莫弗定理

De Moivre’s theorem states that for any integer n, (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ). When combined with the exponential form this is simply (e^(iθ))ⁿ = e^(inθ). The theorem is a powerful tool for deriving trigonometric identities, finding powers of complex numbers, and proving results about roots of unity.

棣莫弗定理指出，对于任意整数 n，有 (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)。结合指数形式，这就是简单的 (e^(iθ))ⁿ = e^(inθ)。该定理是推导三角恒等式、求复数乘方以及证明单位根相关结论的有力工具。

(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)

In IB problems, De Moivre’s theorem is frequently used in both directions. For instance, you might be asked to express cos 3θ in terms of cos θ by expanding (cos θ + i sin θ)³ using the binomial theorem and equating real parts. Conversely, the theorem can simplify expressions like (1 + i)⁶ by first converting to polar form.

在 IB 问题中，棣莫弗定理常常被双向使用。例如，可能会要求你用二项式定理展开 (cos θ + i sin θ)³，并比较实部，从而将 cos 3θ 用 cos θ 表达出来。反过来，该定理也能用于化简像 (1 + i)⁶ 这样的表达式，只需先转化为极坐标形式即可。

8. Integer Powers of Complex Numbers | 复数的整数次幂

To raise a complex number to an integer power, the most efficient method is to write it in polar or exponential form, then apply De Moivre’s theorem. For z = r e^(iθ), we have zⁿ = rⁿ e^(inθ) = rⁿ [cos(nθ) + i sin(nθ)], provided n is an integer. This avoids lengthy algebraic multiplication.

要计算复数的整数次幂，最有效的方法是先将其写成极坐标或指数形式，再应用棣莫弗定理。对于 z = r e^(iθ)，有 zⁿ = rⁿ e^(inθ) = rⁿ [cos(nθ) + i sin(nθ)]，其中 n 为整数。这样做可以避免繁琐的代数乘法。

Be careful when n is negative: write z⁻¹ = (1/r) e^(−iθ) first, then apply the power rule. For example, (1 + i)⁻³ = (√2 e^(iπ/4))⁻³ = (1/(√2)³) e^(−i3π/4) = (1/(2√2)) [cos(−3π/4) + i sin(−3π/4)].

当 n 为负数时要当心：可先将 z⁻¹ = (1/r) e^(−iθ) 写出，再应用乘方法则。例如，(1 + i)⁻³ = (√2 e^(iπ/4))⁻³ = (1/(√2)³) e^(−i3π/4) = (1/(2√2)) [cos(−3π/4) + i sin(−3π/4)]。

9. Roots of Unity and Complex Roots | 单位根与复根

An nth root of unity is a complex solution to zⁿ = 1. There are exactly n distinct nth roots of unity, given by ωₖ = e^(i·2πk/n) = cos(2πk/n) + i sin(2πk/n) for k = 0, 1, 2, …, n−1. Geometrically, these roots are equally spaced points on the unit circle, forming a regular n‑gon.

n 次单位根是方程 zⁿ = 1 的复数解。恰好有 n 个不同的 n 次单位根，由 ωₖ = e^(i·2πk/n) = cos(2πk/n) + i sin(2πk/n)，k = 0, 1, 2, …, n−1 给出。从几何上看，这些根是单位圆上等间距分布的点，构成正 n 边形。

zⁿ = 1 ⇒ zₖ = cos(2πk/n) + i sin(2πk/n), k = 0, 1, … , n−1

More generally, the equation zⁿ = w, where w = ρ e^(iφ) ≠ 0, has n distinct solutions: zₖ = ρ^(1/n) e^(i(φ+2πk)/n) for k = 0, 1, …, n−1. In IB exams, you must be able to find all roots, show them on an Argand diagram, and use geometric relations such as the sum of all nth roots of unity being zero.

更一般地，方程 zⁿ = w，其中 w = ρ e^(iφ) ≠ 0，具有 n 个不同的解：zₖ = ρ^(1/n) e^(i(φ+2πk)/n)，k = 0, 1, …, n−1。在 IB 考试中，你需要能够求出所有根，在阿冈图上标出它们，并利用一些几何关系，例如所有 n 次单位根之和为零。

10. Solving Equations with Complex Numbers | 解包含复数的方程

IB questions frequently combine complex numbers with quadratic and polynomial equations. For a quadratic equation with real coefficients, if the discriminant is negative, the roots are a conjugate pair of complex numbers given by the quadratic formula. For example, x² + 2x + 5 = 0 yields x = −1 ± 2i.

IB 考题经常将复数与二次方程及多项式方程结合起来。对于实系数二次方程，若判别式为负，其根是一对共轭复数，可以通过求根公式得到。例如，x² + 2x + 5 = 0 给出 x = −1 ± 2i。

When complex coefficients are involved, the same algebraic rules apply. To solve an equation such as (1 + i)z − 2 = 3i, isolate z and simplify: z = (3i + 2)/(1 + i). Multiply numerator and denominator by the conjugate 1 − i, expand and simplify to obtain z in algebraic form.

当方程含有复数系数时，同样的代数规则依旧适用。要解如 (1 + i)z − 2 = 3i 的方程，先分离出 z，再化简：z = (3i + 2)/(1 + i)。分子分母同乘以共轭 1 − i，展开并化简即可得到 z 的代数形式。

Higher‑degree real polynomials can be factorised using known complex roots. If one complex root α is given, then ᾱ is also a root, and (z − α)(z − ᾱ) is a real quadratic factor that can be divided out to reduce the degree.

高次实系数多项式可利用已知的复数根进行因式分解。若已知一个复数根 α，则 ᾱ 亦为根，(z − α)(z − ᾱ) 构成一个实二次因式，可以除出去以降低多项式的次数。

11. Loci in the Complex Plane | 复平面上的轨迹

A locus is a set of points in the complex plane satisfying a given condition. Common loci tested in IB Mathematics include circles, perpendicular bisectors, rays, and lines. The equation |z − z₀| = r describes a circle with centre z₀ and radius r. The condition |z − z₁| = |z − z₂| describes the perpendicular bisector of the segment joining z₁ and z₂.

轨迹是复平面上满足给定条件的点集。IB 数学中常考的轨迹包括圆、垂直平分线、射线和直线。方程 |z − z₀| = r 表示以 z₀ 为圆心、半径为 r 的圆。条件 |z − z₁| = |z − z₂| 则描述了连接 z₁ 与 z₂ 的线段的垂直平分线。

Another important locus is the ray: arg(z − z₀) = θ represents a half‑line starting from z₀ (but not including z₀ itself) at an angle θ to the positive real axis. Inequalities combine these sets: |z − z₀| < r defines the interior of a disc, while |z − z₀| ≤ |z − z₁| defines a half‑plane bounded by the perpendicular bisector.

另一类重要轨迹是射线：arg(z − z₀) = θ 表示以 z₀ 为起点（不含 z₀ 本身）、与正实轴成角度 θ 的射线。不等式将这些集合组合起来：|z − z₀| < r 定义了一个圆盘内部，而 |z − z₀| ≤ |z − z₁| 则定义了以垂直平分线为边界的半平面。

When sketching loci, it is helpful to substitute z = x + iy and interpret the condition in terms of Cartesian equations, or to use geometric reasoning directly on the Argand diagram. Both methods are accepted in IB answers.

在绘制轨迹时，既可以代入 z = x + iy 将条件转化为笛卡尔方程进行理解，也可以直接在阿冈图上使用几何推理。这两种方法在 IB 答案中均被接受。

12. Key Exam Techniques and Common Mistakes | 考试技巧与常见错误

In IB exams, complex number questions often carry high marks and require clear logical steps. Always check the principal argument range: many students lose marks by not enforcing −π < Arg(z) ≤ π. When finding roots of a complex number, remember to add 2πk to the argument before dividing by the root index, and list all n distinct roots.

在 IB 考试中，复数题目往往分值较高，需要清晰的逻辑步骤。务必检查主辐角的取值范围：许多学生因为没有强制 −π < Arg(z) ≤ π 而丢分。求复数的根时，记得在辐角上加上 2πk 再除以开方次数，并列出所有 n 个不同的根。

Another common pitfall is forgetting that |z|² = z·z̄, which is useful when simplifying expressions or proving identities. When solving locus problems, carefully distinguish between “less than” and “less than or equal to” to decide whether boundaries are included. Finally, for questions that ask for all solutions to an equation, make sure you have not omitted any; a quick check using the Fundamental Theorem of Algebra can confirm the expected number of roots.

另一个常见陷阱是忘记 |z|² = z·z̄，而这一关系在化简表达式或证明恒等式时非常有用。求解轨迹问题时，要仔细区分“小于”和“小于等于”，以判断边界是否包含在内。最后，对于要求找出方程所有解的题目，确保没有遗漏任何一个；可以用代数学基本定理快速检验预期根的个数。

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IB Maths: Complex Numbers Key Points | IB 数学：复数 考点精讲