CAIE 9709 P3 复数 模与辐角 共轭根 轨迹
复数是 A-Level 数学里两极分化最明显的一块：会做的学生觉得它只是把 i 当字母算，不会做的学生在辐角符号和轨迹画图上反复丢分。这里不按课本目录走，按考纲条目和真题里真正出现的形式过一遍，术语中英对照，每类题给一道算完的例子。
术语先对齐，英文题干一出现这些词，考的就是同一套动作：实部 real part、虚部 imaginary part、共轭 conjugate、模 modulus、辐角 argument、极坐标形式 modulus-argument form。
Complex numbers split every A Level cohort into two groups: those who treat i as a letter and finish in three minutes, and those who lose marks on argument signs and locus sketches. What follows is organised by syllabus statements and by the shapes those statements take in real papers, with one worked example per question type and the English terminology kept beside the Chinese.
先确认你的考纲里有没有复数
英国本土的 A-Level Mathematics 三大考纲（AQA 7357、OCR H240、Edexcel 9MA0）都不含复数，复数属于 Further Mathematics。OCR H240 的考纲原文在二次不等式那一条下面直接写了一句 "Quadratic equations with complex roots are excluded"，AQA 把复数放在 Further Maths 7367 的 B 部分（B1 到 B10），Edexcel 放在 Further Maths 的 Core Pure。真正在「数学」而不是「进阶数学」里考复数的是 CAIE 9709：Paper 3 的 3.9 Complex numbers，卷面 75 分、1 小时 50 分钟，占 A Level 的 30%。2026 至 2027 版考纲和 2028 至 2030 版考纲里，3.9 的编号和条目都没有变。所以如果你读的是英国本土 A-Level Maths，看到复数题先确认是不是 Further Maths 的卷子。
If you sit AQA 7357, OCR H240 or Edexcel 9MA0, complex numbers are not on your papers; the topic belongs to Further Mathematics. OCR states it plainly in H240: quadratic equations with complex roots are excluded. AQA places the topic in 7367 Further Mathematics, statements B1 to B10, and Edexcel places it in Core Pure. Cambridge 9709 is the exception, with complex numbers in Paper 3 section 3.9, inside a 75-mark paper lasting 1 hour 50 minutes and worth 30 per cent of the A Level. The 2028 to 2030 syllabus keeps the same numbering as the 2026 to 2027 version, so material written for one works for the other.
i 的定义与实部虚部的读法
整章的起点只有一条定义：i² = -1，其余都是推论。z = x + iy 里 x 是实部 Re z，y 是虚部 Im z，注意 Im z = y 是一个实数，不是 iy。很多学生以为 Im(3 - 4i) 应该写成 -4i，卷面上写成 -4i 就丢分。两个复数相等的条件是实部与虚部分别相等，一条复数方程等于两条实数方程，这个转换是后面待定系数和轨迹题的底层工具。
The whole topic rests on one definition, i² = -1, and everything else is a consequence of it. For z = x + iy the real part is x and the imaginary part is y, and y is a real number rather than iy, which is why writing Im(3 - 4i) = -4i is a familiar slip that costs a mark even when the rest of the working is sound. Two complex numbers are equal only when both parts match separately, so a single complex equation carries two real equations; that translation is the tool behind coefficient matching and locus questions later on.
加减与乘法：i 当字母展开再换掉 i²
加减是实部对实部、虚部对虚部。乘法按分配律展开，展开的过程中 i 照常当字母，收尾时把每一处 i² 统一换成 -1。以 (2 + 3i)(1 - 4i) 为例，展开得 2 - 8i + 3i - 12i²，换掉 i² 后是 2 - 5i + 12，结果是 14 - 5i。常见的错误是展开到一半就把 i² 当 -1 又同时留着一个 i，或者把 3i 乘 (-4i) 算成 -12i。这一步最容易在选择题里白送分。
Addition and subtraction work component by component. Multiplication is an expansion in which i behaves like a letter until the end, when every i² becomes -1: (2 + 3i)(1 - 4i) = 2 - 8i + 3i - 12i² = 2 - 5i + 12 = 14 - 5i. Two errors show up again and again, namely replacing i² halfway through while leaving a stray i behind, and treating 3i times -4i as -12i instead of +12. Both are avoidable by finishing the expansion before any substitution.
i 的正整数次幂按四个一循环：i¹ = i、i² = -1、i³ = -i、i⁴ = 1，所以 i 的 2027 次方等于 i³ = -i（2027 除以 4 余 3）。化简大指数时先做这一步，比逐项展开靠谱得多。
Positive powers of i run in a cycle of four: i, -1, -i and 1. Any large power is therefore reduced by the remainder on division by 4, so i to the power 2027 is i cubed, which is -i. Reducing first also keeps careless sign errors out of longer expansions.
除法：把分母乘成实数
除法的动作只有一个，分子分母同乘分母的共轭。以 (3 + 4i)/(1 - 2i) 为例，分子乘 (1 + 2i) 得 3 + 6i + 4i + 8i² = -5 + 10i，分母乘 (1 + 2i) 得 1 + 4 = 5，结果是 -1 + 2i。CAIE 9709 的 3.9 对这类计算写明 full details of the working should be shown，只给答案拿不到过程分。检查习惯是把结果乘回原分母：(-1 + 2i)(1 - 2i) = -1 + 2i + 2i - 4i² = 3 + 4i，正好回到分子。
Division has exactly one move: multiply top and bottom by the conjugate of the denominator. Taking (3 + 4i)/(1 - 2i), the numerator becomes 3 + 6i + 4i + 8i² = -5 + 10i, the denominator becomes 1 + 4 = 5, and the answer is -1 + 2i. Cambridge asks for full working in these calculations, so a bare answer gives away the method marks. A quick check is to multiply back: (-1 + 2i)(1 - 2i) = 3 + 4i, which is the numerator you started from.
再看一个分母带有 i 的：(2 - i)/(3 + i)，分子分母同乘 (3 - i)，分子得 6 - 2i - 3i + i² = 5 - 5i，分母得 9 + 1 = 10，结果是 1/2 - i/2。分母是纯虚数时（比如 4i），共轭是 -4i，乘上去能把分母变成 16。
A second division follows the same routine: (2 - i)/(3 + i) becomes (2 - i)(3 - i)/10, the numerator collapses to 5 - 5i, and the answer is 1/2 - i/2. A purely imaginary denominator such as 4i is handled with its conjugate -4i, which turns the denominator into 16.
共轭与 z 乘 z* 得 x 平方加 y 平方
z = x + iy 的共轭是 x - iy，CAIE 的记号是 z*，OCR 和 Edexcel 的卷面写成 z 上加横线，在 Argand 图上就是关于实轴做镜像。z 乘 z* 得 x² + y²，这一个恒等式同时撑起三件事：|z|² = z z*、|z| = √(x² + y²)、以及把复分母清成实数。它还有一个反向用法：题目直接给出 z z* 的值，等于给了模长的平方，不必再分别求 x 和 y。
The conjugate of z = x + iy is x - iy, written z* in Cambridge papers and with an overbar elsewhere; on the Argand diagram it is a reflection in the real axis. The product z z* equals x² + y², and that one identity supports three standard results at once: the squared modulus, the modulus formula itself, and the technique for clearing a complex denominator. Read the other way round, a question that hands you z z* has handed you the squared modulus, so there is no need to hunt for x and y separately.
模与辐角：象限决定符号
模是点到原点的距离，|z| = √(x² + y²)，永远非负。辐角是从正实轴逆时针转到该点的角度，先由 tan θ = y/x 求参考角，再按象限修正。CAIE 9709 一般取 (-π, π] 这个区间，题目没指定时 0 到 2π 也接受。以 z = -1 + 根号3 i 为例，点落在第二象限，tan θ = -根号3，参考角是 π/3，于是 arg z = 2π/3，|z| = 2。很多学生把 arctan 弹出的那个数直接抄成答案，第二、三象限的题一半错在这里。
第三象限同理，只是符号反过来：z = -2 - 2i 的 tan θ = 1，参考角 π/4，但点在第三象限，arg z = -3π/4（写 5π/4 也接受），|z| = 2√2。凡是 tan 算出正值而实部为负的情形，答案必须落在 (-π, -π/2) 这一段，画个草图再落笔比背口诀稳。
The third quadrant works the same way with signs flipped. For z = -2 - 2i, tan θ = 1 gives a reference angle of π/4, but the point sits in the third quadrant, so arg z = -3π/4, with 5π/4 accepted as well, and |z| = 2√2. Whenever tan comes out positive while the real part is negative, the answer has to sit between -π and -π/2.
The modulus is the distance from the origin and is never negative: |z| = √(x² + y²). The argument is the anticlockwise angle from the positive real axis, found by taking the reference angle from tan θ = y/x and then adjusting for the quadrant. Cambridge normally works in the interval (-π, π], and an answer between 0 and 2π is accepted unless the question says otherwise, with the quadrant check deciding the rest. For z = -1 + √3 i the point lies in the second quadrant, tan θ = -√3 gives a reference angle of π/3, so arg z = 2π/3 with |z| = 2. Copying the raw arctan value straight from the calculator is where second-quadrant questions are lost.
极坐标形式：模相乘、辐角相加
写成 r(cos θ + i sin θ) 或 r e 的 iθ 次方之后，乘除规则变得很短：|z1 z2| = |z1||z2|，arg(z1 z2) = arg z1 + arg z2，除法对应相除与相减。设 z1 = 2e 的 iπ/3 次方、z2 = 4e 的 iπ/4 次方，乘积模长 8、辐角 7π/12，商模长 1/2、辐角 π/12。几何上这是放大加旋转：乘 2e 的 iπ/3 次方，等于把图形拉长两倍再逆时针转 60 度。CAIE 3.9 明确要求掌握乘除的几何效果，只算数值不够。
从笛卡尔形式换到极坐标这一步要熟到不用想：z = 1 + i 的 |z| = √2，arg z = π/4，所以写成 √2(cos π/4 + i sin π/4) 或 √2 e 的 iπ/4 次方。换算式子本身只有两条，模和辐角，但落笔前一定先看象限，否则 π/4 会变成 5π/4 这类错误。
Converting from Cartesian form should be automatic. For z = 1 + i the modulus is √2 and the argument is π/4, so the polar form is √2(cos π/4 + i sin π/4), also written √2 e^(iπ/4). Only two quantities change, yet the quadrant check has to happen before writing them down.
Once a number is written as r(cos θ + i sin θ) or r e^(iθ), multiplication becomes short: moduli multiply and arguments add, with division doing the reverse. With z1 = 2e^(iπ/3) and z2 = 4e^(iπ/4), the product has modulus 8 and argument 7π/12, while the quotient has modulus 1/2 and argument π/12. The geometric reading is an enlargement combined with a rotation, because multiplying by 2e^(iπ/3) stretches any figure by 2 and turns it 60 degrees anticlockwise. Cambridge wants that reading, not only the arithmetic.
实系数方程：非实根成对出现
实系数多项式方程如果有非实根 1 + 2i，就一定有 1 - 2i，两条根合成一个实系数二次因子 z² - 2z + 5。三次方程给了一个复根，就用这个因子做一次多项式除法，剩下的必是一次因子，第三个根是实数。以 z³ - 5z² + 11z - 15 = 0 为例，除以 z² - 2z + 5 得 z - 3，三个根就是 1 ± 2i 和 3。条件里「实系数」三个字不能省：系数本身是复数时，这个配对结论直接失效。
If a polynomial with real coefficients has the non-real root 1 + 2i, then 1 - 2i is a root too, and the pair multiplies out to the real quadratic factor z² - 2z + 5. Given one complex root of a cubic, divide by that factor; what remains is linear and its root is real. For z³ - 5z² + 11z - 15 = 0, division by z² - 2z + 5 leaves z - 3, so the three roots are 1 ± 2i and 3. The real coefficients condition does the whole job, since the pairing argument fails as soon as the coefficients are complex.
轨迹：三种式子对应三条图形
|z - a| = k 是以 a 为圆心、k 为半径的圆；|z - a| = |z - b| 是 a、b 两点的中垂线；arg(z - a) = α 是从 a 出发、方向角 α 的射线，起点 a 不含在内。求 |z - 2| = |z - 4i| 时，把 z = x + iy 代入两边平方，得 (x - 2)² + y² = x² + (y - 4)²，展开后化简为 y = x/2 + 3/2。常见的错误是把两点的坐标直接塞进圆方程，或者把中垂线画成连接 a、b 的那条直线。
不等式对应的是区域而不是线条：|z - a| ≤ k 是实线边界的圆盘；arg(z - a) = α 配合模的范围，常用来问最大与最小模长，做法是过点 a 作方向角 α 的直线，取它与圆的两个交点，较远的给最大模、较近的给最小模。画图时把圆心、半径、这条直线三样标出来，剩下的只是量线段长度。
Inequalities describe regions rather than curves: |z - a| ≤ k is a disc with a solid boundary. Questions asking for the greatest or least modulus are answered by drawing the line through a at the given angle and reading its two intersections with the circle, the farther one giving the maximum and the nearer one the minimum. Marking centre, radius and that line on the sketch reduces the rest to measuring lengths.
Three shapes cover most loci questions. |z - a| = k is a circle with centre a and radius k; |z - a| = |z - b| is the perpendicular bisector of the two points; arg(z - a) = α is a half-line from a at angle α, with a itself excluded. For |z - 2| = |z - 4i|, substitute z = x + iy and square both sides to get (x - 2)² + y² = x² + (y - 4)², which simplifies to y = x/2 + 3/2. The two slips that recur are dropping the coordinates into a circle equation, and drawing the line joining a and b instead of its bisector.
开平方：待定系数比套公式可靠
求 5 + 12i 的两个平方根，设 (a + bi)² = 5 + 12i，展开并比较实虚部得 a² - b² = 5、2ab = 12。由第二式 b = 6/a，代入第一式变成 a² - 36/a² = 5，令 u = a² 解 u² - 5u - 36 = 0，得 u = 9 或 u = -4；a² 不能为负，于是 a = ±3，b 与 a 同号取 ±2，两个根是 ±(3 + 2i)。CAIE 3.9 把这条单列，并要求 exact Cartesian form，也就是保留根号与整数的写法，不给小数。
最后一步别省：把 (3 + 2i) 平方回去检验，9 + 12i + 4i² = 5 + 12i，回到原数才算做完。负号那一支也要写，两个平方根是成对出现的，只写 +3 + 2i 少一半分。
One habit closes the question: square (3 + 2i) back and check that 9 + 12i + 4i² returns 5 + 12i. Both signs belong in the answer, since a square root question always has two roots and giving only +3 + 2i loses half the marks.
To find the two square roots of 5 + 12i, set (a + bi)² = 5 + 12i and compare parts: a² - b² = 5 and 2ab = 12. The second equation gives b = 6/a, and substituting turns the first into a² - 36/a² = 5. Writing u = a², the quadratic u² - 5u - 36 = 0 gives u = 9 or u = -4, and only a² = 9 is usable, so a = ±3 with b carrying the same sign, namely ±2. The two roots are ±(3 + 2i), and Cambridge expects exact Cartesian form rather than decimals.
复数方程：一条式子拆两条实数式
z 以一次形式出现时，最快的一条路是把除法做完直接解。以 (1 + i)z = 3 + i 为例，z = (3 + i)/(1 + i)，分子分母同乘 (1 - i)，分子得 3 - 3i + i - i² = 4 - 2i，分母得 1 + 1 = 2，于是 z = 2 - i，代回检验 (1 + i)(2 - i) = 3 + i 成立。若题目要求待定的量本身是实数（比如求实数 p、q，使 p + qi 满足某个方程），就走比较实虚部那条路：把方程两边整理成 A + Bi 的形式，再分别令实部与虚部相等。
When z appears to the first power, finish the division and solve. For (1 + i)z = 3 + i, write z = (3 + i)/(1 + i), multiply by (1 - i), and the numerator gives 4 - 2i over a denominator of 2, so z = 2 - i, which checks back to 3 + i. When the unknowns are declared real, such as finding real p and q that make p + qi satisfy an equation, compare parts instead: put each side in the form A + Bi and equate real with real and imaginary with imaginary.
卷面上的先后顺序与验算方式
Paper 3 是 75 分、9 到 11 道结构题（9709 考纲的评述部分），复数多数时候只占其中一道的一两小问，题型固定，适合放在前面先做。顺序上先处理纯计算（乘法、除法、模、辐角），再做方程和轨迹，因为后者的中间结果都来自前者。验算最有效的一步是代回原式：除法乘回分母，平方根平方回去，轨迹取两个特殊点各查一次。de Moivre 定理不必在这里花时间，它属于 Further Maths，9709 的 3.9 只要求到极坐标形式的乘除与开平方。
写完答案再花十秒看形式：模是否非负、辐角是否落在题目指定的区间、分母是否已经清成实数、平方根是否两支齐全。这四问不占时间，考场上真正丢掉的粗心错基本都落在这四条上。
Paper 3 carries 75 marks across 9 to 11 structured questions according to the syllabus assessment outline, and complex numbers normally occupy one part of one question. The pattern is stable, which makes the topic a good one to bank early in the paper. Do the pure calculation first, multiplication, division, modulus and argument, then equations and loci, because their intermediate values come from the earlier work. Substitution is the check that catches most errors: multiply a division back by its denominator, square a square root answer, and test a locus with two convenient points. De Moivre's theorem is not needed here; it belongs to Further Mathematics, and section 3.9 stops at multiplication, division and square roots in polar form.
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