📚 IB Maths: Complex Numbers Essential Revision | IB 数学：复变函数 考点精讲
Complex numbers form a fundamental extension of the real numbers, enabling solutions to equations such as x² + 1 = 0 and providing deep connections between algebra, geometry, and trigonometry. In the IB Analysis and Approaches HL course, mastery of complex numbers is essential for success in both the core and option topics. This article covers every key concept you need: from Cartesian and polar forms to De Moivre’s theorem, roots of unity, and geometric transformations.
复数是实数的基本扩展，它不仅使 x² + 1 = 0 这类方程有解，还在代数、几何与三角学之间建立了深刻联系。在 IB 数学分析与方法（AA HL）课程中，熟练掌握复数知识是攻克核心内容和选修专题的关键。本文涵盖所有必考概念：从直角坐标形式与极坐标形式，到棣莫弗定理、单位根以及几何变换。
1. Introduction to Complex Numbers | 复数简介
A complex number is a number of the form z = a + bi, where a, b ∈ ℝ and i is the imaginary unit with the property i² = −1. The set of all complex numbers is denoted by ℂ. The real part Re(z) = a, and the imaginary part Im(z) = b (note: Im(z) is the real number b, not bi).
复数是指具有 z = a + bi 形式的数，其中 a, b ∈ ℝ，i 是虚数单位，满足 i² = −1。全体复数的集合记作 ℂ。实部 Re(z) = a，虚部 Im(z) = b（注意 Im(z) 是实数 b，而不是 bi）。
Complex numbers arise naturally when solving polynomial equations. For example, the quadratic x² − 4x + 13 = 0 yields complex roots because the discriminant is negative. The IB requires you to work comfortably with such numbers in algebraic manipulations.
复数在解多项式方程时自然出现。例如二次方程 x² − 4x + 13 = 0 的判别式为负，因此得到复数根。IB 要求你能熟练地对这类数进行代数运算。
2. Cartesian Form and Operations | 直角坐标形式与运算
In Cartesian form, z = a + bi. 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.
在直角坐标形式中，z = a + bi。加减运算按分量执行：(a + bi) ± (c + di) = (a ± c) + (b ± d)i。乘法使用分配律以及 i² = −1 这一性质：(a + bi)(c + di) = ac + adi + bci + bdi² = (ac − bd) + (ad + bc)i。
Division is achieved by multiplying the numerator and denominator by the complex conjugate of the denominator. For z = a + bi and w = c + di ≠ 0, z / w = (a + bi)(c − di) / (c² + d²). This eliminates the imaginary part from the denominator.
除法是通过将分子和分母同乘以分母的共轭复数实现。对于 z = a + bi 和 w = c + di ≠ 0，有 z / w = (a + bi)(c − di) / (c² + d²)。这样可消去分母中的虚部。
Equality of complex numbers: a + bi = c + di if and only if a = c and b = d. This is frequently used to find real and imaginary parts or to solve for unknowns.
复数相等条件：a + bi = c + di 当且仅当 a = c 且 b = d。这一性质常用来求实部和虚部，或解未知数。
3. Complex Conjugate and Modulus | 共轭与模
The complex conjugate of z = a + bi is denoted by z̅ = a − bi (or z*). Geometrically, conjugation reflects a point across the real axis. Conjugation distributes over addition, subtraction, multiplication, and division: (z ± w)̅ = z̅ ± w̅, (zw)̅ = z̅ w̅, (z/w)̅ = z̅ / w̅.
复数 z = a + bi 的共轭记作 z̅ = a − bi（或 z*）。几何上，共轭运算将点关于实轴做反射。共轭对加减乘除满足分配律：(z ± w)̅ = z̅ ± w̅，(zw)̅ = z̅ w̅，(z/w)̅ = z̅ / w̅。
The modulus (or absolute value) of z, denoted |z|, is defined as |z| = √(a² + b²). It represents the distance from the origin to the point (a, b) in the complex plane. The modulus is always a non‑negative real number.
复数 z 的模（或绝对值）记作 |z|，定义为 |z| = √(a² + b²)。它表示复平面上原点到点 (a, b) 的距离。模总是非负实数。
Important properties: |z|² = z z̅. For any two complex numbers, |zw| = |z| |w| and |z/w| = |z| / |w| (w ≠ 0). The triangle inequality: |z + w| ≤ |z| + |w|.
重要性质：|z|² = z z̅。对任意两个复数，有 |zw| = |z| |w|，且 |z/w| = |z| / |w|（w ≠ 0）。三角不等式：|z + w| ≤ |z| + |w|。
4. Argument and Polar Form | 辐角与极坐标形式
The argument of a non‑zero complex number z, denoted arg(z), is the angle θ measured from the positive real axis to the line segment joining the origin to the point, usually measured in radians. The principal argument Arg(z) lies in the interval (−π, π] or [0, 2π) depending on convention; IB typically uses (−π, π]. For z = 0, the argument is undefined.
非零复数 z 的辐角记作 arg(z)，是从正实轴逆时针量到原点到该点连线的角度 θ，通常以弧度为单位。主辐角 Arg(z) 的取值范围是 (−π, π] 或 [0, 2π)，具体取决于约定；IB 通常使用 (−π, π]。对于 z = 0，辐角无定义。
Using modulus r and argument θ, any non‑zero complex number can be written in polar form: z = r(cos θ + i sin θ), where r = |z| and θ = arg(z). This representation is invaluable for multiplication, division, and exponentiation.
利用模 r 和辐角 θ，任意非零复数都可以写成极坐标形式：z = r(cos θ + i sin θ)，其中 r = |z|，θ = arg(z)。这种表示在乘、除和乘方运算中极为有用。
To convert from Cartesian form: r = √(a² + b²), θ = arctan(b/a) adjusted to the correct quadrant. Care must be taken with the signs of a and b. To convert back: a = r cos θ, b = r sin θ.
从直角坐标形式转换：r = √(a² + b²)，θ = arctan(b/a) 并根据象限进行调整。必须注意 a 和 b 的符号。逆向转换：a = r cos θ, b = r sin θ。
5. Multiplication and Division in Polar Form | 极坐标形式的乘除
When multiplying two complex numbers in polar form, multiply their moduli and add their arguments: z₁ z₂ = r₁ r₂ [cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)]. Geometrically, multiplication by a complex number scales and rotates the original vector.
在极坐标形式下，两个复数相乘时，模相乘、辐角相加：z₁ z₂ = r₁ r₂ [cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)]。从几何角度看，乘上一个复数相当于对原向量进行伸缩和旋转。
Division corresponds to dividing the moduli and subtracting the arguments: z₁ / z₂ = (r₁ / r₂) [cos(θ₁ − θ₂) + i sin(θ₁ − θ₂)], provided r₂ ≠ 0. These operations become remarkably simple once the numbers are expressed in polar form.
除法对应模相除、辐角相减：z₁ / z₂ = (r₁ / r₂) [cos(θ₁ − θ₂) + i sin(θ₁ − θ₂)]，其中 r₂ ≠ 0。一旦用极坐标形式表示，这些运算便变得异常简单。
This property is the foundation for understanding De Moivre’s theorem and for finding roots of complex numbers, both of which are heavily examined in IB.
这一性质是理解棣莫弗定理和求复数根的基础，而这两部分内容在 IB 考试中占比很大。
6. 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). Combined with the polar form, it gives zⁿ = rⁿ [cos(nθ) + i sin(nθ)] for any integer n.
棣莫弗定理指出：对于任意整数 n，(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)。结合极坐标形式，可得对于任意整数 n，zⁿ = rⁿ [cos(nθ) + i sin(nθ)]。
This theorem is extremely powerful: it can be used to derive trigonometric identities (e.g., expressions for sin 3θ and cos 3θ), to find powers of complex numbers, and to solve equations. For negative integer powers, simply apply the theorem with a negative exponent, which ends up dividing the modulus and rotating in the opposite direction.
这一定理极为强大：可用于推导三角恒等式（如 sin 3θ 和 cos 3θ 的表达式）、求复数的幂以及解方程。对负整数次幂，只需应用带有负指数的定理，结果是模取倒数并向反方向旋转。
Proof for positive integers uses induction and the angle‑addition formulas. While IB does not require you to reproduce the full proof, understanding its reliance on trigonometric addition formulas is helpful.
对于正整数的证明使用归纳法和角度加法公式。虽然 IB 不要求完整写出证明过程，但理解其依赖于三角加法公式非常有帮助。
7. Roots of Complex Numbers | 复数的根
To find the nth roots of a complex number z = r(cos θ + i sin θ), we use the formula: z^(1/n) = r^(1/n) [cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)], for k = 0, 1, 2, …, n−1. This yields n distinct roots equally spaced around a circle of radius r^(1/n).
求复数 z = r(cos θ + i sin θ) 的 n 次方根，使用公式：z^(1/n) = r^(1/n) [cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)]，其中 k = 0, 1, 2, …, n−1。这给出 n 个不同的根，它们等间距地分布在半径为 r^(1/n) 的圆上。
In particular, the nth roots of unity (solutions to zⁿ = 1) are given by z = cos(2kπ/n) + i sin(2kπ/n). They lie on the unit circle at the vertices of a regular n‑gon. Their sum is zero, a useful fact for certain proof questions.
特别地，n 次单位根（方程 zⁿ = 1 的解）由 z = cos(2kπ/n) + i sin(2kπ/n) 给出。它们位于单位圆上正 n 边形的顶点。这些根之和为零，这在某些证明题中很有用。
When solving equations like z³ = i, write i in polar form: i = 1(cos(π/2) + i sin(π/2)), then apply the root formula. This systematic approach avoids mistakes.
解 z³ = i 这类方程时，先将 i 写成极坐标形式：i = 1(cos(π/2) + i sin(π/2))，然后套用求根公式。这一系统化方法可避免错误。
8. Euler’s Formula and Exponential Form | 欧拉公式与指数形式
Euler’s formula connects complex exponentials to trigonometric functions: e^(iθ) = cos θ + i sin θ. Using this, any complex number can be expressed in exponential form: z = r e^(iθ), where r = |z| and θ = arg(z).
欧拉公式将复指数与三角函数联系起来：e^(iθ) = cos θ + i sin θ。利用该公式，任意复数都可以表示为指数形式：z = r e^(iθ)，其中 r = |z|，θ = arg(z)。
In exponential form, multiplication, division, and exponentiation become straightforward applications of the laws of exponents: z₁ z₂ = r₁ r₂ e^(i(θ₁+θ₂)), z₁/z₂ = (r₁/r₂) e^(i(θ₁−θ₂)), zⁿ = rⁿ e^(i nθ). This compact notation is especially useful in HL and in topics such as differential equations.
在指数形式下，乘法、除法和幂运算可简单套用指数运算律：z₁ z₂ = r₁ r₂ e^(i(θ₁+θ₂))，z₁/z₂ = (r₁/r₂) e^(i(θ₁−θ₂))，zⁿ = rⁿ e^(i nθ)。这种紧凑记号在 HL 以及微分方程等专题中特别有用。
Euler’s identity, e^(iπ) + 1 = 0, is a beautiful consequence that often appears in enrichment contexts. It also provides an easy way to remember the polar form.
欧拉恒等式 e^(iπ) + 1 = 0 是一个优美的推论，常出现在拓展内容中。它也为记忆极坐标形式提供了简单途径。
9. Complex Numbers and Polynomials | 复数与多项式
Complex numbers are vital in the study of polynomials. The Fundamental Theorem of Algebra guarantees that every non‑constant polynomial with complex coefficients has at least one complex root. Consequently, a degree n polynomial has exactly n complex roots (counting multiplicity).
复数在多项式研究中至关重要。代数基本定理保证，任何具有复数系数的非常数多项式至少有一个复根。由此，n 次多项式恰好有 n 个复根（计重数）。
For polynomials with real coefficients, complex roots occur in conjugate pairs. That is, if a + bi is a root, then a − bi is also a root. This property is frequently used to find one root from another, or to form quadratic factors from a pair of conjugates.
对于实系数多项式，复根以共轭对的形式出现。也就是说，如果 a + bi 是一个根，那么 a − bi 也是根。这一性质常被用来由一个根求另一个根，或由一对共轭根构造二次因式。
When a polynomial evaluated at a complex number gives zero, we can use synthetic division or long division to factorise it. The complex conjugate root theorem helps reduce the degree by two at a time.
当多项式在某个复数处的值为零，我们可用综合除法或长除法进行因式分解。共轭复根定理有助于一次将次数降低两阶。
10. Geometric Interpretations and Transformations | 几何解释与变换
The complex plane (Argand diagram) allows us to visualise numbers as points or vectors. Addition of complex numbers corresponds to vector addition. Multiplication by a real number is scaling; multiplication by i is a 90° counterclockwise rotation.
复平面（阿根图）让我们能将数字可视化为点或向量。复数加法对应向量加法。乘以实数相当于伸缩；乘以 i 则是逆时针旋转 90°。
Transformations of the plane can be described using complex functions. For example, the map z ↦ z + w is a translation by w. The map z ↦ kz (k real) is a dilation by factor k. The map z ↦ e^(iθ) z is a rotation by θ about the origin. The map z ↦ z̅ is reflection in the real axis.
平面上的变换可用复变函数来描述。例如映射 z ↦ z + w 是平移 w。映射 z ↦ kz（k 为实数）是缩放变换，因子为 k。映射 z ↦ e^(iθ) z 是绕原点的旋转，角度为 θ。映射 z ↦ z̅ 是关于实轴的反射。
Understanding these geometric mappings is crucial for questions involving loci, such as |z − a| = r (circle) or |z − a| = |z − b| (perpendicular bisector). Combining transformations allows you to infer the effect of functions like f(z) = 2z + i or f(z) = i z̅.
理解这些几何映射对于处理轨迹问题至关重要，例如 |z − a| = r（圆）或 |z − a| = |z − b|（垂直平分线）。组合变换可推断如 f(z) = 2z + i 或 f(z) = i z̅ 等函数的作用效果。
11. Solving Equations with Complex Numbers | 解复数方程
IB exam questions often ask you to solve polynomial equations in ℂ. Strategies include: (1) spotting a real root, then reducing the degree; (2) using the complex conjugate root theorem; (3) equating real and imaginary parts; (4) using substitution such as w = z²; (5) applying De Moivre or the root formula for pure power equations.
IB 考试常要求解复数范围内的多项式方程。策略包括：(1) 找出一个实根，然后降低次数；(2) 使用共轭复根定理；(3) 令实部和虚部分别相等；(4) 使用如 w = z² 的换元；(5) 对纯幂次方程应用棣莫弗定理或求根公式。
For equations like (z − 2)³ = 8i, write the right side in polar form and take the cube root, remembering to add 2 to each result. For equations mixing z and its conjugate, substituting z = x + iy and z̅ = x − iy is a reliable method.
对于 (z − 2)³ = 8i 这类方程，先将右边写成极坐标形式并开三次方，注意每个结果都要加上 2。对于混合了 z 及其共轭的方程，代入 z = x + iy 和 z̅ = x − iy 是一种可靠方法。
Systems of equations with complex variables can be solved by treating them similarly to real systems, but care is needed when dividing or multiplying. Always check answers by substitution.
含有复变量的方程组可用类似实方程组的方法求解，但在乘除时需格外小心。最后务必用代入法验证答案。
12. Summary and Exam Tips | 总结与考试技巧
Mastery of complex numbers requires fluency in switching between Cartesian, polar, and exponential forms. Always choose the form that makes the operation simplest: Cartesian for addition/subtraction, polar/exponential for multiplication, division, powers, and roots.
掌握复数需要熟练在直角坐标、极坐标和指数形式之间切换。总是选择使运算最简便的形式：加减用直角坐标形式，乘除、幂和方根用极坐标/指数形式。
When sketching Argand diagrams, clearly label axes (Re and Im) and key points. For loci, identify the geometric shape described by the condition and provide a clear sketch with important features marked (centre, radius, vertices).
在绘制阿根图时，清楚标明坐标轴（Re 和 Im）和关键点。对于轨迹，要识别出条件所描述的几何形状，并画出清晰草图，标出重要特征（圆心、半径、顶点等）。
Be meticulous with the principal argument range. If a question asks for Arg(z) in (−π, π], ensure your angle falls in that interval. Subtract or add 2π if necessary. For roots, list all k values explicitly.
对主辐角范围要一丝不苟。如果题目要求 Arg(z) 在 (−π, π] 内，务必确保角度落在该区间。必要时减去或加上 2π。求根时，要明确列出所有 k 值。
Proofs involving De Moivre’s theorem often rely on induction or trigonometric identities. Practise the standard derivation of sin nθ and cos nθ as polynomials in sin θ and cos θ, a common HL question.
涉及棣莫弗定理的证明通常依赖归纳法或三角恒等式。练习从 sin θ 和 cos θ 的多项式推导 sin nθ 和 cos nθ 的标准过程，这是 HL 常见题。
Finally, never forget that complex numbers are an extension of the real numbers. All algebraic rules (commutative, associative, distributive) still apply. Keep this in mind and you will find complex numbers a logical and elegant topic.
最后，始终记住复数是实数的扩展。所有代数运算律（交换律、结合律、分配律）仍然成立。牢记这一点，你会发现复数是一个合乎逻辑且优美的话题。
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