Unit A4: Complex
Numbers单元 A4：复数
The fourth sub-unit of Topic 1 (Number & Algebra). Master Cartesian, polar, and Euler forms; De Moivre's theorem; nth roots and roots of unity; and the conjugate root theorem for real polynomials. HL-heavy: most of this unit does not appear on the SL syllabus.Topic 1（数与代数）的第四个子单元。掌握笛卡尔形式（Cartesian form）、极坐标形式（polar form）与欧拉形式（Euler form）；棣莫弗定理（De Moivre's theorem）；n 次根与单位根（roots of unity）；以及实系数多项式的共轭根定理（conjugate root theorem）。本单元偏 HL：绝大部分内容不在 SL 大纲中。
How to use this guide本指南使用说明
This guide is built to serve two students at once:本指南同时为两类学生设计：
Read only the dashed-gold "Cram-Mode Cheat" box at the top of each section, plus the formula boxes. Skim one worked example per section. Skip the ▸ expandable details. Take the practice quiz. That's a 35-minute pass.
只看每节顶端的金色虚线 "Cram-Mode Cheat" 速记框和公式框（formula box）。每节挑一道 worked example（例题）扫一眼，跳过 ▸ 折叠的"深入"部分，做一次练习测验。约 35 分钟过完一遍。
Read straight through. Open every ▸ Going deeper details block — that's where the De Moivre induction proof, the convergence of roots-of-unity sums, and the fundamental theorem of algebra context live. Re-do the quiz with no hints. Owning the geometry of multiplication in the Argand plane is what separates a 5 from a 7.
逐字读完。把每个 ▸ Going deeper（深入）折叠块都打开——棣莫弗定理（De Moivre's theorem）的归纳法证明、单位根求和的收敛、以及代数基本定理（fundamental theorem of algebra）的背景都在那里。再做一次测验，不看提示。真正吃透阿尔冈平面（Argand plane）上"乘法 = 旋转 + 缩放"的几何意义，就是 5 分与 7 分的分水岭。
complex number）。单元内标有 HL+ 的提示框表示更深层的内容（由棣莫弗定理推导的多倍角公式、代数基本定理），它们在 Paper 1 上较少出现，但在 Paper 2 / Paper 3 上很常见。
conjugate）做除法、"漂亮"辐角（$\pi/6, \pi/4, \pi/3, \pi/2$）下的极坐标形式、小整数指数下的棣莫弗定理。Paper 2（可用计算器）允许使用 GDC 的复数模式，但你仍要把题目以正确的形式列出来。永远要核对 GDC 的辐角约定（大多返回 $(-\pi, \pi]$ 内的值）。
Cartesian Form笛卡尔形式 AHL 1.12
imaginary unit）$i$。每个复数（complex number）都有唯一的笛卡尔形式（Cartesian form）$z = a + bi$，其中 $a, b \in \mathbb{R}$。加法按分量相加，乘法用 FOIL 展开后把 $i^2 = -1$ 收掉；除法则把分子分母同乘共轭（conjugate）$\bar{z} = a - bi$。关键恒等式是
$$ z\bar{z} = a^2 + b^2 = |z|^2. $$
凭这一条就能一步有理化任何分母。
imaginary unit）是满足 $i^2 = -1$ 的符号 $i$。复数（complex number）是形如 $z = a + bi$ 的表达式，其中 $a, b \in \mathbb{R}$。称 $a = \operatorname{Re}(z)$ 为实部，$b = \operatorname{Im}(z)$ 为虚部（注意：$\operatorname{Im}(z)$ 本身是一个实数——即 $i$ 的系数，而不是 $bi$）。
Equality and arithmetic相等与四则运算
(2) Addition. $(a + bi) + (c + di) = (a + c) + (b + d)i$ — componentwise.
(3) Multiplication. $(a + bi)(c + di) = (ac - bd) + (ad + bc)i$ — FOIL, then collapse $i^2 = -1$.
（2）加法。$(a + bi) + (c + di) = (a + c) + (b + d)i$——按分量相加。
（3）乘法。$(a + bi)(c + di) = (ac - bd) + (ad + bc)i$——FOIL 展开后把 $i^2 = -1$ 收掉。
Cartesian form）。这是 $\dfrac{1}{\sqrt{2} + 1}$ 用 $\sqrt{2} - 1$ 有理化在复数下的对应。
Worked Example — Arithmetic in Cartesian form例题——笛卡尔形式下的运算
Problem: Let $z_1 = 3 + 2i$ and $z_2 = 1 - 4i$. Compute (a) $z_1 + z_2$, (b) $z_1 z_2$, (c) $\dfrac{z_1}{z_2}$, (d) $|z_1|$.
(a) Sum. $z_1 + z_2 = (3 + 1) + (2 - 4)i = 4 - 2i$.
(b) Product.
$$ z_1 z_2 = (3 + 2i)(1 - 4i) = 3 - 12i + 2i - 8i^2 = 3 - 10i + 8 = 11 - 10i. $$(c) Quotient. Multiply top and bottom by $\bar{z}_2 = 1 + 4i$:
$$ \frac{z_1}{z_2} = \frac{(3 + 2i)(1 + 4i)}{(1 - 4i)(1 + 4i)} = \frac{3 + 12i + 2i + 8i^2}{1 + 16} = \frac{-5 + 14i}{17} = -\tfrac{5}{17} + \tfrac{14}{17} i. $$(d) Modulus. $|z_1| = \sqrt{3^2 + 2^2} = \sqrt{13}$.
题目：设 $z_1 = 3 + 2i$，$z_2 = 1 - 4i$。计算 (a) $z_1 + z_2$，(b) $z_1 z_2$，(c) $\dfrac{z_1}{z_2}$，(d) $|z_1|$。
(a) 和。$z_1 + z_2 = (3 + 1) + (2 - 4)i = 4 - 2i$。
(b) 积。
$$ z_1 z_2 = (3 + 2i)(1 - 4i) = 3 - 12i + 2i - 8i^2 = 3 - 10i + 8 = 11 - 10i. $$(c) 商。分子分母同乘 $\bar{z}_2 = 1 + 4i$：
$$ \frac{z_1}{z_2} = \frac{(3 + 2i)(1 + 4i)}{(1 - 4i)(1 + 4i)} = \frac{3 + 12i + 2i + 8i^2}{1 + 16} = \frac{-5 + 14i}{17} = -\tfrac{5}{17} + \tfrac{14}{17} i. $$(d) 模（modulus）。$|z_1| = \sqrt{3^2 + 2^2} = \sqrt{13}$。
Conjugate identities共轭恒等式
conjugate）保持 $+$ 与 $\times$"——所以对任何实系数多项式 $p$，都有 $\overline{p(z)} = p(\bar z)$。这正是 §A4.6 中共轭根定理（conjugate root theorem）的发动机。
▸ Going deeper — Why "$\mathbb{C}$ is a field" matters▸ 深入——"$\mathbb{C}$ 是一个域"为什么重要
The point of insisting that every non-zero complex number has a multiplicative inverse is that $\mathbb{C}$ is closed under division: every quotient lives in $\mathbb{C}$ again. The formula is
$$ \frac{1}{a + bi} = \frac{a - bi}{a^2 + b^2}, \qquad (a, b) \ne (0, 0). $$This guarantees that all the algebra you know — polynomial long division, partial fractions, factoring — works over $\mathbb{C}$ exactly as it does over $\mathbb{R}$. The bonus is that $\mathbb{C}$ is also algebraically closed (every non-constant polynomial has a root) — that's the fundamental theorem of algebra. We'll see one consequence in §A4.6: a real polynomial of degree $n$ has exactly $n$ roots when complex roots are allowed.
One thing $\mathbb{C}$ loses compared to $\mathbb{R}$: there is no consistent ordering "$z < w$" that respects arithmetic. So inequalities only ever apply to real quantities — modulus, real part, imaginary part — not to complex numbers themselves.
之所以坚持每个非零复数都有乘法逆元，是因为 $\mathbb{C}$ 对除法封闭：任何商仍属于 $\mathbb{C}$。具体公式是
$$ \frac{1}{a + bi} = \frac{a - bi}{a^2 + b^2}, \qquad (a, b) \ne (0, 0). $$这保证了你熟悉的所有代数——多项式长除法、部分分式、因式分解——在 $\mathbb{C}$ 上的运作方式与 $\mathbb{R}$ 上完全一致。额外的好处是 $\mathbb{C}$ 还是代数封闭的（任何非常数多项式都有根）——这就是代数基本定理（fundamental theorem of algebra）。§A4.6 中会看到一个推论：$n$ 次实系数多项式在允许复根时恰好有 $n$ 个根。
$\mathbb{C}$ 相对于 $\mathbb{R}$ 也失去了一样东西：不存在与算术兼容的统一序"$z < w$"。所以不等式只能用于实量——模、实部、虚部——而不能用于复数本身。
Worked Example — Solving $z^2 = w$ in Cartesian form例题——在笛卡尔形式下解 $z^2 = w$
Problem: Find all $z \in \mathbb{C}$ such that $z^2 = 3 + 4i$.
Set up. Let $z = a + bi$. Then $z^2 = (a^2 - b^2) + 2abi$. Equate real and imaginary parts:
$$ a^2 - b^2 = 3, \qquad 2ab = 4 \;\Rightarrow\; ab = 2. $$Solve. From $ab = 2$: $b = 2/a$. Substitute:
$$ a^2 - \tfrac{4}{a^2} = 3 \;\Rightarrow\; a^4 - 3a^2 - 4 = 0 \;\Rightarrow\; (a^2 - 4)(a^2 + 1) = 0. $$Real solutions: $a^2 = 4$, so $a = \pm 2$. Then $b = 2/a$, giving $z = 2 + i$ or $z = -2 - i$.
Check. $(2 + i)^2 = 4 + 4i + i^2 = 3 + 4i$ ✓. Polar form (next section) gives the same answer in one line.
题目：求所有满足 $z^2 = 3 + 4i$ 的 $z \in \mathbb{C}$。
设元。设 $z = a + bi$。则 $z^2 = (a^2 - b^2) + 2abi$。让实部、虚部分别相等：
$$ a^2 - b^2 = 3, \qquad 2ab = 4 \;\Rightarrow\; ab = 2. $$求解。由 $ab = 2$ 得 $b = 2/a$。代入：
$$ a^2 - \tfrac{4}{a^2} = 3 \;\Rightarrow\; a^4 - 3a^2 - 4 = 0 \;\Rightarrow\; (a^2 - 4)(a^2 + 1) = 0. $$实数解：$a^2 = 4$，所以 $a = \pm 2$，再由 $b = 2/a$ 得 $z = 2 + i$ 或 $z = -2 - i$。
验证。$(2 + i)^2 = 4 + 4i + i^2 = 3 + 4i$ ✓。下一节的极坐标形式一行就能给出同样答案。
Argand Diagram & Polar Form阿尔冈图与极坐标形式 AHL 1.13
polar form / modulus-argument form）为
$$ z = r(\cos\theta + i\sin\theta) = r\operatorname{cis}\theta, \qquad r = |z|,\; \theta = \arg(z). $$
互换公式：$a = r\cos\theta$，$b = r\sin\theta$，$r = \sqrt{a^2 + b^2}$，$\tan\theta = b/a$（必须留意象限）。
polar form）才能看清。
Argand diagram）：$z = a + bi$ 位于 $(a, b)$，模为 $r$，辐角为 $\theta$。| Quadrant | Sign of $a, b$ | $\arg(z)$ |
|---|---|---|
| I | $a > 0, b \ge 0$ | $\arctan(b/a) \in [0, \pi/2)$ |
| II | $a < 0, b > 0$ | $\arctan(b/a) + \pi \in (\pi/2, \pi)$ |
| III | $a < 0, b < 0$ | $\arctan(b/a) - \pi \in (-\pi, -\pi/2)$ |
| IV | $a > 0, b < 0$ | $\arctan(b/a) \in (-\pi/2, 0)$ |
principal argument）$\arg(z)$ 通常取在 $(-\pi, \pi]$（IB 约定；有些书用 $[0, 2\pi)$）。捷径 $\theta = \arctan(b/a)$ 只在第 I、IV 象限（$a > 0$）成立。在第 II、III 象限要加或减 $\pi$：
| 象限 | $a, b$ 的符号 | $\arg(z)$ |
|---|---|---|
| I | $a > 0, b \ge 0$ | $\arctan(b/a) \in [0, \pi/2)$ |
| II | $a < 0, b > 0$ | $\arctan(b/a) + \pi \in (\pi/2, \pi)$ |
| III | $a < 0, b < 0$ | $\arctan(b/a) - \pi \in (-\pi, -\pi/2)$ |
| IV | $a > 0, b < 0$ | $\arctan(b/a) \in (-\pi/2, 0)$ |
Worked Example — Cartesian to polar例题——笛卡尔形式转极坐标形式
Problem: Express $z = -1 + i\sqrt{3}$ in polar form with $\theta \in (-\pi, \pi]$.
Modulus. $|z| = \sqrt{(-1)^2 + (\sqrt 3)^2} = \sqrt{1 + 3} = 2$.
Argument. $a = -1 < 0$, $b = \sqrt{3} > 0$ — quadrant II. Reference angle $\arctan(\sqrt{3}/1) = \pi/3$, so
$$ \theta = \pi - \pi/3 = \tfrac{2\pi}{3}. $$Polar form. $z = 2\bigl(\cos\tfrac{2\pi}{3} + i\sin\tfrac{2\pi}{3}\bigr) = 2\operatorname{cis}\tfrac{2\pi}{3}$.
Sanity check. $2\cos(2\pi/3) = 2(-\tfrac{1}{2}) = -1$ ✓, $2\sin(2\pi/3) = 2(\tfrac{\sqrt 3}{2}) = \sqrt 3$ ✓.
题目：将 $z = -1 + i\sqrt{3}$ 写成极坐标形式，$\theta \in (-\pi, \pi]$。
模。$|z| = \sqrt{(-1)^2 + (\sqrt 3)^2} = \sqrt{1 + 3} = 2$。
辐角。$a = -1 < 0$，$b = \sqrt{3} > 0$——第 II 象限。参考角 $\arctan(\sqrt{3}/1) = \pi/3$，所以
$$ \theta = \pi - \pi/3 = \tfrac{2\pi}{3}. $$极坐标形式。$z = 2\bigl(\cos\tfrac{2\pi}{3} + i\sin\tfrac{2\pi}{3}\bigr) = 2\operatorname{cis}\tfrac{2\pi}{3}$。
验证。$2\cos(2\pi/3) = 2(-\tfrac{1}{2}) = -1$ ✓，$2\sin(2\pi/3) = 2(\tfrac{\sqrt 3}{2}) = \sqrt 3$ ✓。
Worked Example — Polar to Cartesian例题——极坐标形式转笛卡尔形式
Problem: Write $z = 4\operatorname{cis}(-\tfrac{\pi}{6})$ in Cartesian form.
$$ z = 4\cos(-\tfrac{\pi}{6}) + 4i\sin(-\tfrac{\pi}{6}) = 4 \cdot \tfrac{\sqrt 3}{2} - 4i \cdot \tfrac{1}{2} = 2\sqrt 3 - 2i. $$题目：将 $z = 4\operatorname{cis}(-\tfrac{\pi}{6})$ 写成笛卡尔形式。
$$ z = 4\cos(-\tfrac{\pi}{6}) + 4i\sin(-\tfrac{\pi}{6}) = 4 \cdot \tfrac{\sqrt 3}{2} - 4i \cdot \tfrac{1}{2} = 2\sqrt 3 - 2i. $$▸ Going deeper — Why "$\tan\theta = b/a$" loses two quadrants▸ 深入——为什么"$\tan\theta = b/a$"会漏掉两个象限
The function $\arctan$ has range $(-\pi/2, \pi/2)$ — it can only return arguments in quadrants I and IV. But the four pairs $(a, b)$ and $(-a, -b)$ have the same ratio $b/a$, so $\arctan$ can't distinguish them. The fix is to track the sign of $a$ separately:
$$ \arg(z) = \begin{cases} \arctan(b/a) & a > 0 \\ \arctan(b/a) + \pi & a < 0, b \ge 0 \\ \arctan(b/a) - \pi & a < 0, b < 0 \\ \pi/2 & a = 0, b > 0 \\ -\pi/2 & a = 0, b < 0 \end{cases} $$Programmers will recognize this as the atan2(b, a) function. The IB markscheme expects you to state which quadrant the point lies in before writing the argument — points are lost for naked $\arctan(b/a)$ when the answer should have been in quadrant II or III.
函数 $\arctan$ 的值域是 $(-\pi/2, \pi/2)$——只能返回第 I、IV 象限的辐角。但是 $(a, b)$ 与 $(-a, -b)$ 这两对的比值 $b/a$ 相同，所以 $\arctan$ 无法区分它们。解决办法是单独追踪 $a$ 的符号：
$$ \arg(z) = \begin{cases} \arctan(b/a) & a > 0 \\ \arctan(b/a) + \pi & a < 0, b \ge 0 \\ \arctan(b/a) - \pi & a < 0, b < 0 \\ \pi/2 & a = 0, b > 0 \\ -\pi/2 & a = 0, b < 0 \end{cases} $$程序员会一眼认出这就是 atan2(b, a) 函数。IB 评分要求你在写下辐角之前要明确指出点位于哪个象限——若答案本应落在第 II 或第 III 象限却只写了裸的 $\arctan(b/a)$，会被扣分。
principal argument）是：Euler Form & Multiplication欧拉形式与乘法 HL AHL 1.13
Euler's formula）：$e^{i\theta} = \cos\theta + i\sin\theta$。因此每个复数都可以写成欧拉形式（Euler form）：
$$ z = r e^{i\theta}, \qquad r \ge 0,\;\theta \in \mathbb{R}. $$
在欧拉形式下，乘法即旋转 + 伸缩：
$$ z_1 z_2 = r_1 r_2 \, e^{i(\theta_1 + \theta_2)}, \qquad \frac{z_1}{z_2} = \frac{r_1}{r_2} e^{i(\theta_1 - \theta_2)}. $$
乘以 $e^{i\alpha}$ 即旋转 $\alpha$；乘以实数 $r > 0$ 即按 $r$ 倍伸缩。这一张图就占了本单元的一半内容。
| Form | Expression | Best for |
|---|---|---|
| Cartesian | $z = a + bi$ | Adding, subtracting, equating to a real polynomial. |
| Polar (cis) | $z = r\operatorname{cis}\theta$ | Reading off $r$ and $\theta$ by inspection. |
| Euler | $z = re^{i\theta}$ | Multiplying, dividing, raising to powers, taking roots. |
| 形式 | 表达式 | 最适合 |
|---|---|---|
笛卡尔（Cartesian） | $z = a + bi$ | 加法、减法，与实多项式比较系数。 |
极坐标 cis（polar） | $z = r\operatorname{cis}\theta$ | 直接读出 $r$ 与 $\theta$。 |
欧拉（Euler） | $z = re^{i\theta}$ | 乘法、除法、乘方、开根。 |
"Moduli multiply, arguments add.""模相乘，辐角相加。"
conjugate）即关于实轴对称（辐角取反）；倒数则既对称又按 $1/r$ 缩放。若 $|z| = 1$，则 $1/z = \bar z$——这是单位模问题中的常用结论。
Worked Example — Multiplication via polar form例题——用极坐标形式相乘
Problem: Let $z_1 = \sqrt 2 \,\operatorname{cis}\tfrac{\pi}{4}$ and $z_2 = 3\operatorname{cis}\tfrac{\pi}{6}$. Find $z_1 z_2$ in Cartesian form.
Multiply in polar. $r_1 r_2 = 3\sqrt 2$, $\theta_1 + \theta_2 = \tfrac{\pi}{4} + \tfrac{\pi}{6} = \tfrac{5\pi}{12}$. So
$$ z_1 z_2 = 3\sqrt 2 \,\operatorname{cis}\tfrac{5\pi}{12}. $$Convert. Use the exact values $\cos\tfrac{5\pi}{12} = \tfrac{\sqrt 6 - \sqrt 2}{4}$, $\sin\tfrac{5\pi}{12} = \tfrac{\sqrt 6 + \sqrt 2}{4}$. Then
$$ z_1 z_2 = 3\sqrt 2 \cdot \tfrac{\sqrt 6 - \sqrt 2}{4} + 3\sqrt 2 i \cdot \tfrac{\sqrt 6 + \sqrt 2}{4} = \tfrac{3(\sqrt{12} - 2)}{4} + i\tfrac{3(\sqrt{12} + 2)}{4} = \tfrac{6\sqrt 3 - 6}{4} + i\tfrac{6\sqrt 3 + 6}{4}. $$Simplify: $z_1 z_2 = \tfrac{3}{2}(\sqrt 3 - 1) + \tfrac{3}{2}(\sqrt 3 + 1) i$. (You can double-check by multiplying the Cartesian forms $z_1 = 1 + i$ and $z_2 = \tfrac{3\sqrt 3}{2} + \tfrac{3}{2} i$ — same answer.)
题目：设 $z_1 = \sqrt 2 \,\operatorname{cis}\tfrac{\pi}{4}$，$z_2 = 3\operatorname{cis}\tfrac{\pi}{6}$。求 $z_1 z_2$ 的笛卡尔形式。
在极坐标中相乘。$r_1 r_2 = 3\sqrt 2$，$\theta_1 + \theta_2 = \tfrac{\pi}{4} + \tfrac{\pi}{6} = \tfrac{5\pi}{12}$。故
$$ z_1 z_2 = 3\sqrt 2 \,\operatorname{cis}\tfrac{5\pi}{12}. $$转换为笛卡尔形式。使用精确值 $\cos\tfrac{5\pi}{12} = \tfrac{\sqrt 6 - \sqrt 2}{4}$，$\sin\tfrac{5\pi}{12} = \tfrac{\sqrt 6 + \sqrt 2}{4}$：
$$ z_1 z_2 = 3\sqrt 2 \cdot \tfrac{\sqrt 6 - \sqrt 2}{4} + 3\sqrt 2 i \cdot \tfrac{\sqrt 6 + \sqrt 2}{4} = \tfrac{3(\sqrt{12} - 2)}{4} + i\tfrac{3(\sqrt{12} + 2)}{4} = \tfrac{6\sqrt 3 - 6}{4} + i\tfrac{6\sqrt 3 + 6}{4}. $$化简：$z_1 z_2 = \tfrac{3}{2}(\sqrt 3 - 1) + \tfrac{3}{2}(\sqrt 3 + 1) i$。（也可以把笛卡尔形式 $z_1 = 1 + i$ 与 $z_2 = \tfrac{3\sqrt 3}{2} + \tfrac{3}{2} i$ 直接相乘验证——答案相同。）
Worked Example — Quotient as rotation例题——商即旋转
Problem: Let $z = 4 e^{i\,\pi/3}$ and $w = 2 e^{i\,\pi/6}$. Find $z/w$ and $w/z$.
$$ \frac{z}{w} = \frac{4}{2} e^{i(\pi/3 - \pi/6)} = 2 e^{i\,\pi/6}, \qquad \frac{w}{z} = \tfrac{1}{2} e^{i(\pi/6 - \pi/3)} = \tfrac{1}{2} e^{-i\,\pi/6}. $$Note that $w/z = 1/(z/w)$, just as you'd hope: moduli reciprocate, arguments negate.
题目：设 $z = 4 e^{i\,\pi/3}$，$w = 2 e^{i\,\pi/6}$。求 $z/w$ 与 $w/z$。
$$ \frac{z}{w} = \frac{4}{2} e^{i(\pi/3 - \pi/6)} = 2 e^{i\,\pi/6}, \qquad \frac{w}{z} = \tfrac{1}{2} e^{i(\pi/6 - \pi/3)} = \tfrac{1}{2} e^{-i\,\pi/6}. $$注意 $w/z = 1/(z/w)$，正如所期望的：模取倒数，辐角取反。
▸ Going deeper — Where does Euler's formula come from?▸ 深入——欧拉公式从何而来？
One clean route uses the Maclaurin (Taylor at $0$) series:
$$ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}, \qquad \cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}, \qquad \sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}. $$Plug $x = i\theta$ into the exponential series. The powers cycle: $i^0 = 1,\, i^1 = i,\, i^2 = -1,\, i^3 = -i$, and back to $1$. Splitting even-index from odd-index terms:
$$ e^{i\theta} = \underbrace{\sum_{n=0}^{\infty}\frac{(-1)^n \theta^{2n}}{(2n)!}}_{\cos\theta} + i\underbrace{\sum_{n=0}^{\infty}\frac{(-1)^n \theta^{2n+1}}{(2n+1)!}}_{\sin\theta} = \cos\theta + i\sin\theta. $$A second route (more suited to a Paper 3): show that the function $f(\theta) = (\cos\theta + i\sin\theta) \cdot e^{-i\theta}$ has derivative $0$ everywhere (using the chain rule and the trig derivatives), so $f$ is constant; check $f(0) = 1$ to conclude $f \equiv 1$. Either way, the multiplicative-additive translation $e^{i(\alpha + \beta)} = e^{i\alpha} e^{i\beta}$ is exactly the trig product-to-sum identity for cosine and sine — Euler's formula is the angle-addition formula in disguise.
One-line corollary (Euler's identity). Setting $\theta = \pi$: $e^{i\pi} + 1 = 0$. The five most important constants of mathematics in a single equation.
一种简洁的推导是利用麦克劳林（在 $0$ 处的泰勒）级数：
$$ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}, \qquad \cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}, \qquad \sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}. $$把 $x = i\theta$ 代入指数级数。$i$ 的幂呈循环：$i^0 = 1,\, i^1 = i,\, i^2 = -1,\, i^3 = -i$，再回到 $1$。将偶数项与奇数项分开：
$$ e^{i\theta} = \underbrace{\sum_{n=0}^{\infty}\frac{(-1)^n \theta^{2n}}{(2n)!}}_{\cos\theta} + i\underbrace{\sum_{n=0}^{\infty}\frac{(-1)^n \theta^{2n+1}}{(2n+1)!}}_{\sin\theta} = \cos\theta + i\sin\theta. $$第二种路径（更适合 Paper 3）：证明函数 $f(\theta) = (\cos\theta + i\sin\theta) \cdot e^{-i\theta}$ 处处导数为 $0$（用链式法则与三角函数导数），故 $f$ 为常数；再由 $f(0) = 1$ 得 $f \equiv 1$。无论哪种方式，"乘法换成加法" 的恒等式 $e^{i(\alpha + \beta)} = e^{i\alpha} e^{i\beta}$ 恰好就是余弦与正弦的积化和差公式——欧拉公式本质上就是角度加法公式的另一种伪装。
一行推论（欧拉恒等式 Euler's identity）。取 $\theta = \pi$：$e^{i\pi} + 1 = 0$。数学中最重要的五个常数在一个方程里相遇。
De Moivre's Theorem棣莫弗定理 HL AHL 1.14
De Moivre's theorem）——对所有整数 $n$：
$$ \bigl(\cos\theta + i\sin\theta\bigr)^n = \cos(n\theta) + i\sin(n\theta), $$
等价地 $\bigl(re^{i\theta}\bigr)^n = r^n e^{i\,n\theta}$。乘 $n$ 次方时辐角乘以 $n$、模取 $n$ 次方。两大用途：一行算出 $(1 + i)^{10}$；推导 $\cos(n\theta)$ 与 $\sin(n\theta)$ 的多倍角恒等式。
▸ Going deeper — Proof of De Moivre by induction▸ 深入——用数学归纳法证明棣莫弗定理 HL+
Statement. For every positive integer $n$, $\bigl(\cos\theta + i\sin\theta\bigr)^n = \cos(n\theta) + i\sin(n\theta)$.
Base case ($n = 1$). Both sides equal $\cos\theta + i\sin\theta$. ✓
Inductive step. Assume the identity holds for $n = k$. Then
$$ \begin{aligned} \bigl(\cos\theta + i\sin\theta\bigr)^{k+1} &= \bigl(\cos\theta + i\sin\theta\bigr)^k \cdot \bigl(\cos\theta + i\sin\theta\bigr) \\ &\stackrel{(IH)}{=} \bigl(\cos(k\theta) + i\sin(k\theta)\bigr) \cdot \bigl(\cos\theta + i\sin\theta\bigr) \\ &= \cos(k\theta)\cos\theta - \sin(k\theta)\sin\theta + i\bigl(\sin(k\theta)\cos\theta + \cos(k\theta)\sin\theta\bigr) \\ &= \cos\bigl((k+1)\theta\bigr) + i\sin\bigl((k+1)\theta\bigr), \end{aligned} $$using the angle-addition formulas for cosine and sine in the last step. By induction the identity holds for all $n \ge 1$. (For $n = 0$ both sides equal $1$; for $n < 0$, write $\bigl(\cos\theta + i\sin\theta\bigr)^{-1} = \cos\theta - i\sin\theta = \cos(-\theta) + i\sin(-\theta)$ and apply the positive case.)
Why the proof "feels right." In Euler form the statement is $\bigl(e^{i\theta}\bigr)^n = e^{i\,n\theta}$, which is just the law of exponents — once you accept Euler's formula, De Moivre is one line. The induction proof is what you write when the question asks you to prove it without assuming Euler.
命题。对每个正整数 $n$，$\bigl(\cos\theta + i\sin\theta\bigr)^n = \cos(n\theta) + i\sin(n\theta)$。
基础步骤（$n = 1$）。两边都等于 $\cos\theta + i\sin\theta$。 ✓
归纳步骤。假设当 $n = k$ 时恒等式成立。则
$$ \begin{aligned} \bigl(\cos\theta + i\sin\theta\bigr)^{k+1} &= \bigl(\cos\theta + i\sin\theta\bigr)^k \cdot \bigl(\cos\theta + i\sin\theta\bigr) \\ &\stackrel{(IH)}{=} \bigl(\cos(k\theta) + i\sin(k\theta)\bigr) \cdot \bigl(\cos\theta + i\sin\theta\bigr) \\ &= \cos(k\theta)\cos\theta - \sin(k\theta)\sin\theta + i\bigl(\sin(k\theta)\cos\theta + \cos(k\theta)\sin\theta\bigr) \\ &= \cos\bigl((k+1)\theta\bigr) + i\sin\bigl((k+1)\theta\bigr), \end{aligned} $$最后一步用了余弦与正弦的角度加法公式。由归纳原理，恒等式对所有 $n \ge 1$ 成立。（$n = 0$ 时两边均为 $1$；$n < 0$ 时，写 $\bigl(\cos\theta + i\sin\theta\bigr)^{-1} = \cos\theta - i\sin\theta = \cos(-\theta) + i\sin(-\theta)$，再套用正整数情形。）
为什么这个证明"自然"。在欧拉形式下，命题就是 $\bigl(e^{i\theta}\bigr)^n = e^{i\,n\theta}$，这只不过是指数律——一旦承认欧拉公式，棣莫弗定理就只剩一行。归纳法证明是不假设欧拉公式时该写的版本。
Worked Example — Compute $(1 + i)^{10}$例题——计算 $(1 + i)^{10}$
Problem: Find $(1 + i)^{10}$ in Cartesian form.
Polar form. $|1 + i| = \sqrt 2$, $\arg(1+i) = \pi/4$, so $1 + i = \sqrt 2 \, e^{i\pi/4}$.
Apply De Moivre.
$$ (1 + i)^{10} = \bigl(\sqrt 2\bigr)^{10} e^{i\cdot 10\pi/4} = 32 \, e^{i\,5\pi/2}. $$Reduce the argument. $5\pi/2 = 2\pi + \pi/2$, so $e^{i\,5\pi/2} = e^{i\pi/2} = i$.
Answer. $(1 + i)^{10} = 32i$.
Try this with the binomial theorem and you'll see why De Moivre is worth the polar conversion.
题目：求 $(1 + i)^{10}$ 的笛卡尔形式。
极坐标形式。$|1 + i| = \sqrt 2$，$\arg(1+i) = \pi/4$，故 $1 + i = \sqrt 2 \, e^{i\pi/4}$。
应用棣莫弗定理（De Moivre's theorem）。
化简辐角。$5\pi/2 = 2\pi + \pi/2$，所以 $e^{i\,5\pi/2} = e^{i\pi/2} = i$。
答案。$(1 + i)^{10} = 32i$。
试着用二项式定理算一遍，就明白为何"换到极坐标再算"是值得的。
Worked Example — Triple-angle formula例题——三倍角公式 HL+
Problem: Use De Moivre to derive $\cos(3\theta)$ and $\sin(3\theta)$ in terms of $\cos\theta$ and $\sin\theta$.
Expand both sides of De Moivre with $n = 3$. Left side: by the binomial theorem,
$$ (\cos\theta + i\sin\theta)^3 = \cos^3\theta + 3i\cos^2\theta\sin\theta + 3i^2\cos\theta\sin^2\theta + i^3\sin^3\theta. $$Using $i^2 = -1$, $i^3 = -i$:
$$ = \bigl(\cos^3\theta - 3\cos\theta\sin^2\theta\bigr) + i\bigl(3\cos^2\theta\sin\theta - \sin^3\theta\bigr). $$Right side is $\cos(3\theta) + i\sin(3\theta)$. Match real and imaginary parts:
$$ \cos(3\theta) = \cos^3\theta - 3\cos\theta\sin^2\theta, \qquad \sin(3\theta) = 3\cos^2\theta\sin\theta - \sin^3\theta. $$Using $\sin^2\theta = 1 - \cos^2\theta$, the first identity becomes $\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$. Likewise, $\sin(3\theta) = 3\sin\theta - 4\sin^3\theta$. Two classical identities falling out of De Moivre in three lines.
题目：用棣莫弗定理推出 $\cos(3\theta)$ 与 $\sin(3\theta)$ 用 $\cos\theta$、$\sin\theta$ 表示的公式。
对 $n = 3$ 展开两边。左边——由二项式定理（binomial theorem）：
代入 $i^2 = -1$、$i^3 = -i$：
$$ = \bigl(\cos^3\theta - 3\cos\theta\sin^2\theta\bigr) + i\bigl(3\cos^2\theta\sin\theta - \sin^3\theta\bigr). $$右边为 $\cos(3\theta) + i\sin(3\theta)$。比较实部与虚部：
$$ \cos(3\theta) = \cos^3\theta - 3\cos\theta\sin^2\theta, \qquad \sin(3\theta) = 3\cos^2\theta\sin\theta - \sin^3\theta. $$由 $\sin^2\theta = 1 - \cos^2\theta$，第一个恒等式化为 $\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$；同理 $\sin(3\theta) = 3\sin\theta - 4\sin^3\theta$。两条经典恒等式三行从棣莫弗定理推出。
Roots of Complex Numbers复数的根 HL AHL 1.14
n-th roots）为
$$ w_k = r^{1/n} \, e^{i(\theta + 2\pi k)/n}, \qquad k = 0, 1, 2, \ldots, n - 1. $$
它们位于半径为 $r^{1/n}$ 的圆上，按 $2\pi/n$ 等间距分布。$n$ 次单位根（roots of unity）（即 $z = 1$ 的情形）为 $e^{2\pi i k/n}$，$k = 0, \ldots, n-1$——它们构成单位圆上的正 $n$ 边形。
$n$ distinct roots, equally spaced by $2\pi/n$, all of modulus $r^{1/n}$.$n$ 个不同的根，按 $2\pi/n$ 等间距分布，模都等于 $r^{1/n}$。
Worked Example — Fourth roots of $-16$例题——$-16$ 的四次方根
Problem: Find all $w \in \mathbb{C}$ with $w^4 = -16$.
Polar form of $-16$. Modulus $16$, argument $\pi$, so $-16 = 16 e^{i\pi}$.
Apply the roots formula with $n = 4$.
$$ w_k = 16^{1/4} \, e^{i(\pi + 2\pi k)/4} = 2 \, e^{i(\pi/4 + \pi k/2)}, \quad k = 0, 1, 2, 3. $$List.
| $k$ | Argument | Cartesian |
|---|---|---|
| $0$ | $\pi/4$ | $\sqrt 2 + i\sqrt 2$ |
| $1$ | $3\pi/4$ | $-\sqrt 2 + i\sqrt 2$ |
| $2$ | $5\pi/4 \;(\equiv -3\pi/4)$ | $-\sqrt 2 - i\sqrt 2$ |
| $3$ | $7\pi/4 \;(\equiv -\pi/4)$ | $\sqrt 2 - i\sqrt 2$ |
The four roots form a square of "radius" $2$ centred at the origin, rotated $45°$ from the axes. ✓
题目：求所有 $w \in \mathbb{C}$ 使 $w^4 = -16$。
$-16$ 的极坐标形式。模为 $16$，辐角为 $\pi$，故 $-16 = 16 e^{i\pi}$。
取 $n = 4$ 应用 $n$ 次根公式。
$$ w_k = 16^{1/4} \, e^{i(\pi + 2\pi k)/4} = 2 \, e^{i(\pi/4 + \pi k/2)}, \quad k = 0, 1, 2, 3. $$逐项列出。
| $k$ | 辐角 | 笛卡尔形式 |
|---|---|---|
| $0$ | $\pi/4$ | $\sqrt 2 + i\sqrt 2$ |
| $1$ | $3\pi/4$ | $-\sqrt 2 + i\sqrt 2$ |
| $2$ | $5\pi/4 \;(\equiv -3\pi/4)$ | $-\sqrt 2 - i\sqrt 2$ |
| $3$ | $7\pi/4 \;(\equiv -\pi/4)$ | $\sqrt 2 - i\sqrt 2$ |
四个根构成以原点为中心、"半径"为 $2$、相对坐标轴旋转 $45°$ 的正方形。 ✓
Roots of unity单位根
roots of unity）
$z^n = 1$ 的 $n$ 个解为
$$ \omega_k = e^{2\pi i k/n}, \qquad k = 0, 1, \ldots, n - 1. $$
记 $\omega = e^{2\pi i/n}$（即"本原根"primitive root）。则全体根为 $1, \omega, \omega^2, \ldots, \omega^{n-1}$——$\omega$ 的幂——它们在单位圆上构成正 $n$ 边形，其中一个顶点位于 $z = 1$。
▸ Going deeper — Sum of the $n$-th roots of unity is $0$ (for $n \ge 2$)▸ 深入——$n$ 次单位根之和为 $0$（$n \ge 2$） HL+
Let $\omega = e^{2\pi i/n}$. The roots of unity are $1, \omega, \omega^2, \ldots, \omega^{n-1}$. Their sum is a finite geometric series with first term $1$ and ratio $\omega$:
$$ S = 1 + \omega + \omega^2 + \cdots + \omega^{n-1} = \frac{\omega^n - 1}{\omega - 1}. $$But $\omega^n = e^{2\pi i} = 1$, so the numerator is $0$. Since $\omega \ne 1$ (we assumed $n \ge 2$), the denominator is non-zero, giving $S = 0$.
Geometric reading. The roots of unity are the vertices of a regular $n$-gon centred at the origin — by symmetry their centroid (= average position) is the centre, $0$. So their sum (= $n \times$ centroid) is also $0$. The algebraic and geometric proofs say the same thing.
Consequence. For the cube roots of unity ($n = 3$), writing $\omega = e^{2\pi i/3}$, we get $1 + \omega + \omega^2 = 0$, i.e. $\omega^2 = -1 - \omega$. This single identity is the entire content of "the cube roots of unity" on Paper 1 — memorize it.
设 $\omega = e^{2\pi i/n}$。单位根为 $1, \omega, \omega^2, \ldots, \omega^{n-1}$。它们的和是首项为 $1$、公比为 $\omega$ 的有限等比数列：
$$ S = 1 + \omega + \omega^2 + \cdots + \omega^{n-1} = \frac{\omega^n - 1}{\omega - 1}. $$但 $\omega^n = e^{2\pi i} = 1$，所以分子为 $0$。因为 $\omega \ne 1$（已假设 $n \ge 2$），分母非零，故 $S = 0$。
几何解读。单位根是以原点为中心的正 $n$ 边形的顶点——由对称性，它们的形心（即平均位置）就是中心 $0$。所以它们的和（$= n \times$ 形心）也是 $0$。代数证明与几何证明殊途同归。
推论。对三次单位根（$n = 3$），记 $\omega = e^{2\pi i/3}$，则 $1 + \omega + \omega^2 = 0$，即 $\omega^2 = -1 - \omega$。这条恒等式就是 Paper 1 上"三次单位根"题目的全部内容——务必记住。
Worked Example — Cube roots of $-8i$例题——$-8i$ 的三次方根
Problem: Find all $z \in \mathbb{C}$ with $z^3 = -8i$.
Polar of $-8i$. $|-8i| = 8$, $\arg(-8i) = -\pi/2$. So $-8i = 8 e^{-i\pi/2}$.
Apply formula with $n = 3$.
$$ z_k = 8^{1/3} e^{i(-\pi/2 + 2\pi k)/3} = 2 e^{i(-\pi/6 + 2\pi k/3)}, \quad k = 0, 1, 2. $$List the arguments. $-\pi/6$, $-\pi/6 + 2\pi/3 = \pi/2$, $-\pi/6 + 4\pi/3 = 7\pi/6 \equiv -5\pi/6$. So
$$ z_0 = 2 e^{-i\pi/6} = \sqrt 3 - i, \quad z_1 = 2 e^{i\pi/2} = 2i, \quad z_2 = 2 e^{-i\,5\pi/6} = -\sqrt 3 - i. $$Three roots on the circle $|z| = 2$, equally spaced by $2\pi/3$. ✓
题目：求所有 $z \in \mathbb{C}$ 使 $z^3 = -8i$。
$-8i$ 的极坐标。$|-8i| = 8$，$\arg(-8i) = -\pi/2$。故 $-8i = 8 e^{-i\pi/2}$。
取 $n = 3$ 套公式。
$$ z_k = 8^{1/3} e^{i(-\pi/2 + 2\pi k)/3} = 2 e^{i(-\pi/6 + 2\pi k/3)}, \quad k = 0, 1, 2. $$列出各辐角。$-\pi/6$，$-\pi/6 + 2\pi/3 = \pi/2$，$-\pi/6 + 4\pi/3 = 7\pi/6 \equiv -5\pi/6$。故
$$ z_0 = 2 e^{-i\pi/6} = \sqrt 3 - i, \quad z_1 = 2 e^{i\pi/2} = 2i, \quad z_2 = 2 e^{-i\,5\pi/6} = -\sqrt 3 - i. $$三个根都在圆 $|z| = 2$ 上，按 $2\pi/3$ 等间距分布。 ✓
Polynomials over ℂ复数域 ℂ 上的多项式 HL AHL 1.14
conjugate root theorem）。若 $p(x)$ 是实系数多项式且 $z_0$ 是它的非实根，则 $\bar z_0$ 也是根。所以实系数多项式的复根成对出现（共轭对），每对给出一个实二次因子
$$ (x - z_0)(x - \bar z_0) = x^2 - 2\operatorname{Re}(z_0)\,x + |z_0|^2. $$
因此奇次实系数多项式至少有一个实根。
fundamental theorem of algebra）说：每个 $n$ 次非常数多项式 $p(x) \in \mathbb{C}[x]$ 在 $\mathbb{C}$ 中恰好有 $n$ 个根（按重数计）。所以
$$ p(x) = a_n (x - r_1)(x - r_2)\cdots(x - r_n), $$
其中 $r_1, \ldots, r_n \in \mathbb{C}$ 是这些根（可以有重复）。在实数域上分解时，会用实二次因子代替每对共轭复根。
▸ Going deeper — Proof of the conjugate root theorem▸ 深入——共轭根定理的证明 HL+
Let $p(x) = \sum_{k=0}^{n} a_k x^k$ with every $a_k \in \mathbb{R}$, and suppose $p(z_0) = 0$. Apply conjugation to both sides:
$$ 0 = \overline{0} = \overline{p(z_0)} = \overline{\sum_{k=0}^{n} a_k z_0^k} = \sum_{k=0}^{n} \overline{a_k z_0^k} = \sum_{k=0}^{n} \bar a_k \, \bar z_0^{\,k} = \sum_{k=0}^{n} a_k \, \bar z_0^{\,k} = p(\bar z_0), $$where we used (i) $\overline{\sum} = \sum \overline{(\cdot)}$ (conjugation respects addition), (ii) $\overline{ab} = \bar a\, \bar b$ (respects multiplication, hence powers), and (iii) $\bar a_k = a_k$ because $a_k$ is real. So $\bar z_0$ is also a root. ∎
Why the "real coefficients" assumption matters. If $p(x) = x - i$, the only root is $z_0 = i$ and its conjugate $\bar z_0 = -i$ is not a root: $p(-i) = -2i \ne 0$. The conjugate root theorem fails the moment a single coefficient leaves $\mathbb{R}$.
Consequence — quadratic factor. If $z_0 = a + bi$ ($b \ne 0$) is a root of a real polynomial, then so is $\bar z_0 = a - bi$, and
$$ (x - z_0)(x - \bar z_0) = x^2 - (z_0 + \bar z_0)x + z_0 \bar z_0 = x^2 - 2a x + (a^2 + b^2) $$is a real quadratic factor of $p(x)$.
设 $p(x) = \sum_{k=0}^{n} a_k x^k$，其中每个 $a_k \in \mathbb{R}$，又假设 $p(z_0) = 0$。两边取共轭（conjugate）：
这里用了 (i) $\overline{\sum} = \sum \overline{(\cdot)}$（共轭保持加法），(ii) $\overline{ab} = \bar a\, \bar b$（保持乘法，因此保持乘方），以及 (iii) $\bar a_k = a_k$（因为 $a_k$ 是实数）。所以 $\bar z_0$ 也是根。∎
"实系数"假设为什么不可少。若 $p(x) = x - i$，唯一的根是 $z_0 = i$，而它的共轭 $\bar z_0 = -i$ 不是根：$p(-i) = -2i \ne 0$。只要有一个系数离开 $\mathbb{R}$，共轭根定理就立刻失效。
推论——二次因子。若 $z_0 = a + bi$（$b \ne 0$）是实系数多项式的根，则 $\bar z_0 = a - bi$ 也是根，且
$$ (x - z_0)(x - \bar z_0) = x^2 - (z_0 + \bar z_0)x + z_0 \bar z_0 = x^2 - 2a x + (a^2 + b^2) $$是 $p(x)$ 的一个实二次因子。
Worked Example — Build a polynomial from a complex root例题——由复根构造多项式
Problem: Find a monic polynomial $p(x)$ with real coefficients of smallest possible degree such that $p(2 + 3i) = 0$ and $p(-1) = 0$.
Force the conjugate. Real coefficients $\Rightarrow$ $2 - 3i$ is also a root.
Three roots, three factors.
$$ p(x) = (x + 1)\bigl(x - (2 + 3i)\bigr)\bigl(x - (2 - 3i)\bigr) = (x + 1)\bigl(x^2 - 4x + 13\bigr). $$Here $z_0 + \bar z_0 = 4$ and $z_0 \bar z_0 = 4 + 9 = 13$.
Expand.
$$ p(x) = (x + 1)(x^2 - 4x + 13) = x^3 - 4x^2 + 13x + x^2 - 4x + 13 = x^3 - 3x^2 + 9x + 13. $$Degree $3$ — odd — so $p$ must have at least one real root, and indeed $p(-1) = 0$.
题目：求次数尽可能小的首一实系数多项式 $p(x)$，使 $p(2 + 3i) = 0$ 且 $p(-1) = 0$。
强制共轭。实系数 $\Rightarrow$ $2 - 3i$ 也是根。
三根，三因子。
$$ p(x) = (x + 1)\bigl(x - (2 + 3i)\bigr)\bigl(x - (2 - 3i)\bigr) = (x + 1)\bigl(x^2 - 4x + 13\bigr). $$这里 $z_0 + \bar z_0 = 4$，$z_0 \bar z_0 = 4 + 9 = 13$。
展开。
$$ p(x) = (x + 1)(x^2 - 4x + 13) = x^3 - 4x^2 + 13x + x^2 - 4x + 13 = x^3 - 3x^2 + 9x + 13. $$次数 $3$——奇数——所以 $p$ 必有至少一个实根，确实 $p(-1) = 0$。
Worked Example — Recovering all roots from one例题——由一个根恢复其余所有根
Problem: $p(x) = x^4 - 4x^3 + 14x^2 - 4x + 13$ has $i$ as a root. Find the remaining three roots.
Conjugate. By the conjugate root theorem, $-i$ is also a root. So $(x - i)(x + i) = x^2 + 1$ divides $p$.
Polynomial long division.
$$ p(x) \div (x^2 + 1) = x^2 - 4x + 13. $$(Verify by multiplying back: $(x^2 + 1)(x^2 - 4x + 13) = x^4 - 4x^3 + 13x^2 + x^2 - 4x + 13 = x^4 - 4x^3 + 14x^2 - 4x + 13$. ✓)
Solve the quadratic.
$$ x = \frac{4 \pm \sqrt{16 - 52}}{2} = \frac{4 \pm \sqrt{-36}}{2} = 2 \pm 3i. $$Four roots: $i,\, -i,\, 2 + 3i,\, 2 - 3i$ — two conjugate pairs, as required.
题目：已知 $p(x) = x^4 - 4x^3 + 14x^2 - 4x + 13$ 以 $i$ 为根。求其余三个根。
共轭。由共轭根定理（conjugate root theorem），$-i$ 也是根。所以 $(x - i)(x + i) = x^2 + 1$ 整除 $p$。
多项式长除法。
$$ p(x) \div (x^2 + 1) = x^2 - 4x + 13. $$（反向验算：$(x^2 + 1)(x^2 - 4x + 13) = x^4 - 4x^3 + 13x^2 + x^2 - 4x + 13 = x^4 - 4x^3 + 14x^2 - 4x + 13$。✓）
解二次方程。
$$ x = \frac{4 \pm \sqrt{16 - 52}}{2} = \frac{4 \pm \sqrt{-36}}{2} = 2 \pm 3i. $$四个根：$i,\, -i,\, 2 + 3i,\, 2 - 3i$——两对共轭对，正如所料。
Exam Strategy & Common Pitfalls考试策略与常见陷阱
- $i^2 = -1$, $\bar z = a - bi$, $|z|^2 = z\bar z = a^2 + b^2$
- Divide-by-conjugate recipe
- Quadrant rules for $\arg(z)$
- $e^{i\theta} = \cos\theta + i\sin\theta$ (Euler) & $z_1 z_2 = r_1 r_2 e^{i(\theta_1 + \theta_2)}$
- De Moivre: $(re^{i\theta})^n = r^n e^{i n\theta}$ for $n \in \mathbb{Z}$
- $n$-th roots formula: spacing $2\pi/n$, modulus $r^{1/n}$
- Conjugate root theorem & real-quadratic factor formula
- $i^2 = -1$，$\bar z = a - bi$，$|z|^2 = z\bar z = a^2 + b^2$
- "乘以共轭（
conjugate）做除法"的套路 - $\arg(z)$ 的象限规则
- 欧拉公式（
Euler's formula）$e^{i\theta} = \cos\theta + i\sin\theta$ 与 $z_1 z_2 = r_1 r_2 e^{i(\theta_1 + \theta_2)}$ - 棣莫弗定理（
De Moivre's theorem）：$(re^{i\theta})^n = r^n e^{i n\theta}$，$n \in \mathbb{Z}$ - $n$ 次根公式：间距 $2\pi/n$，模 $r^{1/n}$
- 共轭根定理与实二次因子公式
- Multiplication = rotation + scaling in the Argand plane
- Why Cartesian/polar/Euler are the same number, different forms
- The induction proof of De Moivre
- Why $\arctan(b/a)$ "loses" quadrants II and III
- Sum of $n$-th roots of unity is $0$ — algebraic and geometric arguments
- Why the conjugate root theorem needs real coefficients
- Why every odd-degree real polynomial has a real root
- 在阿尔冈平面（
Argand plane）上，乘法 = 旋转 + 缩放 - 为什么笛卡尔（
Cartesian）、极坐标（polar）、欧拉（Euler）三种形式是同一个数 - 棣莫弗定理的归纳法证明
- 为什么 $\arctan(b/a)$ 会"漏掉"第 II、III 象限
- $n$ 次单位根（
roots of unity）之和为 $0$——代数论证与几何论证 - 为什么共轭根定理必须以实系数为前提
- 为什么每个奇次实系数多项式必有实根
Common Pitfalls常见陷阱
2. Forgetting that $|z|$ is defined to be the positive square root: $\sqrt{a^2 + b^2}$, not $\pm$.
3. Mixing up $\operatorname{Im}(z) = b$ (a real number) with $bi$.
4. Applying $\sqrt{ab} = \sqrt a \sqrt b$ across negative reals — that identity fails: $\sqrt{(-1)(-1)} = 1$, not $i \cdot i = -1$.
5. Forgetting to reduce a De Moivre argument modulo $2\pi$ — leaving $e^{i\,15\pi/4}$ instead of $e^{-i\pi/4}$.
6. Listing only one $n$-th root when the question asks for all $n$ — the wording "find all $z$" is non-negotiable.
7. Forgetting the conjugate when you're handed a single complex root of a real polynomial.
8. Treating the conjugate root theorem as if it applied to complex-coefficient polynomials.
2. 忘了 $|z|$ 按定义是正的平方根：$\sqrt{a^2 + b^2}$，不带 $\pm$。
3. 把 $\operatorname{Im}(z) = b$（一个实数）与 $bi$ 混为一谈。
4. 在负实数之间套用 $\sqrt{ab} = \sqrt a \sqrt b$——此恒等式失效：$\sqrt{(-1)(-1)} = 1$，而 $i \cdot i = -1$。
5. 用棣莫弗定理后忘了把辐角对 $2\pi$ 取模——把 $e^{i\,15\pi/4}$ 留在那里而不化为 $e^{-i\pi/4}$。
6. 当题目要全部 $n$ 个 $n$ 次根时只列出一个——"find all $z$"（求所有 $z$）的措辞不可商量。
7. 拿到实系数多项式的一个复根却忘了它的共轭。
8. 把共轭根定理（
conjugate root theorem）当成对复系数多项式也成立——它不成立。
Paper 2 (calc): longer factoring problems, larger De Moivre powers, $n$-th roots displayed on a numerical Argand plot. You can use the GDC's complex mode, but write your set-up in proper notation — the marks are for the method, not the final number.
Paper 3 (HL extension): derivations using De Moivre (multiple-angle identities, sums of $\cos n\theta$), proofs of the conjugate root theorem and the roots-of-unity sum.
Paper 2（可用计算器）：更长的因式分解、更大的棣莫弗指数、把 $n$ 次根画在数值阿尔冈图（
Argand plot）上。可用 GDC 的复数模式，但解题过程要按标准记号写出来——分数给的是方法，不是最终数字。Paper 3（HL 延伸）：用棣莫弗定理推导（多倍角恒等式、$\sum \cos n\theta$ 等）、共轭根定理与单位根求和的证明。
Flashcards闪卡
Unit A4 — Practice Quiz单元 A4——练习测验
Ten mixed-difficulty items. Your score updates in real time at the top of the page. Aim for 8/10 before exam day.十道难度不一的题。得分实时显示在页面顶部。考前目标 8/10。
Readiness Checklist备考清单
Click each item you've mastered. Aim for 100% before exam day. Items marked HL+ are deeper-tier mastery (Paper 3 / top-mark territory).点击你已经掌握的条目。考前目标 100%。标有 HL+ 的为更深层掌握（Paper 3 / 拉分题）。
- Add, subtract, multiply, and divide complex numbers in Cartesian form能在笛卡尔形式下对复数加、减、乘、除
- Use $z \bar z = |z|^2$ to rationalize denominators用 $z \bar z = |z|^2$ 把复数分母实化
- Solve $z^2 = w$ by equating real and imaginary parts通过比较实部、虚部解 $z^2 = w$
- Plot complex numbers on the Argand diagram & read off modulus and argument能在阿尔冈图上画出复数，并读取模与辐角
- Convert between Cartesian and polar (or Euler) form with correct quadrant能在笛卡尔与极坐标 / 欧拉形式间互转，并选对象限
- State and apply Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$写出并应用欧拉公式 $e^{i\theta} = \cos\theta + i\sin\theta$
- Multiply and divide in polar form by adding/subtracting arguments在极坐标形式下通过辐角加 / 减做乘除
- Apply De Moivre's theorem to compute integer powers用棣莫弗定理计算整数次幂
- HL+ Reproduce the inductive proof of De Moivre能复现棣莫弗的归纳证明
- HL+ Derive multiple-angle identities ($\cos n\theta$, $\sin n\theta$) using De Moivre用棣莫弗推导多倍角恒等式 $\cos n\theta$、$\sin n\theta$
- Find all $n$ distinct $n$-th roots of a given complex number求出给定复数的全部 $n$ 个 $n$ 次方根
- HL+ Show that the $n$-th roots of unity sum to $0$ (algebraically and geometrically)证明 $n$ 次单位根之和为 $0$（代数与几何两种）
- State and apply the conjugate root theorem to factor real polynomials写出并应用共轭根定理来因式分解实多项式
- Build the real quadratic factor $x^2 - 2ax + (a^2+b^2)$ from a conjugate pair $a \pm bi$由共轭对 $a \pm bi$ 构造实二次因式 $x^2 - 2ax + (a^2+b^2)$