📚 Complex Functions | 复变函数考点精讲
Complex functions build directly on the idea of extending real numbers to the complex plane. In CCEA A‑Level Mathematics, you are expected to work fluently with complex numbers in Cartesian, polar and exponential forms, apply De Moivre’s theorem to find powers and roots, and solve polynomial equations whose coefficients are real but whose roots may be complex. This revision guide covers every key topic, pairing conceptual explanations with worked‑style reasoning to help you master the exam.
复变函数建立在将实数扩展到复平面的基础之上。在 CCEA A‑Level 数学中，你需要熟练掌握复数的直角坐标、极坐标与指数形式，运用棣莫弗定理求幂与方根，并能求解系数为实数但根可能为复数的多项式方程。这份考点精讲覆盖每一个核心主题，将概念讲解与推演过程配对呈现，帮助你彻底掌握考试要点。
1. Introduction to Complex Numbers | 复数简介
A complex number is written as z = x + iy, where x and y are real numbers, and i is the imaginary unit satisfying i² = –1. The real part Re(z) = x and the imaginary part Im(z) = y. Complex numbers allow us to solve equations such as z² + 1 = 0, which have no real solutions.
复数写作 z = x + iy，其中 x 与 y 为实数，i 是满足 i² = –1 的虚数单位。实部 Re(z) = x，虚部 Im(z) = y。复数使我们能够求解诸如 z² + 1 = 0 这样无实数解的方程。
Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. Addition and subtraction are performed component‑wise: (a + ib) ± (c + id) = (a ± c) + i(b ± d).
两个复数相等当且仅当它们的实部相等且虚部相等。加减法按分量分别进行：(a + ib) ± (c + id) = (a ± c) + i(b ± d)。
2. Argand Diagram and Geometric Representation | Argand图与几何表示
An Argand diagram represents the complex number z = x + iy as a point (x, y) in a plane, with the x‑axis as the real axis and the y‑axis as the imaginary axis. It transforms algebraic operations into geometric movements.
Argand 图将复数 z = x + iy 表示为平面上的点 (x, y)，x 轴为实轴，y 轴为虚轴。它把代数运算转化为几何移动。
For example, addition corresponds to vector addition using the parallelogram law. The negative of z is a reflection through the origin, and the complex conjugate z̅ = x – iy is a reflection in the real axis.
例如，加法对应向量加法的平行四边形法则。z 的相反数是通过原点的反射，而复共轭 z̅ = x – iy 是关于实轴的反射。
3. Modulus and Argument | 模与辐角
The modulus |z| of a complex number z = x + iy is its distance from the origin: |z| = √(x² + y²). The argument arg(z) is the angle θ that the line joining the origin to (x, y) makes with the positive real axis, usually measured in radians between –π and π (principal argument).
复数 z = x + iy 的模 |z| 是它到原点的距离：|z| = √(x² + y²)。辐角 arg(z) 是从原点到 (x, y) 的连线与正实轴之间的夹角 θ，通常以弧度为单位，取 –π 到 π 之间的主值。
Always draw a quick Argand sketch to determine the correct quadrant for θ. The argument satisfies tan θ = y/x, but direct use of arctan requires careful adjustment of ±π for Quadrants II and III.
务必快速画出 Argand 草图来确定 θ 的正确象限。辐角满足 tan θ = y/x，但直接使用 arctan 时需注意在第二、三象限作 ±π 的调整。
4. Polar Form of Complex Numbers | 复数的极坐标形式
A complex number can be expressed in polar form as z = r(cos θ + i sin θ), where r = |z| and θ = arg(z). This representation separates the magnitude from the direction, making multiplication and division far simpler.
复数可用极坐标形式表示为 z = r(cos θ + i sin θ)，其中 r = |z|，θ = arg(z)。这种表示将大小与方向分离，使乘除法变得极其简便。
| Cartesian form | Polar form |
|---|---|
| z = x + iy | z = r(cos θ + i sin θ) |
| x, y real | r ≥ 0, θ ∈ ℝ (usually –π < θ ≤ π) |
To convert from Cartesian to polar, compute r = √(x² + y²) and θ = arctan(y/x) with quadrant checks. Opposite conversion uses x = r cos θ, y = r sin θ.
从直角坐标转换到极坐标时，计算 r = √(x² + y²) 和 θ = arctan(y/x) 并检查象限。反向转换则使用 x = r cos θ，y = r sin θ。
5. Euler’s Formula and Exponential Form | 欧拉公式与指数形式
Euler’s formula states that eⁱᶿ = cos θ + i sin θ. This identity allows us to write any complex number in exponential form: z = r eⁱᶿ. The exponential form is compact and reveals why multiplication and division rules work so neatly.
欧拉公式 eⁱᶿ = cos θ + i sin θ 使我们可以将任意复数写成指数形式：z = r eⁱᶿ。指数形式简洁明了，揭示了乘除法规则为何如此整齐。
In CCEA exams, you are often expected to switch between polar, exponential and Cartesian forms fluently. Memorising key values such as eⁱπ/² = i, eⁱπ = –1 and e²ⁱπ = 1 is very useful.
在 CCEA 考试中，常要求你熟练地在极坐标、指数和直角坐标形式之间转换。牢记 eⁱπ/² = i、eⁱπ = –1 和 e²ⁱπ = 1 等关键值非常有帮助。
6. Multiplication and Division in Polar Form | 极坐标形式的乘除法
If z₁ = r₁ (cos θ₁ + i sin θ₁) and z₂ = r₂ (cos θ₂ + i sin θ₂), then:
若 z₁ = r₁ (cos θ₁ + i sin θ₁)，z₂ = r₂ (cos θ₂ + i sin θ₂)，则：
z₁ z₂ = r₁ r₂ [cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)]
z₁ / z₂ = r₁/r₂ [cos(θ₁ – θ₂) + i sin(θ₁ – θ₂)], r₂ ≠ 0
This means multiply the moduli and add the arguments for a product; divide the moduli and subtract the arguments for a quotient. The geometric interpretation — scaling and rotation — often appears in locus problems.
这意味着乘积的模相乘、辐角相加；商的模相除、辐角相减。其几何解释——伸缩与旋转——经常出现在轨迹问题中。
7. De Moivre’s Theorem | 棣莫弗定理
De Moivre’s theorem states that for any integer n, (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ). It is a direct consequence of Euler’s formula and is used to find powers of complex numbers and to derive trigonometric identities.
棣莫弗定理指出，对于任意整数 n，有 (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)。这是欧拉公式的直接推论，用于求复数的幂和推导三角恒等式。
When n is negative, write (cos θ + i sin θ)⁻ⁿ = cos(–nθ) + i sin(–nθ) = cos(nθ) – i sin(nθ). The theorem also extends to rational exponents when finding roots, but careful handling of multiple values is required.
当 n 为负时，可写成 (cos θ + i sin θ)⁻ⁿ = cos(–nθ) + i sin(–nθ) = cos(nθ) – i sin(nθ)。该定理也可推广至有理指数以求根，但需小心处理多值性。
8. Powers of Complex Numbers | 复数的幂运算
To raise a complex number to an integer power, first express it in polar or exponential form, then apply De Moivre’s theorem. For example, to compute (1 + i√3)⁵, write 1 + i√3 = 2(cos(π/3) + i sin(π/3)), then raise the modulus to the fifth power and multiply the argument by 5.
要计算复数的整数次幂，首先将其表示为极坐标或指数形式，然后应用棣莫弗定理。例如，计算 (1 + i√3)⁵，先写成 1 + i√3 = 2(cos(π/3) + i sin(π/3))，再对模进行 5 次方，并将辐角乘以 5。
Bracketing the final argument back into the principal range (–π, π] is often required. Add or subtract multiples of 2π to obtain the principal argument after multiplication.
通常需要将最终辐角调整回主值范围 (–π, π]。乘以指数后可通过加减 2π 的整数倍来得到主辐角。
9. Roots of Complex Numbers | 复数的根
To find the n distinct nth roots of a complex number z = r(cos θ + i sin θ), use the formula:
要求复数 z = r(cos θ + i sin θ) 的 n 个不同的 n 次方根，使用公式：
zk = r1/n [cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)], k = 0, 1, 2, …, n–1
These n roots lie equally spaced on a circle of radius r1/n centred at the origin, forming a regular n‑gon. The principal root corresponds to k = 0.
这 n 个根均匀分布在以原点为中心、半径为 r1/n 的圆上，构成正 n 边形。主根对应 k = 0。
Solving equations like zⁿ = w often reduces to finding all nth roots of w. When w is a real negative number, remember to express it with argument π before applying the formula.
求解 zⁿ = w 类方程通常归结为求 w 的所有 n 次方根。当 w 为负实数时，记住先用辐角 π 表示再代入公式。
10. Complex Conjugates and Equations | 复共轭与方程
The complex conjugate of z = x + iy is z̅ = x – iy. Geometrically it reflects the point across the real axis. Key properties include:
z = x + iy 的复共轭为 z̅ = x – iy，几何上表示将点关于实轴反射。重要性质包括：
- z z̅ = |z|² (always real)
- z + z̅ = 2 Re(z), z – z̅ = 2i Im(z)
- For a polynomial with real coefficients, non‑real roots occur in conjugate pairs.
- z z̅ = |z|²（恒为实数）
- z + z̅ = 2 Re(z), z – z̅ = 2i Im(z)
- 对于实系数多项式，非实根以共轭对出现。
If a polynomial has real coefficients and you know one complex root a + ib, then a – ib is also a root. This fact drastically reduces the work needed to factorise or solve equations.
若一个多项式具有实系数，且已知一个复根 a + ib，则 a – ib 也是根。这一事实可极大地减少因式分解或解方程的工作量。
11. Loci in the Complex Plane | 复平面上的轨迹
Locus problems ask you to sketch or interpret sets of complex numbers satisfying a given condition. Common forms include:
轨迹问题要求你描绘或解释满足给定条件的复数集。常见形式包括：
|z – a| = r describes a circle centred at a with radius r.
|z – a| = r 表示以 a 为圆心、半径为 r 的圆。
|z – a| = |z – b| describes the perpendicular bisector of the line segment joining a and b.
|z – a| = |z – b| 表示连接 a 和 b 的线段的垂直平分线。
arg(z – a) = α describes a half‑line (ray) starting at a but not including a, making angle α with the positive real axis.
arg(z – a) = α 表示从 a 出发（不含 a）、与正实轴夹角为 α 的半射线。
You may be asked to find the Cartesian equation of a locus. Substitute z = x + iy and simplify moduli or arguments using algebraic and trigonometric identities.
你可能会被要求求出轨迹的直角坐标方程。代入 z = x + iy，然后借助代数和三角恒等式简化模或辐角条件。
12. Solving Polynomial Equations with Complex Roots | 解含复数根的多项式方程
CCEA problems frequently involve cubic or quartic equations with real coefficients. If one complex root is given or found, its conjugate is also a root. Use the factor theorem to construct a real quadratic factor from the conjugate pair, then divide the polynomial to find other roots.
CCEA 试题常涉及实系数的三次或四次方程。如果给定或求出了一个复根，则其共轭也是根。利用因式定理，由共轭对构造出一个实二次因式，再对多项式做除法以求出其他根。
For example, if 2 + i is a root of a real cubic, then 2 – i is also a root. Their product (z – (2 + i))(z – (2 – i)) = (z – 2)² + 1 = z² – 4z + 5 is a real quadratic factor. Long division or comparing coefficients yields the remaining real root.
例如，若 2 + i 是一个实系数三次方程的根，则 2 – i 也是根。它们的乘积 (z – (2 + i))(z – (2 – i)) = (z – 2)² + 1 = z² – 4z + 5 是一个实二次因式。通过长除法或比较系数可得到剩下的实根。
When the polynomial has complex coefficients, the conjugate root theorem does not apply. In such cases, you must use algebraic manipulation or substitution and equate real and imaginary parts.
当多项式的系数为复数时，共轭根定理不适用。这种情况下必须使用代数操作或代入后令实部、虚部分别相等来求解。
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