PART I · PAPER 1 SECTION A第一部分 · 第一卷 A 节No calculator · short response · 22 marks不可使用计算器 · 简答题 · 22 分
Section A — Short ResponseA 节 —— 简答题
Show all working in the space below each question. Method marks dominate. Keep results in exact form; do not approximate $\pi$, $\sqrt{n}$, or trig values.在每题下方的空白处写出全部解题过程。方法分（method marks）权重最大。结果保留精确形式 —— 不要近似 $\pi$、$\sqrt{n}$ 或三角函数值。
(b)Find $\arg(z)$ (principal value, $-\pi < \arg z \le \pi$).求 $\arg(z)$（主值，principal value，$-\pi < \arg z \le \pi$）。[2]
(c)Write $z$ in (i) polar form $r(\cos\theta + i\sin\theta)$ and (ii) Euler form $r e^{i\theta}$.用 (i) 极坐标形式 $r(\cos\theta + i\sin\theta)$ 和 (ii) 欧拉形式 $r e^{i\theta}$ 写出 $z$。[2]
Q3MEDIUMPaper 1AAHL 1.14 De Moivre — Power[6 marks]
Use De Moivre's theorem to evaluate $(1 + i)^{8}$, giving your answer in Cartesian form $a + bi$ with $a, b \in \mathbb{Z}$.使用棣莫弗定理（De Moivre's theorem）求 $(1 + i)^{8}$，答案以笛卡尔形式 $a + bi$ 给出（$a, b \in \mathbb{Z}$）。
Expand $(\cos\theta + i\sin\theta)^{3}$ using the binomial theorem, equate real parts on both sides of De Moivre's theorem, and substitute $\sin^{2}\theta = 1 - \cos^{2}\theta$ where needed. Show every step.用二项定理（binomial theorem）展开 $(\cos\theta + i\sin\theta)^{3}$，对棣莫弗等式两边取实部，并在必要处代入 $\sin^{2}\theta = 1 - \cos^{2}\theta$。写出每一步。
PART II · PAPER 1 SECTION B第二部分 · 第一卷 B 节No calculator · extended response · 13 marks不可使用计算器 · 长答题 · 13 分
Section B — Extended ResponseB 节 —— 长答题
Take time to set up cleanly. State De Moivre's theorem (or the root-of-complex-number formula) before using it. Carry expressions in exact form; do not approximate.认真梳理思路再下笔。引用棣莫弗定理（或复数开方公式，roots of a complex number）前先写出公式。表达式保留精确形式，不要近似。
Q5HARDPaper 1BAHL 1.14 Roots of a Complex Number[13 marks]
Consider the equation $z^{5} = 32$ where $z \in \mathbb{C}$.考虑方程 $z^{5} = 32$，其中 $z \in \mathbb{C}$。
(a)State De Moivre's theorem for raising a complex number to an integer power.写出棣莫弗定理（复数的整数次幂形式）。[1]
(b)Write the right-hand side $32$ in polar form, accounting for the periodicity of the argument.将右边 $32$ 写为极坐标形式，注意辐角（argument）的周期性。[2]
(c)Find all five solutions $z_{0}, z_{1}, z_{2}, z_{3}, z_{4}$ in polar form. Use $-\pi < \arg z \le \pi$ for the principal values.求出所有五个解 $z_{0}, z_{1}, z_{2}, z_{3}, z_{4}$ 的极坐标形式，主值取 $-\pi < \arg z \le \pi$。[4]
(d)Sketch the five roots on an Argand diagram, showing the regular pentagon they form. Mark the centre, the radius, and one root explicitly.在 Argand 图（Argand diagram）上画出五个根所构成的正五边形（regular pentagon），并标出圆心、半径以及其中一个根。[3]
(e)Without computing each root in Cartesian form, find the sum $z_{0} + z_{1} + z_{2} + z_{3} + z_{4}$. Justify your answer by symmetry.不必将每个根化为笛卡尔形式，求和 $z_{0} + z_{1} + z_{2} + z_{3} + z_{4}$。用对称性论证。[3]
PART III · PAPER 2第三部分 · 第二卷Calculator · mixed response · 16 marks可使用计算器 · 混合题型 · 16 分
Paper 2 — Calculator Permitted第二卷 —— 允许使用计算器
A graphing calculator is required. Give exact answers (integers, fractions, exact surds) where reasonable; otherwise correct to 3 significant figures.需要图形计算器（GDC）。能给出精确答案（整数、分数、根号形式）的题目就给精确答案；其余保留 3 位有效数字。
Let $P(z) = z^{3} + az^{2} + bz + c$ where $a, b, c \in \mathbb{R}$. It is given that $z = 1 + 2i$ is a root, and that $z = 3$ is also a root.设 $P(z) = z^{3} + az^{2} + bz + c$，其中 $a, b, c \in \mathbb{R}$。已知 $z = 1 + 2i$ 是其一个根，且 $z = 3$ 也是其一个根。
(a)State the third root, justifying your reasoning in one sentence.写出第三个根，并用一句话说明理由。[2]
(b)Find $a$, $b$, $c$ by expanding $P(z) = (z - r_{1})(z - r_{2})(z - r_{3})$.通过展开 $P(z) = (z - r_{1})(z - r_{2})(z - r_{3})$ 求 $a$、$b$、$c$。[4]
(c)Write the complete factorisation of $P(z)$ over $\mathbb{R}$ (real factors only — pair conjugates into a real quadratic).将 $P(z)$ 在 $\mathbb{R}$ 上完全因式分解（仅含实因式 —— 将共轭根合并为实系数二次式）。[1]
Q7HARDPaper 2AHL 1.13 Argand Locus[9 marks]
Consider the locus of points $z \in \mathbb{C}$ satisfying考虑满足下式的复数 $z \in \mathbb{C}$ 的轨迹（locus）：
$$ |z - 2| = |z + 2i|. $$
(a)Interpret each side geometrically as a distance in the Argand plane (one sentence each).在 Argand 平面上从几何角度解释两边各自的含义（每条一句话即可）。[2]
(b)Let $z = x + iy$ with $x, y \in \mathbb{R}$. Square both sides and simplify to show that the locus is the line $y = -x$.设 $z = x + iy$，$x, y \in \mathbb{R}$。对两边平方并化简，证明轨迹是直线 $y = -x$。[4]
(c)Sketch the locus on an Argand diagram, clearly marking the two reference points $2$ and $-2i$.在 Argand 图上画出轨迹，清晰标出两个参考点 $2$ 与 $-2i$。[2]
(d)Find the point on this locus closest to the origin.求该轨迹上距原点最近的点。[1]
PART IV · PAPER 3第四部分 · 第三卷Calculator · HL extended exploration · 16 marks可使用计算器 · HL 长题探究 · 16 分
Paper 3 — HL Extended Problem第三卷 —— HL 长题探究
A graphing calculator is required. Method marks are heavily weighted. Show full reasoning at every step.需要图形计算器（GDC）。方法分权重很高。每一步都要写出完整推理。
Q8HARDPaper 3AHL 1.14 Roots of Unity & Trig Sums[16 marks]
This question develops the $n$th roots of unity and uses them to derive a closed-form value for a sum of cosines.本题构造 $n$ 次单位根（$n$th roots of unity），并用以推导某余弦和的闭合值。
(a)Define $\omega = e^{2\pi i/n}$ for a positive integer $n \ge 2$. Show that $\omega^{n} = 1$, and that $\omega^{k}$ for $k = 0, 1, 2, \ldots, n-1$ are $n$ distinct complex numbers.对正整数 $n \ge 2$，定义 $\omega = e^{2\pi i/n}$。证明 $\omega^{n} = 1$，并证明 $\omega^{k}$（$k = 0, 1, 2, \ldots, n-1$）是 $n$ 个互不相同的复数。[3]
(b)By summing the geometric series $1 + \omega + \omega^{2} + \cdots + \omega^{n-1}$ (and treating the case $\omega = 1$ separately), show that for $n \ge 2$通过对几何级数 $1 + \omega + \omega^{2} + \cdots + \omega^{n-1}$ 求和（并单独处理 $\omega = 1$ 情形），证明当 $n \ge 2$ 时[3]
$$ \sum_{k=0}^{n-1} \omega^{k} \;=\; 0. $$
(c)Take real parts of the identity in (b) to derive the closed-form trigonometric sum对 (b) 中恒等式取实部，推导下列三角求和的闭合形式：[3]
$$ \sum_{k=0}^{n-1} \cos\!\left(\tfrac{2\pi k}{n}\right) \;=\; 0. $$
(d)Take imaginary parts to derive the companion identity for $\sin$.取虚部，推导关于 $\sin$ 的伴随恒等式。[2]
(e)Apply (c) with $n = 5$ to write down the exact value of $\cos\!\left(\tfrac{2\pi}{5}\right) + \cos\!\left(\tfrac{4\pi}{5}\right) + \cos\!\left(\tfrac{6\pi}{5}\right) + \cos\!\left(\tfrac{8\pi}{5}\right)$. Verify on your GDC to 4 decimal places.在 (c) 中取 $n = 5$，写出 $\cos\!\left(\tfrac{2\pi}{5}\right) + \cos\!\left(\tfrac{4\pi}{5}\right) + \cos\!\left(\tfrac{6\pi}{5}\right) + \cos\!\left(\tfrac{8\pi}{5}\right)$ 的精确值，并在 GDC 上验证至 4 位小数。[2]
(f)Using only the result of (e) and the pairing $\cos(2\pi - x) = \cos(x)$, deduce the exact value of $\cos\!\left(\tfrac{2\pi}{5}\right) + \cos\!\left(\tfrac{4\pi}{5}\right)$. (No further numeric work needed.)仅利用 (e) 的结论与配对关系 $\cos(2\pi - x) = \cos(x)$，推出 $\cos\!\left(\tfrac{2\pi}{5}\right) + \cos\!\left(\tfrac{4\pi}{5}\right)$ 的精确值。（不必再做数值计算。）[3]