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Unit A4 · SolutionsUnit A4 · 解析

Complex Numbers — Solutions复数 —— 解析

Companion to the IB-Style Practice SetIB 风格练习题的解析配套

EASY MEDIUM HARD Paper 1A Paper 1B Paper 2 Paper 3

Syllabus AHL 1.12 – 1.14 (HL-only super-topic)考纲 AHL 1.12 – 1.14（HL 专属超主题）AA HL

PART I  ·  PAPER 1 SECTION A — SOLUTIONS第一部分  ·  第一卷 A 节 —— 解析No calculator · 22 marks不可使用计算器 · 22 分

Section A — Worked SolutionsA 节 —— 详细解析

Q1EASYPaper 1AAHL 1.12 Cartesian Arithmetic[5 marks]

$z_1 = 3 + 4i$, $z_2 = 1 - 2i$ — find (a) $z_1 + 2z_2$, (b) $z_1 z_2$, (c) $z_1/z_2$ all in Cartesian form.$z_1 = 3 + 4i$、$z_2 = 1 - 2i$ —— 以笛卡尔形式求 (a) $z_1 + 2z_2$，(b) $z_1 z_2$，(c) $z_1/z_2$。

Answers:答案：  (a) $5$  ·  (b) $11 - 2i$  ·  (c) $-1 + 2i$

(a) Linear combination A1

$$ z_1 + 2z_2 \;=\; (3 + 4i) + 2(1 - 2i) \;=\; (3 + 4i) + (2 - 4i) \;=\; 5 + 0i \;=\; 5. $$

(b) Product M1·A1

Distribute, then use $i^2 = -1$: $$ z_1 z_2 \;=\; (3 + 4i)(1 - 2i) \;=\; 3 - 6i + 4i - 8i^{2} \;=\; 3 - 2i + 8 \;=\; 11 - 2i. $$

(c) Quotient via conjugate multiplication M1·A1

Multiply numerator and denominator by the conjugate of the denominator $\overline{z_2} = 1 + 2i$: $$ \frac{z_1}{z_2} \;=\; \frac{3 + 4i}{1 - 2i} \cdot \frac{1 + 2i}{1 + 2i} \;=\; \frac{(3+4i)(1+2i)}{|z_2|^{2}} \;=\; \frac{3 + 6i + 4i + 8i^{2}}{1 + 4} \;=\; \frac{-5 + 10i}{5} \;=\; -1 + 2i. $$
Why the conjugate trick. $z \bar{z} = |z|^{2}$ is always real and non-negative — so multiplying top and bottom by $\overline{z_2}$ converts a complex denominator into a real one. This is the complex-number analogue of rationalising a surd denominator (multiply by the conjugate). It works for any $z_2 \ne 0$ and is the canonical IB write-up for division.

(a) 线性组合 A1

$$ z_1 + 2z_2 \;=\; (3 + 4i) + 2(1 - 2i) \;=\; (3 + 4i) + (2 - 4i) \;=\; 5 + 0i \;=\; 5. $$

(b) 乘积 M1·A1

分配律展开，再代入 $i^2 = -1$： $$ z_1 z_2 \;=\; (3 + 4i)(1 - 2i) \;=\; 3 - 6i + 4i - 8i^{2} \;=\; 3 - 2i + 8 \;=\; 11 - 2i. $$

(c) 用共轭相乘求商 M1·A1

分子分母同时乘以分母的共轭 $\overline{z_2} = 1 + 2i$： $$ \frac{z_1}{z_2} \;=\; \frac{3 + 4i}{1 - 2i} \cdot \frac{1 + 2i}{1 + 2i} \;=\; \frac{(3+4i)(1+2i)}{|z_2|^{2}} \;=\; \frac{3 + 6i + 4i + 8i^{2}}{1 + 4} \;=\; \frac{-5 + 10i}{5} \;=\; -1 + 2i. $$
共轭技巧为何有效。$z \bar{z} = |z|^{2}$ 恒为非负实数 —— 所以分子分母同乘 $\overline{z_2}$ 后，分母即为实数。这正是"复数除法"对应于"无理分母有理化"（分子分母乘共轭）的做法。只要 $z_2 \ne 0$ 就成立，是 IB 复数除法的标准格式。
Q2MEDIUMPaper 1AAHL 1.13 Modulus, Argument, Polar[5 marks]

$z = \sqrt{3} - i$ — find (a) $|z|$, (b) $\arg z$, (c) polar + Euler forms.$z = \sqrt{3} - i$ —— 求 (a) $|z|$，(b) $\arg z$，(c) 极坐标与欧拉形式。

Answers:答案：  (a) $|z| = 2$  ·  (b) $\arg z = -\tfrac{\pi}{6}$  ·  (c) $z = 2\!\left[\cos(-\tfrac{\pi}{6}) + i\sin(-\tfrac{\pi}{6})\right] = 2\,e^{-i\pi/6}$

(a) Modulus A1

$|z| = \sqrt{(\sqrt{3})^{2} + (-1)^{2}} = \sqrt{3 + 1} = 2$.

(b) Argument M1·A1

$\Re(z) = \sqrt{3} > 0$ and $\Im(z) = -1 < 0$, so $z$ is in the fourth quadrant. Reference angle: $\tan^{-1}\!\bigl(\tfrac{|{-1}|}{\sqrt{3}}\bigr) = \tan^{-1}\!\bigl(\tfrac{1}{\sqrt{3}}\bigr) = \tfrac{\pi}{6}$. Apply the QIV sign (negative): $\arg z = -\tfrac{\pi}{6}$.

(c) Polar and Euler forms A1·A1

With $r = 2$, $\theta = -\tfrac{\pi}{6}$: $$ z \;=\; 2\!\left[\cos\!\left(-\tfrac{\pi}{6}\right) + i\sin\!\left(-\tfrac{\pi}{6}\right)\right] \;=\; 2\,e^{-i\pi/6}. $$ Check: $2\cos(-\pi/6) = 2 \cdot \tfrac{\sqrt{3}}{2} = \sqrt{3}$; $2 i\sin(-\pi/6) = 2i \cdot (-\tfrac{1}{2}) = -i$. ✓
Quadrant table for argument. Plot the point in the Argand plane first — the quadrant determines the sign of $\theta$. QI: $\theta > 0$ small; QII: $\theta > 0$ close to $\pi$; QIII: $\theta < 0$ close to $-\pi$ (or equivalently positive close to $\pi$); QIV: $\theta < 0$ small. A common slip is to write $\arg(\sqrt{3} - i) = +\tfrac{\pi}{6}$ without checking the quadrant — the calculator's $\tan^{-1}$ only returns the reference angle in $(-\tfrac{\pi}{2}, \tfrac{\pi}{2})$, not the true argument.

(a) 模 A1

$|z| = \sqrt{(\sqrt{3})^{2} + (-1)^{2}} = \sqrt{3 + 1} = 2$。

(b) 辐角 M1·A1

$\Re(z) = \sqrt{3} > 0$，$\Im(z) = -1 < 0$，故 $z$ 在第四象限。参考角：$\tan^{-1}\!\bigl(\tfrac{|{-1}|}{\sqrt{3}}\bigr) = \tan^{-1}\!\bigl(\tfrac{1}{\sqrt{3}}\bigr) = \tfrac{\pi}{6}$。按第四象限取负号：$\arg z = -\tfrac{\pi}{6}$。

(c) 极坐标与欧拉形式 A1·A1

取 $r = 2$、$\theta = -\tfrac{\pi}{6}$： $$ z \;=\; 2\!\left[\cos\!\left(-\tfrac{\pi}{6}\right) + i\sin\!\left(-\tfrac{\pi}{6}\right)\right] \;=\; 2\,e^{-i\pi/6}. $$ 验证：$2\cos(-\pi/6) = 2 \cdot \tfrac{\sqrt{3}}{2} = \sqrt{3}$；$2 i\sin(-\pi/6) = 2i \cdot (-\tfrac{1}{2}) = -i$。✓
辐角的象限表。先在 Argand 平面上标点 —— 象限决定 $\theta$ 的正负。第一象限：$\theta$ 小且为正；第二象限：$\theta$ 接近 $\pi$ 且为正；第三象限：$\theta$ 接近 $-\pi$（或等价于接近 $\pi$ 的正值）；第四象限：$\theta$ 小且为负。常见失误是直接写 $\arg(\sqrt{3} - i) = +\tfrac{\pi}{6}$ 而不检查象限 —— 计算器的 $\tan^{-1}$ 只返回 $(-\tfrac{\pi}{2}, \tfrac{\pi}{2})$ 内的参考角，并非真正辐角。
Q3MEDIUMPaper 1AAHL 1.14 De Moivre — Power[6 marks]

Evaluate $(1+i)^{8}$ via De Moivre.用棣莫弗定理求 $(1+i)^{8}$。

Answer:答案：  $(1+i)^{8} = 16$

(a) Polar form of $1 + i$ M1·A1

$|1+i| = \sqrt{1+1} = \sqrt{2}$. The point $(1, 1)$ is in QI with reference angle $\tan^{-1}(1) = \tfrac{\pi}{4}$. So $$ 1 + i \;=\; \sqrt{2}\!\left[\cos\!\left(\tfrac{\pi}{4}\right) + i\sin\!\left(\tfrac{\pi}{4}\right)\right] \;=\; \sqrt{2}\,e^{i\pi/4}. $$

(b) Apply De Moivre M1·M1·A1·A1

De Moivre's theorem: $[r\,e^{i\theta}]^{n} = r^{n}\,e^{in\theta}$. With $n = 8$: $$ (1+i)^{8} \;=\; (\sqrt{2})^{8}\,e^{i \cdot 8\pi/4} \;=\; 2^{4}\,e^{i \cdot 2\pi} \;=\; 16\,e^{i \cdot 2\pi} \;=\; 16 \cdot \bigl[\cos(2\pi) + i\sin(2\pi)\bigr] \;=\; 16. $$
Why De Moivre crushes brute-force expansion. Expanding $(1+i)^{8}$ via the binomial theorem requires nine terms with cancellation. De Moivre reduces the whole calculation to "raise modulus, multiply argument" — two simple operations. The pay-off scales: $(1+i)^{20}$ has 21 binomial terms but takes the same one line in polar form. Whenever you see an integer power of a complex number larger than $\sim 4$, switch to polar first.

(a) $1 + i$ 的极坐标形式 M1·A1

$|1+i| = \sqrt{1+1} = \sqrt{2}$。点 $(1, 1)$ 在第一象限，参考角 $\tan^{-1}(1) = \tfrac{\pi}{4}$。故 $$ 1 + i \;=\; \sqrt{2}\!\left[\cos\!\left(\tfrac{\pi}{4}\right) + i\sin\!\left(\tfrac{\pi}{4}\right)\right] \;=\; \sqrt{2}\,e^{i\pi/4}. $$

(b) 套用棣莫弗 M1·M1·A1·A1

棣莫弗定理：$[r\,e^{i\theta}]^{n} = r^{n}\,e^{in\theta}$。取 $n = 8$： $$ (1+i)^{8} \;=\; (\sqrt{2})^{8}\,e^{i \cdot 8\pi/4} \;=\; 2^{4}\,e^{i \cdot 2\pi} \;=\; 16\,e^{i \cdot 2\pi} \;=\; 16 \cdot \bigl[\cos(2\pi) + i\sin(2\pi)\bigr] \;=\; 16. $$
棣莫弗比硬展开省力的原因。用二项定理展开 $(1+i)^{8}$ 要写九项并相消。棣莫弗把整套计算压缩为"模取幂、辐角相乘"两步。优势随幂次增大而放大：$(1+i)^{20}$ 二项展开有 21 项，但极坐标下仍是一行。指数 $\gtrsim 4$ 时，先转极坐标。
Q4HARDPaper 1AAHL 1.14 De Moivre — Trig Identity[6 marks]

Prove $\cos(3\theta) = 4\cos^{3}\theta - 3\cos\theta$ using De Moivre.用棣莫弗定理证明 $\cos(3\theta) = 4\cos^{3}\theta - 3\cos\theta$。

Identity proven; see derivation.恒等式得证；推导见下。

Apply De Moivre at $n = 3$ M1·A1

$$ (\cos\theta + i\sin\theta)^{3} \;=\; \cos(3\theta) + i\sin(3\theta). \tag{$\star$} $$

Expand the LHS via the binomial theorem M1·A1

$$ (\cos\theta + i\sin\theta)^{3} \;=\; \cos^{3}\theta + 3\cos^{2}\theta(i\sin\theta) + 3\cos\theta(i\sin\theta)^{2} + (i\sin\theta)^{3}. $$ Use $i^{2} = -1$ and $i^{3} = -i$: $$ = \cos^{3}\theta + 3i\cos^{2}\theta\sin\theta - 3\cos\theta\sin^{2}\theta - i\sin^{3}\theta. $$ Separate real and imaginary parts: $$ = \bigl[\cos^{3}\theta - 3\cos\theta\sin^{2}\theta\bigr] + i\bigl[3\cos^{2}\theta\sin\theta - \sin^{3}\theta\bigr]. $$

Equate real parts of $(\star)$ M1·A1

$$ \cos(3\theta) \;=\; \cos^{3}\theta - 3\cos\theta\sin^{2}\theta. $$ Substitute $\sin^{2}\theta = 1 - \cos^{2}\theta$: $$ \cos(3\theta) \;=\; \cos^{3}\theta - 3\cos\theta\,(1 - \cos^{2}\theta) \;=\; \cos^{3}\theta - 3\cos\theta + 3\cos^{3}\theta \;=\; 4\cos^{3}\theta - 3\cos\theta. \qquad \blacksquare $$
The Chebyshev pattern. The De Moivre method delivers every multiple-angle identity. Real part gives $\cos(n\theta)$ as a polynomial in $\cos\theta$ (the Chebyshev polynomial $T_{n}(\cos\theta) = \cos(n\theta)$); imaginary part gives $\sin(n\theta)$ as $\sin\theta$ times a polynomial in $\cos\theta$ (related to $U_{n-1}$). The $n = 3$ identity above is $T_{3}(x) = 4x^{3} - 3x$. The $n = 5$ identity $\cos(5\theta) = 16\cos^{5}\theta - 20\cos^{3}\theta + 5\cos\theta$ is the Paper 1B follow-up.

在 $n = 3$ 套用棣莫弗 M1·A1

$$ (\cos\theta + i\sin\theta)^{3} \;=\; \cos(3\theta) + i\sin(3\theta). \tag{$\star$} $$

用二项定理展开左边 M1·A1

$$ (\cos\theta + i\sin\theta)^{3} \;=\; \cos^{3}\theta + 3\cos^{2}\theta(i\sin\theta) + 3\cos\theta(i\sin\theta)^{2} + (i\sin\theta)^{3}. $$ 利用 $i^{2} = -1$ 与 $i^{3} = -i$： $$ = \cos^{3}\theta + 3i\cos^{2}\theta\sin\theta - 3\cos\theta\sin^{2}\theta - i\sin^{3}\theta. $$ 分离实部与虚部： $$ = \bigl[\cos^{3}\theta - 3\cos\theta\sin^{2}\theta\bigr] + i\bigl[3\cos^{2}\theta\sin\theta - \sin^{3}\theta\bigr]. $$

对 $(\star)$ 两边取实部 M1·A1

$$ \cos(3\theta) \;=\; \cos^{3}\theta - 3\cos\theta\sin^{2}\theta. $$ 代入 $\sin^{2}\theta = 1 - \cos^{2}\theta$： $$ \cos(3\theta) \;=\; \cos^{3}\theta - 3\cos\theta\,(1 - \cos^{2}\theta) \;=\; \cos^{3}\theta - 3\cos\theta + 3\cos^{3}\theta \;=\; 4\cos^{3}\theta - 3\cos\theta. \qquad \blacksquare $$
切比雪夫规律。棣莫弗方法可推出所有倍角恒等式。实部给出 $\cos(n\theta)$ 关于 $\cos\theta$ 的多项式（切比雪夫多项式（Chebyshev polynomial）$T_{n}(\cos\theta) = \cos(n\theta)$）；虚部给出 $\sin(n\theta)$ 等于 $\sin\theta$ 乘以一个关于 $\cos\theta$ 的多项式（与 $U_{n-1}$ 相关）。上面 $n = 3$ 的恒等式即 $T_{3}(x) = 4x^{3} - 3x$。$n = 5$ 的恒等式 $\cos(5\theta) = 16\cos^{5}\theta - 20\cos^{3}\theta + 5\cos\theta$ 是 Paper 1B 的进阶练习。
PART II  ·  PAPER 1 SECTION B — SOLUTIONS第二部分  ·  第一卷 B 节 —— 解析No calculator · 13 marks不可使用计算器 · 13 分

Section B — Worked SolutionsB 节 —— 详细解析

Q5HARDPaper 1BAHL 1.14 Roots of a Complex Number[13 marks]

$z^{5} = 32$ — (a) De Moivre statement; (b) polar form of RHS with periodicity; (c) all five roots; (d) Argand sketch; (e) sum of roots by symmetry.$z^{5} = 32$ —— (a) 棣莫弗定理；(b) 右边的极坐标周期化；(c) 五个根；(d) Argand 图；(e) 用对称性求和。

Answers:答案：  (c) $z_k = 2\,e^{i \cdot 2\pi k/5},\; k = 0, 1, 2, 3, 4$  ·  (e) $\sum z_k = 0$

(a) De Moivre A1

$$ \bigl[r(\cos\theta + i\sin\theta)\bigr]^{n} \;=\; r^{n}(\cos n\theta + i\sin n\theta), \qquad n \in \mathbb{Z}. $$

(b) Polar form of $32$ with periodicity M1·A1

$32$ is a positive real, so $|32| = 32$ and $\arg(32) = 0$. Including the $2\pi$-periodicity of the argument: $$ 32 \;=\; 32\,\bigl[\cos(0 + 2\pi k) + i\sin(0 + 2\pi k)\bigr] \;=\; 32\,e^{i \cdot 2\pi k}, \qquad k \in \mathbb{Z}. $$

(c) Solve for the five roots M1·M1·A1·A1

Let $z = r\,e^{i\theta}$. Then $z^{5} = r^{5}\,e^{i \cdot 5\theta} = 32\,e^{i \cdot 2\pi k}$. Match modulus and argument: $$ r^{5} = 32 \;\Longrightarrow\; r = 2, \qquad 5\theta = 2\pi k \;\Longrightarrow\; \theta = \tfrac{2\pi k}{5}. $$ Choose $k = 0, 1, 2, 3, 4$ to get the five distinct principal-value roots (further $k$ values repeat):
  • $z_{0} = 2\,e^{i \cdot 0} = 2$
  • $z_{1} = 2\,e^{i \cdot 2\pi/5}$
  • $z_{2} = 2\,e^{i \cdot 4\pi/5}$
  • $z_{3} = 2\,e^{i \cdot 6\pi/5} \;=\; 2\,e^{-i \cdot 4\pi/5}$ (subtract $2\pi$ for principal value)
  • $z_{4} = 2\,e^{i \cdot 8\pi/5} \;=\; 2\,e^{-i \cdot 2\pi/5}$

(d) Argand sketch M1·A1·A1

All five roots lie on a circle of radius $2$ centered at the origin. Adjacent roots are separated by argument $\tfrac{2\pi}{5} = 72°$, so they form a regular pentagon inscribed in the circle. Mark the positive real axis intersection (the root $z_{0} = 2$); then rotate by $72°$ counter-clockwise to find $z_{1}, z_{2}, z_{3}, z_{4}$ in order.

(e) Sum by symmetry M1·R1·A1

The five roots are the vertices of a regular pentagon centred at the origin. The centroid of the vertices of any regular $n$-gon is its centre — here the origin. So $$ \frac{1}{5}\sum_{k=0}^{4} z_{k} \;=\; 0 \quad \Longrightarrow \quad \sum_{k=0}^{4} z_{k} \;=\; 0. $$ Equivalently, $\sum z_{k} = 2 \sum_{k=0}^{4} \omega^{k}$ where $\omega = e^{2\pi i/5}$ is a primitive 5th root of unity; the geometric series $1 + \omega + \omega^{2} + \omega^{3} + \omega^{4} = (1 - \omega^{5})/(1 - \omega) = 0/(1 - \omega) = 0$.
Two routes, both rigorous. The "centroid" argument is intuitive and visual; the "geometric series" argument is algebraic and generalises. Both are mark-scheme legitimate — the IB asks for a justification, not a specific framing. The geometric-series route generalises immediately: for any $n \ge 2$, the sum of the $n$th roots of $w$ is zero. This pattern (which we develop fully in Q8) is the key tool for evaluating closed-form trig sums.

(a) 棣莫弗定理 A1

$$ \bigl[r(\cos\theta + i\sin\theta)\bigr]^{n} \;=\; r^{n}(\cos n\theta + i\sin n\theta), \qquad n \in \mathbb{Z}. $$

(b) $32$ 的极坐标周期化 M1·A1

$32$ 是正实数，故 $|32| = 32$、$\arg(32) = 0$。考虑辐角的 $2\pi$ 周期性： $$ 32 \;=\; 32\,\bigl[\cos(0 + 2\pi k) + i\sin(0 + 2\pi k)\bigr] \;=\; 32\,e^{i \cdot 2\pi k}, \qquad k \in \mathbb{Z}. $$

(c) 求五个根 M1·M1·A1·A1

设 $z = r\,e^{i\theta}$。则 $z^{5} = r^{5}\,e^{i \cdot 5\theta} = 32\,e^{i \cdot 2\pi k}$。模与辐角分别匹配： $$ r^{5} = 32 \;\Longrightarrow\; r = 2, \qquad 5\theta = 2\pi k \;\Longrightarrow\; \theta = \tfrac{2\pi k}{5}. $$ 取 $k = 0, 1, 2, 3, 4$ 得到五个不同的主值根（更大 $k$ 会重复）：
  • $z_{0} = 2\,e^{i \cdot 0} = 2$
  • $z_{1} = 2\,e^{i \cdot 2\pi/5}$
  • $z_{2} = 2\,e^{i \cdot 4\pi/5}$
  • $z_{3} = 2\,e^{i \cdot 6\pi/5} \;=\; 2\,e^{-i \cdot 4\pi/5}$（减去 $2\pi$ 取主值）
  • $z_{4} = 2\,e^{i \cdot 8\pi/5} \;=\; 2\,e^{-i \cdot 2\pi/5}$

(d) Argand 图 M1·A1·A1

五个根都在以原点为中心、半径为 $2$ 的圆上。相邻根之间辐角差 $\tfrac{2\pi}{5} = 72°$，构成内接于该圆的正五边形（regular pentagon）。先标出正实轴上的根 $z_{0} = 2$；再逆时针每次旋转 $72°$ 依次得到 $z_{1}, z_{2}, z_{3}, z_{4}$。

(e) 用对称性求和 M1·R1·A1

五个根是以原点为中心的正五边形的顶点。任意正 $n$ 边形顶点的形心就是其中心 —— 此处为原点。故 $$ \frac{1}{5}\sum_{k=0}^{4} z_{k} \;=\; 0 \quad \Longrightarrow \quad \sum_{k=0}^{4} z_{k} \;=\; 0. $$ 或者，记 $\omega = e^{2\pi i/5}$（一个 5 次单位本原根），则 $\sum z_{k} = 2 \sum_{k=0}^{4} \omega^{k}$；该几何级数 $1 + \omega + \omega^{2} + \omega^{3} + \omega^{4} = (1 - \omega^{5})/(1 - \omega) = 0/(1 - \omega) = 0$。
两种路径，皆严谨。"形心"论证直观、可视化；"几何级数"论证代数化、可推广。两者都给方法分 —— IB 要的是合理说明，而非特定形式。几何级数路径可立即推广：对任意 $n \ge 2$，$w$ 的 $n$ 次根之和为 $0$。这一规律（在 Q8 中完整展开）是闭合形式三角求和的关键工具。
PART III  ·  PAPER 2 — SOLUTIONS第三部分  ·  第二卷 —— 解析Calculator · 16 marks可使用计算器 · 16 分

Paper 2 — Worked Solutions第二卷 —— 详细解析

Q6MEDIUMPaper 2AHL 1.14 Real-Coefficient Polynomial[7 marks]

$P(z) = z^{3} + az^{2} + bz + c$, $a,b,c \in \mathbb{R}$, with roots $1+2i$ and $3$ — (a) third root, (b) find $a,b,c$, (c) factor over $\mathbb{R}$.$P(z) = z^{3} + az^{2} + bz + c$，$a,b,c \in \mathbb{R}$，已知 $1+2i$ 与 $3$ 为根 —— (a) 第三根，(b) 求 $a,b,c$，(c) $\mathbb{R}$ 上因式分解。

Answers:答案：  (a) $1 - 2i$  ·  (b) $a = -5,\, b = 11,\, c = -15$  ·  (c) $P(z) = (z - 3)(z^{2} - 2z + 5)$

(a) Third root via conjugate-pair theorem A1·R1

Real-coefficient polynomials have non-real roots occurring in complex-conjugate pairs. Since $1 + 2i$ is a root and the coefficients are real, $\overline{1 + 2i} = 1 - 2i$ must also be a root. With $3$ already given and the polynomial cubic ($3$ roots total), the third root is $\boxed{1 - 2i}$.

(b) Find $a$, $b$, $c$ by expansion M1·M1·A1·A1

$$ P(z) \;=\; (z - 3)\bigl(z - (1+2i)\bigr)\bigl(z - (1-2i)\bigr). $$ Combine the conjugate pair into a real quadratic first — this is the calculator-free step: $$ \bigl(z - (1+2i)\bigr)\bigl(z - (1-2i)\bigr) \;=\; z^{2} - \bigl((1+2i)+(1-2i)\bigr)z + (1+2i)(1-2i) \;=\; z^{2} - 2z + (1 - (2i)^{2}) \;=\; z^{2} - 2z + 5. $$ Now multiply by $(z - 3)$: $$ P(z) \;=\; (z - 3)(z^{2} - 2z + 5) \;=\; z^{3} - 2z^{2} + 5z - 3z^{2} + 6z - 15 \;=\; z^{3} - 5z^{2} + 11z - 15. $$ Matching coefficients: $a = -5$, $b = 11$, $c = -15$.

(c) Factorisation over $\mathbb{R}$ A1

$$ P(z) \;=\; (z - 3)(z^{2} - 2z + 5). $$ (The quadratic factor has discriminant $4 - 20 = -16 < 0$, so it's irreducible over $\mathbb{R}$ — exactly as expected from a conjugate-pair of complex roots.)
Conjugate-pair shortcut. Whenever a real polynomial has complex roots, they pair into real quadratic factors of the form $(z - \alpha)(z - \bar{\alpha}) = z^{2} - 2\Re(\alpha)z + |\alpha|^{2}$. Memorise this template — it lets you write the real quadratic factor directly from one complex root, no expansion needed. Here $\alpha = 1+2i$ gives $z^{2} - 2(1)z + (1^{2}+2^{2}) = z^{2} - 2z + 5$ in one line.

(a) 由共轭根定理得第三根 A1·R1

实系数多项式（real-coefficient polynomial）的非实数根必成共轭对（complex-conjugate pair）出现。既然 $1 + 2i$ 是根且系数为实数，$\overline{1 + 2i} = 1 - 2i$ 也必是根。加上已给的 $3$，且该三次多项式恰有 $3$ 根，第三根即 $\boxed{1 - 2i}$。

(b) 通过展开求 $a$、$b$、$c$ M1·M1·A1·A1

$$ P(z) \;=\; (z - 3)\bigl(z - (1+2i)\bigr)\bigl(z - (1-2i)\bigr). $$ 先把共轭根对合并为一个实系数二次式 —— 这一步无需计算器： $$ \bigl(z - (1+2i)\bigr)\bigl(z - (1-2i)\bigr) \;=\; z^{2} - \bigl((1+2i)+(1-2i)\bigr)z + (1+2i)(1-2i) \;=\; z^{2} - 2z + (1 - (2i)^{2}) \;=\; z^{2} - 2z + 5. $$ 再乘 $(z - 3)$： $$ P(z) \;=\; (z - 3)(z^{2} - 2z + 5) \;=\; z^{3} - 2z^{2} + 5z - 3z^{2} + 6z - 15 \;=\; z^{3} - 5z^{2} + 11z - 15. $$ 对应系数：$a = -5$、$b = 11$、$c = -15$。

(c) $\mathbb{R}$ 上的因式分解 A1

$$ P(z) \;=\; (z - 3)(z^{2} - 2z + 5). $$ （二次因式判别式 $4 - 20 = -16 < 0$，在 $\mathbb{R}$ 上不可约 —— 与共轭复根对应。）
共轭对模板。实系数多项式的复数根必成共轭对，构成实二次因式 $(z - \alpha)(z - \bar{\alpha}) = z^{2} - 2\Re(\alpha)z + |\alpha|^{2}$。把这个模板记住 —— 只要给出一个复根，无需展开即可直接写出对应实二次因式。本题 $\alpha = 1+2i$，立得 $z^{2} - 2(1)z + (1^{2}+2^{2}) = z^{2} - 2z + 5$。
Q7HARDPaper 2AHL 1.13 Argand Locus[9 marks]

$|z - 2| = |z + 2i|$ — (a) geometric interpretation, (b) algebraic derivation that it's $y = -x$, (c) sketch, (d) closest point to origin.$|z - 2| = |z + 2i|$ —— (a) 几何意义，(b) 代数推导其为 $y = -x$，(c) 草图，(d) 距原点最近的点。

Answers:答案：  (b) Line $y = -x$  ·  (d) $z = 0$ (the origin itself)

(a) Geometric meaning R1·R1

$|z - 2|$ is the distance in the Argand plane from the point $z$ to the point $2$ (the real number $2 + 0i$, plotted at $(2, 0)$). $|z + 2i| = |z - (-2i)|$ is the distance from $z$ to the point $-2i$, plotted at $(0, -2)$. The equation $|z - 2| = |z + 2i|$ says: $z$ is equidistant from $(2, 0)$ and $(0, -2)$ — i.e., $z$ lies on the perpendicular bisector of the segment joining these two points.

(b) Algebraic derivation M1·M1·A1·A1

Set $z = x + iy$ with $x, y \in \mathbb{R}$. Square both sides (both sides are non-negative, so this is reversible): $$ |z - 2|^{2} \;=\; (x - 2)^{2} + y^{2}, \qquad |z + 2i|^{2} \;=\; x^{2} + (y + 2)^{2}. $$ Equate: $$ (x - 2)^{2} + y^{2} \;=\; x^{2} + (y + 2)^{2}. $$ Expand both sides: $$ x^{2} - 4x + 4 + y^{2} \;=\; x^{2} + y^{2} + 4y + 4. $$ Cancel $x^{2} + y^{2} + 4$: $$ -4x \;=\; 4y \quad \Longrightarrow \quad y \;=\; -x. \qquad \square $$

(c) Sketch A1·A1

Plot the line $y = -x$ on the Argand plane (slope $-1$, passes through origin). Mark the two reference points $(2, 0)$ and $(0, -2)$ with circles. The midpoint of the segment joining them is $(1, -1)$, which lies on the line (good sanity check).

(d) Closest point to the origin A1

The line $y = -x$ passes through the origin. So the closest point on the line to the origin is the origin itself: $z = 0$. (Equivalently, the perpendicular from $(0,0)$ to the line has length $0$.) Sanity check: $|0 - 2| = 2$ and $|0 + 2i| = 2$. ✓
Locus — equidistance equals perpendicular bisector. The equation $|z - \alpha| = |z - \beta|$ is the IB-canonical setup for a line — specifically the perpendicular bisector of the segment from $\alpha$ to $\beta$ in the Argand plane. Slight variants give other loci: $|z - \alpha| = k$ is a circle (radius $k$, centre $\alpha$); $|z - \alpha| + |z - \beta| = k$ is an ellipse; $|z - \alpha| - |z - \beta| = k$ is a hyperbola branch. Recognising the locus type from the equation form is the Paper 2 speed-up.

(a) 几何含义 R1·R1

$|z - 2|$ 表示在 Argand 平面上 $z$ 与点 $2$（即实数 $2 + 0i$，对应 $(2, 0)$）之间的距离。 $|z + 2i| = |z - (-2i)|$ 表示 $z$ 与点 $-2i$（对应 $(0, -2)$）之间的距离。 $|z - 2| = |z + 2i|$ 即：$z$ 到 $(2, 0)$ 与到 $(0, -2)$ 的距离相等 —— 故 $z$ 在两点连线段的垂直平分线（perpendicular bisector）上。

(b) 代数推导 M1·M1·A1·A1

设 $z = x + iy$，$x, y \in \mathbb{R}$。两边平方（两边均非负，可逆）： $$ |z - 2|^{2} \;=\; (x - 2)^{2} + y^{2}, \qquad |z + 2i|^{2} \;=\; x^{2} + (y + 2)^{2}. $$ 令其相等： $$ (x - 2)^{2} + y^{2} \;=\; x^{2} + (y + 2)^{2}. $$ 展开： $$ x^{2} - 4x + 4 + y^{2} \;=\; x^{2} + y^{2} + 4y + 4. $$ 消去 $x^{2} + y^{2} + 4$： $$ -4x \;=\; 4y \quad \Longrightarrow \quad y \;=\; -x. \qquad \square $$

(c) 草图 A1·A1

在 Argand 平面上画出直线 $y = -x$（斜率 $-1$，过原点）。用圆圈标出两个参考点 $(2, 0)$ 与 $(0, -2)$。连线段中点 $(1, -1)$ 在直线上（自我验证用）。

(d) 距原点最近的点 A1

直线 $y = -x$ 过原点，故离原点最近的点就是原点本身：$z = 0$。（等价：从 $(0,0)$ 到该直线的垂线长度为 $0$。）验证：$|0 - 2| = 2$，$|0 + 2i| = 2$。 ✓
轨迹 — 等距即垂直平分线。$|z - \alpha| = |z - \beta|$ 是 IB 经典"直线"型轨迹方程 —— 即 Argand 平面上 $\alpha$ 到 $\beta$ 的垂直平分线。略加变形得其它轨迹：$|z - \alpha| = k$ 是圆（半径 $k$，圆心 $\alpha$）；$|z - \alpha| + |z - \beta| = k$ 是椭圆；$|z - \alpha| - |z - \beta| = k$ 是双曲线分支。看方程形式直接识别轨迹类型，是 Paper 2 提速的关键。
PART IV  ·  PAPER 3 — SOLUTIONS第四部分  ·  第三卷 —— 解析Calculator · HL extended exploration · 16 marks可使用计算器 · HL 长题探究 · 16 分

Paper 3 — Worked Solutions第三卷 —— 详细解析

Q8HARDPaper 3AHL 1.14 Roots of Unity & Trig Sums[16 marks]

$\omega = e^{2\pi i/n}$, $n \ge 2$ — develop $\omega^n = 1$ and $\sum \omega^k = 0$, then derive $\sum \cos(2\pi k/n) = 0$, evaluate at $n=5$, and use pairing to find $\cos(2\pi/5) + \cos(4\pi/5)$.$\omega = e^{2\pi i/n}$，$n \ge 2$ —— 推出 $\omega^n = 1$ 与 $\sum \omega^k = 0$，再推 $\sum \cos(2\pi k/n) = 0$，在 $n=5$ 处求值，并用配对求 $\cos(2\pi/5) + \cos(4\pi/5)$。

Answers:答案：  (b)/(c)/(d) Sums all $0$  ·  (e) $-1$  ·  (f) $\cos(2\pi/5) + \cos(4\pi/5) = -\tfrac{1}{2}$

(a) $\omega^{n} = 1$ and distinctness M1·A1·R1

$\omega^{n} = (e^{2\pi i/n})^{n} = e^{2\pi i} = \cos(2\pi) + i\sin(2\pi) = 1$. $\checkmark$ For distinctness: $\omega^{j} = \omega^{k}$ (with $0 \le j, k \le n-1$) iff $e^{2\pi i j/n} = e^{2\pi i k/n}$ iff $\tfrac{2\pi (j - k)}{n} \in 2\pi \mathbb{Z}$ iff $j - k \in n\mathbb{Z}$. Since $|j - k| < n$, the only solution is $j = k$. So the $n$ values $\omega^{0}, \omega^{1}, \ldots, \omega^{n-1}$ are all distinct.

(b) $\sum_{k=0}^{n-1} \omega^{k} = 0$ M1·A1·R1

For $n \ge 2$, $\omega \ne 1$ (since $\tfrac{2\pi}{n} \in (0, 2\pi)$ for $n \ge 2$). Apply the finite geometric series formula: $$ \sum_{k=0}^{n-1} \omega^{k} \;=\; \frac{1 - \omega^{n}}{1 - \omega} \;=\; \frac{1 - 1}{1 - \omega} \;=\; \frac{0}{1 - \omega} \;=\; 0. \qquad \square $$ (The $\omega = 1$ case is excluded by the assumption $n \ge 2$.)

(c) Real part — cosine sum M1·A1·A1

Write each term as $\omega^{k} = e^{2\pi i k/n} = \cos(2\pi k/n) + i\sin(2\pi k/n)$. Taking real parts of $\sum \omega^{k} = 0$: $$ \Re\!\left(\sum_{k=0}^{n-1} \omega^{k}\right) \;=\; \sum_{k=0}^{n-1} \cos\!\left(\tfrac{2\pi k}{n}\right) \;=\; 0. \qquad \square $$

(d) Imaginary part — sine sum M1·A1

$$ \Im\!\left(\sum_{k=0}^{n-1} \omega^{k}\right) \;=\; \sum_{k=0}^{n-1} \sin\!\left(\tfrac{2\pi k}{n}\right) \;=\; 0. $$

(e) Apply at $n = 5$ A1·A1

From (c), $\sum_{k=0}^{4} \cos(\tfrac{2\pi k}{5}) = 0$. The $k = 0$ term is $\cos(0) = 1$, so $$ \cos\!\left(\tfrac{2\pi}{5}\right) + \cos\!\left(\tfrac{4\pi}{5}\right) + \cos\!\left(\tfrac{6\pi}{5}\right) + \cos\!\left(\tfrac{8\pi}{5}\right) \;=\; -1. $$ GDC verification: $0.3090 + (-0.8090) + (-0.8090) + 0.3090 = -1.0000$. ✓

(f) Use the pairing $\cos(2\pi - x) = \cos(x)$ M1·M1·A1

Apply pairing to the higher-argument terms: $$ \cos\!\left(\tfrac{8\pi}{5}\right) \;=\; \cos\!\left(2\pi - \tfrac{8\pi}{5}\right) \;=\; \cos\!\left(\tfrac{2\pi}{5}\right), $$ $$ \cos\!\left(\tfrac{6\pi}{5}\right) \;=\; \cos\!\left(2\pi - \tfrac{6\pi}{5}\right) \;=\; \cos\!\left(\tfrac{4\pi}{5}\right). $$ Substitute into the $n = 5$ identity from (e): $$ -1 \;=\; 2\cos\!\left(\tfrac{2\pi}{5}\right) + 2\cos\!\left(\tfrac{4\pi}{5}\right) \quad \Longrightarrow \quad \cos\!\left(\tfrac{2\pi}{5}\right) + \cos\!\left(\tfrac{4\pi}{5}\right) \;=\; -\tfrac{1}{2}. $$
Why this is famous. The exact value $\cos(2\pi/5) = (\sqrt{5} - 1)/4$ and $\cos(4\pi/5) = -(\sqrt{5} + 1)/4$ are special — they're the algebraic ingredients in the regular pentagon and the golden ratio $\varphi = (1 + \sqrt{5})/2$. The sum $-\tfrac{1}{2}$ comes from collapsing the four-term Q8(e) identity using pairing — a strategy that generalises: any trig sum over a set of arguments closed under $x \mapsto 2\pi - x$ collapses to a sum over half the terms. Paper 3 explorations often build to a $-1/2$ or a $(\sqrt{5} \pm 1)/4$ punchline; recognising the roots-of-unity route is the key unlock.

(a) $\omega^{n} = 1$ 与互不相同 M1·A1·R1

$\omega^{n} = (e^{2\pi i/n})^{n} = e^{2\pi i} = \cos(2\pi) + i\sin(2\pi) = 1$。 $\checkmark$ 互不相同：$\omega^{j} = \omega^{k}$（$0 \le j, k \le n-1$）$\Longleftrightarrow e^{2\pi i j/n} = e^{2\pi i k/n} \Longleftrightarrow \tfrac{2\pi (j - k)}{n} \in 2\pi \mathbb{Z} \Longleftrightarrow j - k \in n\mathbb{Z}$。因 $|j - k| < n$，唯一解是 $j = k$。所以 $\omega^{0}, \omega^{1}, \ldots, \omega^{n-1}$ 这 $n$ 个值彼此不同。

(b) $\sum_{k=0}^{n-1} \omega^{k} = 0$ M1·A1·R1

当 $n \ge 2$ 时 $\omega \ne 1$（因为 $n \ge 2$ 时 $\tfrac{2\pi}{n} \in (0, 2\pi)$）。套用有限项几何级数求和公式： $$ \sum_{k=0}^{n-1} \omega^{k} \;=\; \frac{1 - \omega^{n}}{1 - \omega} \;=\; \frac{1 - 1}{1 - \omega} \;=\; \frac{0}{1 - \omega} \;=\; 0. \qquad \square $$ （$\omega = 1$ 情形被假设 $n \ge 2$ 排除。）

(c) 取实部 —— 余弦求和 M1·A1·A1

把每项写为 $\omega^{k} = e^{2\pi i k/n} = \cos(2\pi k/n) + i\sin(2\pi k/n)$。对 $\sum \omega^{k} = 0$ 取实部： $$ \Re\!\left(\sum_{k=0}^{n-1} \omega^{k}\right) \;=\; \sum_{k=0}^{n-1} \cos\!\left(\tfrac{2\pi k}{n}\right) \;=\; 0. \qquad \square $$

(d) 取虚部 —— 正弦求和 M1·A1

$$ \Im\!\left(\sum_{k=0}^{n-1} \omega^{k}\right) \;=\; \sum_{k=0}^{n-1} \sin\!\left(\tfrac{2\pi k}{n}\right) \;=\; 0. $$

(e) 在 $n = 5$ 上应用 A1·A1

由 (c)，$\sum_{k=0}^{4} \cos(\tfrac{2\pi k}{5}) = 0$。$k = 0$ 项为 $\cos(0) = 1$，故 $$ \cos\!\left(\tfrac{2\pi}{5}\right) + \cos\!\left(\tfrac{4\pi}{5}\right) + \cos\!\left(\tfrac{6\pi}{5}\right) + \cos\!\left(\tfrac{8\pi}{5}\right) \;=\; -1. $$ GDC 验证：$0.3090 + (-0.8090) + (-0.8090) + 0.3090 = -1.0000$。 ✓

(f) 用配对 $\cos(2\pi - x) = \cos(x)$ M1·M1·A1

对高辐角项做配对： $$ \cos\!\left(\tfrac{8\pi}{5}\right) \;=\; \cos\!\left(2\pi - \tfrac{8\pi}{5}\right) \;=\; \cos\!\left(\tfrac{2\pi}{5}\right), $$ $$ \cos\!\left(\tfrac{6\pi}{5}\right) \;=\; \cos\!\left(2\pi - \tfrac{6\pi}{5}\right) \;=\; \cos\!\left(\tfrac{4\pi}{5}\right). $$ 代入 (e) 中 $n = 5$ 的恒等式： $$ -1 \;=\; 2\cos\!\left(\tfrac{2\pi}{5}\right) + 2\cos\!\left(\tfrac{4\pi}{5}\right) \quad \Longrightarrow \quad \cos\!\left(\tfrac{2\pi}{5}\right) + \cos\!\left(\tfrac{4\pi}{5}\right) \;=\; -\tfrac{1}{2}. $$
该结果的著名地位。精确值 $\cos(2\pi/5) = (\sqrt{5} - 1)/4$、$\cos(4\pi/5) = -(\sqrt{5} + 1)/4$ 极为特殊 —— 它们是正五边形与黄金分割比（golden ratio）$\varphi = (1 + \sqrt{5})/2$ 的代数素材。和 $-\tfrac{1}{2}$ 来自 Q8(e) 四项恒等式经配对压缩 —— 该思路可推广：任意关于 $x \mapsto 2\pi - x$ 封闭的辐角集的三角求和都可压缩到一半项。Paper 3 探究题常以 $-1/2$ 或 $(\sqrt{5} \pm 1)/4$ 为收尾结果；识别"单位根路径"是关键。