📚 AS CCEA Further Mathematics: Formula & Theorem Quick Reference | AS CCEA 进阶数学：公式定理速查手册
This quick-reference handbook brings together the formulas and theorems that matter most for the CCEA AS Further Mathematics course. Each section states the core results precisely, shows how they are actually used in exam questions, and pairs every English statement with a Chinese equivalent so that you can revise efficiently in either language.
本速查手册汇总了 CCEA AS 进阶数学课程最核心的公式与定理。每节都精确陈述关键结论，说明它们在考题中的实际用法，并把每一条英文表述配以中文对照，方便你以任意一种语言高效复习。
Use it as a checklist rather than a textbook: cover the Chinese column, reproduce the English result from memory, then check yourself. The final section lists the mistakes that cost candidates the most marks.
请把它当作检查清单而非教科书使用：遮住中文部分，凭记忆复现英文结论，再自行核对。最后一节列出考生失分最多的典型错误。
1. Complex Numbers: Cartesian Form and the Conjugate | 复数：笛卡儿形式与共轭
A complex number can be expressed in Cartesian form as z = x + iy, where x and y are real numbers and i² = −1. The real part is Re(z) = x and the imaginary part is Im(z) = y.
复数可表示为笛卡儿形式 z = x + iy，其中 x, y 为实数，且 i² = −1。实部 Re(z) = x，虚部 Im(z) = y。
₂ + iy₂ implies x₁ = x₂ and y₁ = y₂. Addition and subtraction work component by component.
两个复数相等，当且仅当实部相等且虚部相等，因此 x₁ + iy₁ = x₂ + iy₂ 意味着 x₁ = x₂ 且 y₁ = y₂。加法与减法按分量分别进行。
The complex conjugate is z* = x − iy. The product z z* = x² + y² is always real and gives |z| distributes over addition and multiplication: (z₁ + z₂)* = z₁* + z₂* and (z₁z₂)* = z₁*z₂*.
共轭复数为 z* = x − iy。乘积 z z* = x² + y² 恒为实数，即为 |z|²。共轭对加法与乘法都可分配：(z₁ + z₂)* = z₁* + z₂*，(z₁z₂)* = z₁*z₂*。
Multiplication is carried out by expanding brackets and replacing i² with −1. For division, multiply numerator and denominator by the conjugate of the denominator so that the denominator becomes real.
乘法通过展开括号并把 i² 替换为 −1 完成。除法时，把分子与分母同乘分母的共轭，使分母化为实数。
- Equality: real parts equal and imaginary parts equal. / 相等：实部相等且虚部相等。
- Conjugate: z* = x − iy and (z*)* = z. / 共轭：z* = x − iy，且 (z*)* = z。
- Division: multiply top and bottom by the conjugate of the denominator. / 除法：分子分母同乘分母的共轭。
- Powers of i cycle with period 4: i¹ = i, i² = −1, i³ = −i, i⁴ = 1. / i 的幂以 4 为周期循环：i¹ = i，i² = −1，i³ = −i，i⁴ = 1。
2. Modulus, Argument and the Argand Diagram | 模、辐角与阿冈图
The modulus of z = x + iy is |z| = √x² + y², the distance of the point (x, y) from the origin on an Argand diagram. It satisfies |z₁z₂| = |z₁||z₂| and |z₁ ÷ z₂| = |z₁| ÷ |z₂| for z₂ ≠ 0.
z = x + iy 的模为 |z| = √x² + y²，即阿冈图上点 (x, y) 到原点的距离。它满足 |z₁z₂| = |z₁||z₂|，且当 z₂ ≠ 0 时 |z₁ ÷ z₂| = |z₁| ÷ |z₂|。
The argument is θ = arg(z) with tan θ = y/x aken as −π < θ ≤ π, and you must check the quadrant rather than trusting the calculator's inverse tangent.
辐角为 θ = arg(z)，满足 tan θ = y/x，并需根据象限调整。主辐角通常取 −π < θ ≤ π；必须自行判断象限，而不能直接照搬计算器的反正切值。
Polar form writes the same number as z = r(cos θ + i sin θ) with r = |z| he moduli and adds the arguments; division divides the moduli and subtracts the arguments.
极坐标形式把同一个数写成 z = r(cos θ + i sin θ)，其中 r = |z|，θ = arg(z)。乘法即模相乘、辐角相加；除法即模相除、辐角相减。
modulo 2π). Geometrically, multiplying by i rotates a point through 90° anticlockwise about the origin.
因此 arg(z₁z₂) = arg z₁ + arg z₂（模 2π），arg(z₁ ÷ z₂) = arg z₁ − arg z₂（模 2π）。从几何上看，乘以 i 相当于把点绕原点逆时针旋转 90°。
3. Euler's Formula and De Moivre's Theorem | 欧拉公式与棣莫弗定理
Euler's formula is eiθ = cos θ + i sin θ. Replacing θ by −θ gives e−iθ r subtracting the two results yields the exponential forms of the trigonometric functions.
欧拉公式为 eiθ = cos θ + i sin θ。把 θ 换成 −θ 得 e−iθ = cos θ − i sin θ；两式相加或相减即得三角函数的指数形式。
cos θ = (eiθ + e−iθ) ÷ 2 and sin θ = (eiθ − e−iθ) ÷ (2i). These are the standard tools for expressing powers of cos θ and sin θ as sums of multiple angles, and vice versa.
cos θ = (eiθ + e−iθ) ÷ 2，sin θ = (eiθ − e−iθ) ÷ (2i)。这是把 cos θ、sin θ 的幂表示成倍角之和（以及反向操作）的标准工具。
De Moivre's theorem states that (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ for integers n, and the result extends to rational n when the argument is handled with care about multiple values.
棣莫弗定理指出：对整数 n，(cos θ + i sin θ)ⁿ = cos nθ + i sin nθ；当 n 为有理数时，只要妥善处理辐角的多值性，结论同样成立。
For a general complex number the theorem reads zⁿ = rⁿ(cos nθ + i sin nθ). Setting the two sides of (cos θ + i sin θ)ⁿ equal and comparing real and imaginary parts is the fastest route to formulas for cos nθ and sin nθ.
对一般复数，该定理写作 zⁿ = rⁿ(cos nθ + i sin nθ)。把 (cos θ + i sin θ)ⁿ 两边展开并比较实部与虚部，是推导 cos nθ 与 sin nθ 公式最快的途径。
The n distinct nth roots of z = r(cos θ + i sin θ) are given by z1/n = r1/n[cos((θ + 2kπ) ÷ n) + i sin((θ + 2kπ) ÷ n)] for k = 0, 1, 2, …, n − 1. Their images form the vertices of a regular n-gon centred on the origin.
z = r(cos θ + i sin θ) 的 n 个互不相同的 n 次根为 z1/n = r1/n[cos((θ + 2kπ) ÷ n) + i sin((θ + 2kπ) ÷ n)]，其中 k = 0, 1, 2, …, n − 1。它们在阿冈图上构成以原点为中心的正 n 边形顶点。
4. Complex Roots of Polynomial Equations | 多项式方程的复根
also a root, and (z − a − ib)(z − a + ib) = z² − 2az + a² + b² is a real quadratic factor.
对于实系数多项式方程，非实根必以共轭对出现。若 z = a + ib（b ≠ 0）是根，则 z* = a − ib 也是根，且 (z − a − ib)(z − a + ib) = z² − 2az + a² + b² 是实的二次因式。
Consequently a cubic with real coefficients has at least one real root, and a quartic can have four real roots, two real and one conjugate pair, or two conjugate pairs, depending on how many times the curve meets the x-axis.
因此，实系数三次方程至少有一个实根；四次方程可能有四个实根、两个实根加一对共轭复根，或两对共轭复根，取决于曲线与 x 轴相交的次数。
The relationship between roots and coefficients (Vieta's formulas) for az² + bz + c = 0 gives α + β = −b ÷ a and αβ = c ÷ a. For az³ + bz² + cz + d = 0, α + β + γ = −b ÷ a, αβ + βγ + γα = c ÷ a and αβγ = −d ÷ a.
对 az² + bz + c = 0，根与系数的关系（韦达定理）为 α + β = −b ÷ a，αβ = c ÷ a。对 az³ + bz² + cz + d = 0，α + β + γ = −b ÷
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