📚 Complex Numbers: Cartesian, Polar and Exponential Forms | 复数：笛卡尔形式、极形式与指数形式
Complex numbers are a central topic in IB Mathematics: Analysis and Approaches HL. They appear throughout Paper 1 and Paper 2 questions, testing both algebraic manipulation and geometric reasoning. This guide covers the HL 2 content from Cartesian arithmetic and conjugates to De Moivre's theorem, roots of unity, polynomial factors, and Argand loci, with worked examples and exam-focused advice.
复数是 IB 数学分析与进阶（AA）HL 的核心主题，常出现在试卷一和试卷二中，同时考查代数运算与几何推理。本指南涵盖 HL 2 内容：从笛卡尔形式运算、共轭，到棣莫弗定理、单位根、多项式因式与阿尔冈图轨迹，并配有例题与应试建议。
1. Why HL Extends the Number System | 为什么 HL 要扩展数系
n, because no real number squares to give a negative result. The real number line R is simply not large enough to hold the solutions of every polynomial equation.
在 SL 和更早的课程中，像 x² + 1 = 0 这样的方程没有实数解，因为没有实数的平方是负数。实数轴 R 并不足够大，无法容纳每个多项式方程的解。
nd build a larger system in which every polynomial equation has a solution. The new system is called the complex field C.
因此 HL 学生引入虚数单位 i，定义 i² = -1，并构造出一个更大的体系，使每个多项式方程都有解。这个新体系称为复数域 C。
A complex number has the form z = a + bi, where a, b ∈ both Re(z) and Im(z) are real numbers; it is the coefficient b that multiplies the unit i.
复数具有 z = a + bi 的形式，其中 a、b ∈ R 且 i² = -1。称 a 为实部 Re(z)，b 为虚部 Im(z)。要特别注意：Re(z) 与 Im(z) 都是实数，真正与单位 i 相乘的是系数 b。
th powerful tools for algebra and geometry. It is closed under addition, subtraction, multiplication, and division by any non-zero complex number.
复数集 C 包含实数作为 b = 0 的特殊情形，但它是一个更大的体系，为代数和几何提供了强大工具。C 在加法、减法、乘法以及除以任意非零复数下都封闭。
ith complex coefficients has at least one complex root, and therefore exactly n roots when counted with multiplicity. Powers of i cycle with period 4, which makes them easy to simplify:
HL 扩展数系的根本原因是代数基本定理：任何次数 n ≥ 1、系数为复数的多项式至少有一个复根，因此按重数计算恰有 n 个根。i 的幂以 4 为周期循环，因此很容易化简：
i¹ = i, i² = -1, i³ = -i, i⁴ = 1, i⁵ = i, ...
So for any integer k, i4k = 1, i4k+1 = i, i4k+2 = -1 and i4k+3 = -i. This cycle is used constantly when simplifying products and powers.
因此对任意整数 k，有 i4k = 1，i4k+1 = i，i4k+2 = -1，i4k+3 = -i。在化简乘积与幂时，这个循环会被反复使用。
2. Cartesian Form and Arithmetic | 笛卡尔形式与运算
ubtraction are performed component by component: real parts with real parts, imaginary parts with imaginary parts.
对于 z = a + bi 和 w = c + di，加法和减法按分量进行：实部与实部相加，虚部与虚部相加。
z + w = (a + c) + (b + d)i z - w = (a - c) + (b - d)i
Multiplication uses the distributive law exactly as with real brackets, and every occurrence of i² is replaced by -1. Collecting terms gives the standard product formula.
乘法完全按照实数括号的分配律展开，并把出现的每一个 i² 换成 -1，合并同类项后即得标准乘积公式。
zw = (ac - bd) + (ad + bc)i
1 - 4i, then z + w = 4 - 2i, and zw = (3 × 1 - 2 × (-4)) + (3 × (-4) + 2 × 1)i = 11 - 10i.
示例：若 z = 3 + 2i，w = 1 - 4i，则 z + w = 4 - 2i，且 zw = (3 × 1 - 2 × (-4)) + (3 × (-4) + 2 × 1)i = 11 - 10i。
ator, and convert expressions such as √-4 to 2i before further simplification, since √-4 = √4 × -1 = 2i.
最终答案务必写成标准的 a + bi 形式。避免分母中出现 i，并先把 √-4 这类表达式化为 2i 再继续化简，因为 √-4 = √4 × -1 = 2i。
Two useful checks: the sum z + w lies at the fourth vertex of the parallelogram formed by 0, z and w in the Argand diagram, and any product can be verified numerically with a calculator that supports complex arithmetic.
两个有用的检查：在阿尔冈图中，z + w 位于由 0、z、w 构成的平行四边形的第四个顶点；任何乘积都可以用支持复数运算的计算器进行数值验证。
3. Conjugates and Division | 共轭与除法
The complex conjugate of z = a + bi is z* = a - bi, sometimes written as z̄ the point z in the real axis, so the modulus is unchanged while the argument changes sign.
复数 z = a + bi 的共轭是 z* = a - bi，有时写作 z̄。几何上，取共轭相当于把点 z 关于实轴作镜像，因此模不变而辐角变号。
-negative real number, which is exactly why conjugates are used to divide complex numbers: multiplying by the conjugate removes i from the denominator.
乘积 z z* = a² + b² 总是非负实数，这正是共轭被用于复数除法的原因：乘以共轭可以把分母中的 i 消去。
ly top and bottom by (1 + i): the numerator becomes (3 + 2i)(1 + i) = 1 + 5i, and the denominator is 1² + 1² = 2, so the result is 1/2 + (5/2)i.
做除法时，将分子和分母同时乘以分母的共轭，然后化简。例如化简 (3 + 2i)/(1 - i)，把分子分母同乘 (1 + i)：分子变为 (3 + 2i)(1 + i) = 1 + 5i，分母为 1² + 1² = 2，故结果为 1/2 + (5/2)i。
Conjugation behaves well under the standard operations, and these properties are frequently used in proof-style questions:
共轭在标准运算下性质良好，这些性质在证明类题目中经常用到：
- (z + w)* = z* + w*, and (zw)* = z* w*
- (z/w)* = z*/w* for w ≠ 0
- z + z* = 2 Re(z), and z - z* = 2i Im(z)
- z z* = |z|², so z* = |z|²/z for z ≠ 0
- (z + w)* = z* + w*，且 (zw)* = z* w*
- 当 w ≠ 0 时，(z/w)* = z*/w*
- z + z* = 2 Re(z)，且 z - z* = 2i Im(z)
- z z* = |z|²，故当 z ≠ 0 时 z* = |z|²/z
A special case worth remembering: if z has real coefficients in a polynomial, then z is a root if and only if z* is a root. This is developed further in Section 10.
一个值得记住的特殊结论：若多项式的系数为实数，则 z 是根当且仅当 z* 也是根。第 10 节会进一步展开。
4. The Argand Diagram, Modulus and Argument | 阿尔冈图、模与辐角
nted in the Argand diagram by the point (a, b), or equivalently by the position vector from the origin to that point. The horizontal axis is the real axis and the vertical axis is the imaginary axis.
复数 z = a + bi 在阿尔冈图中用点 (a, b) 表示，也可用从原点指向该点的位置向量表示。水平轴是实轴，竖直轴是虚轴。
The modulus of z is its distance from the origin, and the argument of z is the angle measured anticlockwise from the positive real axis to the vector Oz.
复数 z 的模是它到原点的距离，z 的辐角是从正实轴逆时针旋转到向量 Oz 所成的角。
|z| = √a² + b² = √z z* tan(arg z) = b/a
; 2π, depending on the convention used in the question). Always check the quadrant before quoting an argument, because the inverse tangent function only returns angles in (-π/2, π/2).
主辐角满足 -π < arg z ≤ π（有时按题目约定取 0 ≤ arg z < 2π）。在写出辐角前务必先判断象限，因为反正切函数只返回 (-π/2, π/2) 内的角。
| Complex number | |z| | Principal argument |
| 1 + i | √2 | π/4 |
| -1 + i | √2 | 3π/4 |
| -1 - i | √2 | -3π/4 |
| 1 - i | √2 | -π/4 |
| 复数 | 模 |z| | 主辐角 |
| 1 + i | √2 | π/4 |
| -1 + i | √2 | 3π/4 |
| -1 - i | √2 | -3π/4 |
| 1 - i | √2 | -π/4 |
Two identities make modulus questions quick: |z w| = |z| |w| and |z|² = z z*. Also |z - w| ing z and w in the Argand diagram, a fact that underpins all the locus work in Section 11.
两个恒等式可以让模的题目变得快捷：|z w| = |z| |w| 与 |z|² = z z*。此外 |z - w| 正好是阿尔冈图中代表 z 与 w 的两点之间的距离，这一事实是第 11 节全部轨迹问题的基础。
5. Modulus-Argument (Polar) Form | 模-辐角（极）形式
es the complex number in terms of its modulus and argument, giving the polar or modulus-argument form, often abbreviated as cis θ.
令 a = r cos θ、b = r sin θ，就可以用模与辐角表示复数，即极形式（模-辐角形式），常缩写为 cis θ。
z = r(cos θ + i sin θ) = r cis θ, where r = |z| and θ = arg z
To convert from Cartesian to polar form, compute r = √a² + b² rect quadrant. To convert back, expand r cos θ and r sin θ to obtain the real and imaginary parts.
从笛卡尔形式转为极形式：先算 r = √a² + b²，再用 tan θ = b/a 并结合正确象限求出 θ
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