Edexcel Maths: Complex Numbers | Edexcel 数学：复数考点精讲

📚 Edexcel Maths: Complex Numbers | Edexcel 数学：复数考点精讲

Complex numbers extend the real number system by introducing the imaginary unit i, where i² = -1. In the Edexcel A Level Further Mathematics syllabus, mastering complex numbers is essential for topics such as roots of polynomials, trigonometric identities, and coordinate geometry. This revision guide breaks down the key concepts into clear, bilingual sections to help you understand the algebra, geometry, and applications of complex numbers. Each point is explained in English and Chinese to support dual‑language learners at aleveler.com.

复数通过引入虚数单位 i（满足 i² = -1）将实数系进行了扩展。在 Edexcel A Level 进阶数学的考纲中，掌握复数是多项式求根、三角恒等式及坐标几何等内容的基础。本考点精讲将核心概念拆解为清晰的双语小节，帮助学习者理解复数的代数、几何以及应用。每个要点均提供英文与中文对照解释，以服务于 aleveler.com 的双语学习者。

1. Imaginary Unit and Complex Numbers | 虚数单位与复数

The imaginary unit i is defined as √(-1), giving i² = -1. A complex number has the form z = a + bi, where a and b are real numbers. The real part is Re(z) = a, and the imaginary part is Im(z) = b. When b = 0, z is real; when a = 0, z is purely imaginary.

虚数单位 i 定义为 √(-1)，因此 i² = -1。复数具有形式 z = a + bi，其中 a 与 b 为实数。实部记为 Re(z) = a，虚部记为 Im(z) = b。当 b = 0 时 z 为实数；当 a = 0 时 z 为纯虚数。

Equality of complex numbers: a + bi = c + di ⇔ a = c and b = d. Two complex numbers are equal only when both their real and imaginary parts are equal.

复数相等：a + bi = c + di 等价于 a = c 且 b = d。两个复数相等当且仅当实部与虚部分别对应相等。

2. Conjugate and Modulus | 共轭与模

The complex conjugate of z = a + bi is z* = a – bi (also denoted as z̅). The conjugate reflects the number across the real axis. Key properties: (z*)* = z, z + z* = 2a, z – z* = 2bi, and zz* = a² + b², which is always a non‑negative real number.

z = a + bi 的复共轭为 z* = a – bi（也记作 z̅）。共轭在复平面上表示关于实轴的反射。关键性质：(z*)* = z，z + z* = 2a，z – z* = 2bi，zz* = a² + b²，它始终是一个非负实数。

The modulus of z, written |z|, is the distance from the origin: |z| = √(a² + b²). It satisfies |z|² = zz*, |z₁z₂| = |z₁||z₂|, and the triangle inequality |z₁ + z₂| ≤ |z₁| + |z₂|.

z 的模记作 |z|，表示复平面上到原点的距离：|z| = √(a² + b²)。模满足：|z|² = zz*，|z₁z₂| = |z₁||z₂|，以及三角不等式 |z₁ + z₂| ≤ |z₁| + |z₂|。

3. Argument and Principal Argument | 辐角与主辐角

The argument of a non‑zero complex number z = a + bi, denoted arg(z), is the angle θ that the vector from the origin to (a, b) makes with the positive real axis. It is usually measured in radians. The principal argument, Arg(z), lies in the interval (-π, π]. For a > 0, Arg(z) = arctan(b/a); adjust by ±π for a < 0.

非零复数 z = a + bi 的辐角记作 arg(z)，是指从原点到 (a, b) 的向量与正实轴之间的夹角 θ，通常以弧度为单位。主辐角 Arg(z) 的取值范围为 (-π, π]。当 a > 0 时，Arg(z) = arctan(b/a)；当 a < 0 时需加上或减去 π 进行调整。

For example, z = -1 – i has Arg(z) = -3π/4, because the point lies in the third quadrant. Remember that arguments differing by 2π represent the same direction.

例如，z = -1 – i 的主辐角为 -3π/4，因为该点位于第三象限。注意相差 2π 整数倍的辐角表示相同的方向。

4. Polar Form and Euler’s Formula | 极形式与欧拉公式

A complex number can be expressed in polar form as z = r(cos θ + i sin θ), where r = |z| and θ = Arg(z). Using Euler’s formula, e^(iθ) = cos θ + i sin θ, we obtain the compact exponential form z = r e^(iθ). This form is extremely useful for multiplication, division, and powers.

复数可以表示为极形式 z = r(cos θ + i sin θ)，其中 r = |z|，θ = Arg(z)。利用欧拉公式 e^(iθ) = cos θ + i sin θ，可以得到简洁的指数形式 z = r e^(iθ)。这种形式在乘法、除法和乘方运算中极为有用。

When converting between forms, r = √(a² + b²) and θ satisfies tan θ = b/a with the correct quadrant. The exponential form highlights that multiplying a complex number by e^(iθ) rotates it by angle θ.

在进行形式转换时，r = √(a² + b²)，而 θ 满足 tan θ = b/a 并需选取正确的象限。指数形式清楚地表明，将一个复数乘以 e^(iθ) 会使它旋转角度 θ。

5. Multiplication and Division in Polar Form | 极形式乘法与除法

If z₁ = r₁ e^(iθ₁) and z₂ = r₂ e^(iθ₂), then z₁z₂ = r₁r₂ e^(i(θ₁+θ₂)). The moduli multiply and the arguments add. For division, z₁/z₂ = (r₁/r₂) e^(i(θ₁-θ₂)). Moduli divide and arguments subtract.

若 z₁ = r₁ e^(iθ₁) 且 z₂ = r₂ e^(iθ₂)，则 z₁z₂ = r₁r₂ e^(i(θ₁+θ₂))。模相乘，辐角相加。除法时，z₁/z₂ = (r₁/r₂) e^(i(θ₁-θ₂))，模相除，辐角相减。

Geometrically, multiplying by z₂ scales the vector z₁ by |z₂| and rotates it by Arg(z₂). Dividing by z₂ scales by 1/|z₂| and rotates by -Arg(z₂). This provides a powerful way to describe transformations in the complex plane.

几何上，乘以 z₂ 会将向量 z₁ 缩放 |z₂| 倍并旋转 Arg(z₂)；除以 z₂ 则缩放 1/|z₂| 倍并旋转 -Arg(z₂)。这为描述复平面中的变换提供了强有力的工具。

6. de Moivre’s Theorem | 德莫弗定理

de Moivre’s theorem states that for any integer n, (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ). Equivalently, (e^(iθ))ⁿ = e^(i nθ). This theorem is fundamental for finding powers and roots of complex numbers, and for deriving trigonometric identities.

德莫弗定理指出，对任意整数 n，(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ)。等价地，(e^(iθ))ⁿ = e^(i nθ)。该定理是计算复数的乘方与开方、推导三角恒等式的基础。

To compute (1 + i)⁵, first write 1 + i in polar form: √2 (cos π/4 + i sin π/4). Applying the theorem gives (√2)⁵ (cos 5π/4 + i sin 5π/4) = 4√2 (-√2/2 – i√2/2) = -4 – 4i.

要计算 (1 + i)⁵，先将 1 + i 写成极形式 √2 (cos π/4 + i sin π/4)。应用定理得 (√2)⁵ (cos 5π/4 + i sin 5π/4) = 4√2 (-√2/2 – i√2/2) = -4 – 4i。

7. Roots of Complex Numbers | 复数的根

To find the n‑th roots of a complex number z = r e^(iθ), use the formula z^(1/n) = r^(1/n) e^(i(θ + 2kπ)/n) for k = 0, 1, 2, …, n-1. This yields n distinct roots equally spaced on a circle of radius r^(1/n) in the complex plane.

求复数 z = r e^(iθ) 的 n 次方根，可使用公式 z^(1/n) = r^(1/n) e^(i(θ + 2kπ)/n)，其中 k = 0, 1, 2, …, n-1。这样会得到 n 个不同的根，它们均匀分布在复平面上半径为 r^(1/n) 的圆上。

The n‑th roots of unity (zⁿ = 1) have modulus 1 and arguments 2kπ/n. They are e^(i·2kπ/n). The sum of all distinct roots is zero. For cube roots of unity 1, ω, ω², we have ω³ = 1, 1+ω+ω² = 0, and ω = e^(2πi/3).

n 次单位根（满足 zⁿ = 1）的模为 1，辐角为 2kπ/n，即 e^(i·2kπ/n)。所有互异根之和为零。以三次单位根 1, ω, ω² 为例，满足 ω³ = 1，1+ω+ω² = 0，且 ω = e^(2πi/3)。

8. Loci in the Complex Plane | 复平面中的轨迹

Loci are sets of points satisfying a given condition. Common types: |z – a| = r is a circle with centre a (complex constant) and radius r. |z – a| = |z – b| is the perpendicular bisector of the segment joining a and b. arg(z – a) = θ is a half‑line from a (excluding a) at angle θ to the positive real axis.

轨迹是满足给定条件的点集。常见类型：|z – a| = r 表示以 a（复常数）为中心、r 为半径的圆。|z – a| = |z – b| 表示连接 a 与 b 的线段的垂直平分线。arg(z – a) = θ 表示从 a 出发（不含 a）与正实轴夹角为 θ 的射线。

Inequalities define regions. |z – a| < r is the interior of the circle. |z - a| ≤ r includes the boundary. arg(z - a) < θ gives a half‑plane region bounded by a half‑line. Combining conditions, e.g., 1 < |z - 2| ≤ 3 and -π/4 < arg(z - 2) ≤ π/4, describes the intersection of regions.

不等式定义区域。|z – a| < r 为圆的内部，|z - a| ≤ r 包含边界。arg(z - a) < θ 给出以射线为界的半平面区域。组合条件，例如 1 < |z - 2| ≤ 3 且 -π/4 < arg(z - 2) ≤ π/4，描述的是这些区域的交集。

When sketching, always clearly mark open/closed boundaries: a solid line for included boundary, dashed for excluded, and use arrow on half‑lines to indicate direction.

作图时，需清楚标明开/闭边界：实线表示包含边界，虚线表示不含边界，并在射线上用箭头标出方向。

9. Solving Equations with Complex Numbers | 复数方程求解

Quadratic equations with real coefficients but negative discriminant give complex conjugate roots. For ax² + bx + c = 0, if Δ = b² – 4ac < 0, roots are (-b ± i√(-Δ)) / (2a). These are conjugates.

实系数的二次方程当判别式为负时会产生共轭复根。对于 ax² + bx + c = 0，若 Δ = b² – 4ac < 0，则根为 (-b ± i√(-Δ)) / (2a)，它们互为共轭。

Cubic and quartic equations with real coefficients also have roots in conjugate pairs if complex. Knowing one complex root allows you to find other factors by using the conjugate. For polynomial equations, factor theorem and algebraic division remain valid over the complex numbers.

实系数的三次及四次方程如有复根，也以共轭对的形式出现。已知一个复根即可利用其共轭找出其他因式。对于多项式方程，因式定理和长除法在复数域上依然成立。

When solving zⁿ = c, first convert c to polar form, then apply the root formula. Write all roots explicitly and, if required, express them in Cartesian form a + bi.

求解 zⁿ = c 时，先将 c 转化为极形式，然后应用求根公式。显式写出所有根，如有需要再化为直角形式 a + bi。

10. Exam Tips | 考试技巧

1. Always state the principal argument unless a general argument is asked. Be careful with quadrants: sketch the complex number to avoid sign errors.

1. 除非题目要求一般辐角，否则一律写主辐角。注意象限：画草图可避免符号错误。

2. When proving trigonometric identities using de Moivre, expand (cos θ + i sin θ)ⁿ using the binomial theorem, then equate real and imaginary parts.

2. 用德莫弗定理证明三角恒等式时，先用二项式定理展开 (cos θ + i sin θ)ⁿ，再分别令实部和虚部相等。

3. For locus questions, describe the geometric feature (circle, line, half‑line) and shade the region accurately. Always indicate whether boundaries are included.

3. 对于轨迹问题，要描述几何特征（圆、直线、射线）并准确标示区域。始终标明边界是否包含在内。

4. When finding roots, remember to add 2kπ before dividing by n. List all k values to ensure you have all distinct roots.

4. 求根时，记得先加上 2kπ 再除以 n。列出所有 k 值以保证得到全部互异的根。

5. Check your answer: multiply complex conjugates to verify modulus; substitute roots back into the original equation; use calculator in complex mode only after algebraic manipulation.

5. 检查答案：将复共轭相乘来验证模；将根代回原方程；仅在完成代数推导后才使用计算器的复数模式。

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Edexcel Maths: Complex Numbers | Edexcel 数学：复数 考点精讲