Exercise I.2: Complex Numbers | 练习I.2：复数

📚 Exercise I.2: Complex Numbers | 练习I.2：复数

he fundamental operations, conjugates, Argand diagrams, modulus, argument, and polar representation frequently found in IB Mathematics Exercise I.2. It is designed to help you consolidate key skills and avoid common pitfalls in algebraic manipulation, geometric interpretation, and equation solving with complex numbers.

复数将实数系扩展，使之能容纳像 x² = -1 这样没有实数解的方程。这份复习指南涵盖了 IB 数学练习 I.2 中常见的基本运算、共轭、阿尔冈图、模、辐角以及极坐标表示。它旨在帮助你巩固复数代数运算、几何解释和方程求解中的关键技能，同时避免常见错误。

1. What is a Complex Number? | 什么是复数？

m(z). When b = 0, the number is purely real; when a = 0 and b ≠ 0, it is purely imaginary.

复数可表示为 z = a + b i，其中 a 和 b 是实数，i 是虚数单位且满足 i² = -1。a 称为实部，记作 Re(z)；b 称为虚部，记作 Im(z)。当 b = 0 时，该数为实数；当 a = 0 且 b ≠ 0 时，该数为纯虚数。

The set of all complex numbers is denoted by ℂ = d. This property is extremely useful when solving for unknown real and imaginary parts.

所有复数的集合用 ℂ 表示。两个复数 a + b i = c + d i 相等当且仅当 a = c 且 b = d。这一性质在求解未知的实部和虚部时极为有用。

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2. Powers of the Imaginary Unit i | 虚数单位 i 的幂次

. For any integer n, iⁿ can be reduced by dividing n by 4 and taking the remainder: iⁿ = i^{n mod 4}.

虚数单位 i 的幂次呈现循环规律：i¹ = i，i² = -1，i³ = -i，i⁴ = 1，之后每四次循环一次。对任意整数 n，可通过将 n 除以 4 取余数来化简 iⁿ：iⁿ = i^{n mod 4}。

n (mod 4)	in
0	1
1	i
2	-1
3	-i

fications like i²⁷ = i³ = -i, and helps avoid sign errors in longer algebraic expressions.

记住这个循环可以快速完成诸如 i²⁷ = i³ = -i 的化简，并有助于在较长的代数表达式中避免符号错误。

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3. Addition and Subtraction | 复数加减法

i and z₂ = c + d i, then z₁ + z₂ = (a + c) + (b + d) i and z₁ - z₂ = (a - c) + (b - d) i.

复数的加减法只需分别合并实部和虚部。若 z₁ = a + b i，z₂ = c + d i，则 z₁ + z₂ = (a + c) + (b + d) i，z₁ - z₂ = (a - c) + (b - d) i。

This operation is commutative and associative, just like vector addition in two dimensions. Geometrically, addition corresponds to the parallelogram law on the Argand diagram.

该运算满足交换律和结合律，与二维向量加法类似。在几何上，加法对应于阿尔冈图中的平行四边形法则。

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4. Multiplication of Complex Numbers | 复数乘法

hen replace i² with -1. For z₁ = a + b i and z₂ = c + d i: z₁z₂ = (ac - bd) + (ad + bc) i.

两个复数相乘时，将 i 视为变量并使用分配律展开，然后将 i² 替换为 -1。若 z₁ = a + b i，z₂ = c + d i，则有 z₁z₂ = (ac - bd) + (ad + bc) i。

Note that the product' uickly by expanding (a + b i)(c + d i) = ac + ad i + bc i + bd i², and then using i² = -1.

注意，乘积的实部为 ac - bd，虚部为 ad + bc。可通过展开 (a + b i)(c + d i) = ac + ad i + bc i + bd i²，然后利用 i² = -1 快速得到该公式。

z₁ z₂ = (a c − b d) + (a d + b c) i

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5. Complex Conjugates | 复共轭

The complex conjugate of z = a + b i is denoted by z̄ and defined as z̄ jugates are extremely useful for division and for finding the modulus of a complex number.

复数 z = a + b i 的共轭记作 z̄，定义为 z̄ = a - b i。它在复平面上关于实轴作镜像反射。共轭在复数除法和求模时极为有用。

Key properties of conjugates:
(i) (z̄)̄ = z
(ii) z + z̄ = 2 Re(z), z - z̄ = 2 i Im(z)
(iii) z z̄ = a² + b², which is always a non-negative real number.
These identities simplify many algebraic manipulations, especially when rationalising denominators.

共轭的重要性质：
(i) (z̄)̄ = z
(ii) z + z̄ = 2 Re(z)，z - z̄ = 2 i Im(z)
(iii) z z̄ = a² + b²，它总是非负实数。
这些恒等式可以简化许多代数操作，特别是在分母有理化时。

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6. Division of Complex Numbers | 复数除法

To divide two complex numbers, multiply the numerator and the denominator by the conjugate of the denominator. This turns the denominator into a real number, making the expression easy to separate into real and imaginary parts.

两个复数相除时，可将分子分母同时乘以分母的共轭复数。这将分母化为实数，从而使表达式易于拆分为实部与虚部。

For z₁ = a + b i and z₂ = c + d i (with z₂ ≠ 0), we have:
z₁ / z₂ = (a + b i)(c - d i) / (c² + d²) = [(ac + bd) + (bc - ad) i] / (c² + d²).

对于 z₁ = a + b i，z₂ = c + d i (z₂ ≠ 0)，有：
z₁ / z₂ = (a + b i)(c - d i) / (c² + d²) = [(ac + bd) + (bc - ad) i] / (c² + d²)。

z₁ / z₂ = (a c + b d) / (c² + d²) + ((b c − a d) / (c² + d²)) i

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7. Solving Quadratic Equations with Complex Roots | 解含复数根的二次方程

tive. The roots are then given by the quadratic formula and expressed as a conjugate pair.

形如 ax² + bx + c = 0 的二次方程（a, b, c 为实数且 a ≠ 0），当判别式 Δ = b² - 4ac 为负数时会有复数根。此时可利用求根公式将根表示为一对共轭复数。

x = [−b ± √b² − 4ac] / (2a) ⇒ x = (−b / 2a) ± (√4ac − b² / 2a) i

. Notice the roots are complex conjugates, which always occurs when coefficients are real.

举例来说，x² + 2 x + 5 = 0 的判别式 Δ = 4 - 20 = -16，因此解为 x = -1 + 2 i 和 x = -1 - 2 i。注意这两个根互为共轭复数，当系数为实数时，这一结论恒成立。

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8. The Argand Diagram | 阿尔冈图

is the imaginary axis. A complex number z = a + b i is plotted at the coordinates (a, b).

阿尔冈图将复数表示为平面上的点或位置向量。横轴为实轴，纵轴为虚轴。复数 z = a + b i 对应于坐标 (a, b)。

The distance from the origin to the point (a, b) is the modulus |z| = √a² + b². This is the length of the vector representing z. The modulus is always a non-negative real number and equals √z z̄.

原点到点 (a, b) 的距离为模 |z| = √a² + b²，即代表该复数的向量的长度。模总是非负实数，且等于 √z z̄。

Graphical interpretation helps visualise addition (vector sum), subtraction, and the effect of conjugation as a reflection across the real axis.

几何解释有助于直观理解加法（向量和）、减法以及共轭作为关于实轴的反射。

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9. Modulus and Argument | 模与辐角

The modulus of z = a + b i, written |z|, is defined as √a² + b². The argument of z, denoted arg(z), is the angle θ that the line from the origin to (a, b) makes with the positive real axis, measured in radians. The principal argument usually lies in (-π, π].

复数 z = a + b i 的模记为 |z|，定义为 √a² + b²。辐角记作 arg(z)，是从原点到 (a, b) 的线段与正实轴之间的夹角 θ，以弧度度量。主辐角通常落在 (-π, π] 区间内。

d then adjust the quadrant based on the signs of a and b. For example, if a < 0 and b > 0, add π to the arctan result; if a < 0 and b < 0, subtract π (or equivalently add π and then keep within the principal range).

求 arg(z) 时，先计算 θ = arctan(b / a)，再根据 a 和 b 的符号调整象限。例如，若 a < 0 且 b > 0，应加上 π；若 a < 0 且 b < 0，应减去 π（或等效地加上 π 然后调整至主值区间）。

exam questions require the modulus and the principal argument in exact forms involving π.

务必记住，0 的辐角无定义。考试中常要求以包含 π 的精确值形式给出模和主辐角。

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10. Polar Form of a Complex Number | 复数的极坐标形式

Using the modulus r = |z| ntation is particularly useful for multiplication, division, and finding powers (De Moivre's theorem at HL).

利用模 r = |z| 和辐角 θ = arg(z)，复数可写成极坐标形式：z = r (cos θ + i sin θ)。这种表示法对乘法、除法和求幂（高阶课程中的棣莫弗定理）特别有用。

To convert from rectangular form a + b i to polar form, compute r = √a² + b² = arctan(b / a) adjusted to the correct quadrant. Conversely, a = r cos θ and b = r sin θ.

从代数形式 a + b i 转换为极坐标形式时，先计算 r = √a² + b²，再求 θ = arctan(b / a) 并调整到正确象限。反之，a = r cos θ，b = r sin θ。

z = r (cos θ + i sin θ) with r = √a² + b², θ = arg(z)

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11. Multiplication and Division in Polar Form (HL) | 极坐标形式下的乘除法（高级水平）

θ₂) + i sin(θ₁ + θ₂)]. This dramatically simplifies multiplication compared to expanding rectangular forms.

两个复数以极坐标形式表示时，它们的乘积可通过将模相乘、辐角相加得到：z₁ z₂ = r₁ r₂ [cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)]。与展开代数形式相比，这极大地简化了乘法。

₁ - θ₂) + i sin(θ₁ - θ₂)], provided r₂ ≠ 0. The modulus is divided and the arguments are subtracted.

对于除法：z₁ / z₂ = (r₁ / r₂) [cos(θ₁ - θ₂) + i sin(θ₁ - θ₂)]，其中 r₂ ≠ 0。即模相除，辐角相减。

These rules lead naturally to De Moivre' s nθ + i sin nθ) for any integer n, which is essential for finding powers and roots of complex numbers at Higher Level.

这些规则自然引出棣莫弗定理：[r (cos θ + i sin θ)]ⁿ = rⁿ (cos nθ + i sin nθ) 对任意整数 n 成立。该定理在高级水平课程中是进行复数乘幂和求根的基础。

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12. Summary and Common Mistakes | 总结与常见错误

Key takeaways:
- i² = -1, not 1 or -i.
- When adding/subtracting, only like parts can be combined: real with real, imaginary with imaginary.
- The conjugate z̄ = a - b i is indispensable for division and for finding the modulus.
- A negative discriminant yields a pair of complex conjugate roots.
- The modulus is always non-negative; the argument requires careful quadrant checking.

核心要点：
- i² = -1，而非 1 或 -i。
- 加减法时只能合并同类部分：实部加实部，虚部加虚部。
- 共轭 z̄ = a - b i 在除法和求模时不可或缺。
- 判别式为负会产生一对共轭复根。
- 模恒为非负；求辐角时必须仔细判定象限。

Common mistakes include:
- Forgetting to replace i² with -1 during multiplication.
- Incorrectly distributing the negative sign when subtracting complex numbers.
- Using degrees instead of radians for the argument (IB expects radians unless stated otherwise).
- Misidentifying the principal argument, e.g., reporting a positive angle when a negative one is required because the point lies in the third or fourth quadrant.

常见错误包括：
- 在乘法运算中忘记将 i² 替换为 -1。
- 进行复数减法时错误地分配负号。
- 辐角中使用了角度制而非弧度制（IB 若无特殊说明均采用弧度）。
- 误判主辐角，例如当点位于第三或第四象限时，将本应为负的角写成了正角。

Regular practice with Exercise I.2 style problems, coupled with careful algebraic checking and Argand diagram sketching, will build both procedural fluency and conceptual understanding.

经常练习与练习 I.2 类似的题目，并辅以仔细的代数验算和阿尔冈图草图，将有助于培养熟练的计算流程和深刻的概念理解。

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