📚 Complex Numbers | 复数考点精讲
In IGCSE Additional Mathematics, complex numbers extend the real number system by introducing the imaginary unit i, where i² = –1. This topic covers the algebra of complex numbers, their geometrical representation on an Argand diagram, and the use of polar form to simplify multiplication, division and powers. A strong grasp of complex numbers is essential for solving quadratic equations with negative discriminants and for tackling higher‑level pure mathematics.
在 IGCSE 附加数学中，复数通过引入虚数单位 i（满足 i² = –1）扩展了实数系。本专题涵盖复数的代数运算、在阿尔冈图上的几何表示，以及利用极坐标形式简化乘法、除法和乘方运算。牢固掌握复数对于求解带有负判别式的二次方程以及学习更高层次的纯数学至关重要。
1. The Imaginary Unit and Basic Definition | 虚数单位与基本定义
The imaginary unit i is defined such that i² = –1. A complex number is any number that can be written in the form z = a + bi, where a and b are real numbers. Here a is called the real part, Re(z), and b is the imaginary part, Im(z).
虚数单位 i 定义为满足 i² = –1。任何可以写成 z = a + bi 形式的数都称为复数，其中 a 和 b 为实数。a 叫做 实部，记作 Re(z)；b 叫做 虚部，记作 Im(z)。
= 0, the number is purely real; if a = 0 and b ≠ 0, it is purely imaginary. For example, 3 – 2i has real part 3 and imaginary part –2.
若 b = 0，该数为实数；若 a = 0 且 b ≠ 0，则为纯虚数。例如 3 – 2i 的实部为 3，虚部为 –2。
2. Equality, Addition and Subtraction | 相等、加法与减法
f their real parts are equal and their imaginary parts are equal. That is, a + bi = c + di ⟹ a = c and b = d.
两个复数相等当且仅当它们的实部相等且虚部相等。即 a + bi = c + di 意味着 a = c 且 b = d。
Addition and subtraction are performed by treating the real and imaginary parts separately:
加法和减法只需分别处理实部和虚部：
(a + bi) + (c + di) = (a + c) + (b + d)i
(a + bi) – (c + di) = (a – c) + (b – d)i
3. Multiplication of Complex Numbers | 复数的乘法
Multiply complex numbers as you would binomials, using i² = –1 to simplify. For example, (2 + 3i)(1 – i) = 2 – 2i + 3i – 3i² = 2 + i + 3 = 5 + i.
复数相乘时像二项式一样展开，并利用 i² = –1 化简。例如 (2 + 3i)(1 – i) = 2 – 2i + 3i – 3i² = 2 + i + 3 = 5 + i。
In general, (a + bi)(c + di) = (ac – Keep in mind that the product of a complex number and its conjugate yields a real number.
一般地，(a + bi)(c + di) = (ac – bd) + (ad + bc)i。注意复数与其共轭的乘积是一个实数。
4. Complex Conjugate | 共轭复数
The complex conjugate of z = a + bi is denoted as z* or z̄ and is defined as z̄ = a – bi. Graphically, it is the reflection of z in the real axis on an Argand diagram.
复数 z = a + bi 的共轭记作 z* 或 z̄，定义为 z̄ = a – bi。在几何上，它是 z 关于实轴的镜像。
Key properties: z·z̄ = a² + b² (a real number); z + z̄ = 2a (twice the real part); z – z̄ . The conjugate is extremely useful when dividing complex numbers and finding the modulus.
重要性质：z·z̄ = a² + b²（实数）；z + z̄ = 2a（实部的两倍）；z – z̄ = 2bi（纯虚数）。共轭在复数除法与求模时极为有用。
5. Division of Complex Numbers | 复数的除法
To divide two complex numbers, multiply the numerator and denominator by the conjugate of the denominator. This turns the denominator into a real number.
做复数除法时，将分子分母同时乘以分母的共轭，从而将分母化为实数。
(a + bi) / (c + di) = (a + bi)(c – di) / (c² + d²)
For instance, (1 + i) / (2 – i) = (1 + i)(2 + i) / (4 + 1) = (2 + i + 2i – 1)/5 = (1 + 3i)/5 = 0.2 + 0.6i.
比如，(1 + i) / (2 – i) = (1 + i)(2 + i) / (4 + 1) = (2 + i + 2i – 1)/5 = (1 + 3i)/5 = 0.2 + 0.6i。
6. Argand Diagram and Modulus | 阿尔冈图与模
An Argand diagram represents complex numbers as points or vectors on a plane, with the x‑axis as the real axis and the y‑axis as the imaginary axis. The complex number z = a + bi corresponds to the point (a, b).
阿尔冈图将复数表示为平面上的点或向量，x 轴为实轴，y 轴为虚轴。复数 z = a + bi 对应点 (a, b)。
The modulus of z, written |z|, is the distance from the origin to the point (a, b): |z| = √a² + b². It is always a non‑negative real number. For example, | –3 + 4i | = √9 + 16 = 5.
z 的模写作 |z|，是原点到点 (a, b) 的距离：|z| = √a² + b²。它总是一个非负实数。例如 | –3 + 4i | = √9 + 16 = 5。
7. Argument of a Complex Number | 复数的辐角
The argument of a non‑ measured in radians. It satisfies tan θ = b/a, taking care to choose the correct quadrant.
非零复数 z 的辐角记作 arg(z)，是与正实轴的夹角 θ，通常以弧度为单位。满足 tan θ = b/a，并需选择正确的象限。
The principal argument typically lies in the interval (–π, π] or [0, 2π). For z = –1 – i, the point lies in the third quadrant, giving arg(z) = –3π/4 (or 5π/4).
主辐角通常落在区间 (–π, π] 或 [0, 2π) 内。例如 z = –1 – i，该点位于第三象限，arg(z) = –3π/4（或 5π/4）。
8. Polar Form of Complex Numbers | 复数的极坐标形式
A complex number can be expressed in polar form z = r(cos θ + i sin θ), where r = |z| arg(z). This representation links algebra to geometry and is often abbreviated as r cis θ.
复数可以用极坐标形式表示为 z = r(cos θ + i sin θ)，其中 r = |z|，θ = arg(z)。这种表示将代数与几何联系起来，常简写为 r cis θ。
For example, z = 1 + i√3 has modulus 2 and argument π/3, so its polar form is 2(cos π/3 + i sin π/3) or 2 cis π/3.
例如 z = 1 + i√3 的模为 2，辐角为 π/3，因此其极坐标形式为 2(cos π/3 + i sin π/3) 或 2 cis π/3。
9. Multiplication and Division in Polar Form | 极坐标形式下的乘除运算
When multiplying two complex numbers in polar form, multiply their moduli and add their arguments: r₁cisθ₁ × r₂cisθ₂ = (r₁r₂) cis(θ₁ + θ₂). Division follows a similar rule: divide the moduli and subtract the arguments.
极坐标形式下两个复数相乘时，将模相乘并将辐角相加：r₁cisθ₁ × r₂cisθ₂ = (r₁r₂) cis(θ₁ + θ₂)。除法规则类似：模相除，辐角相减。
(r₁cisθ₁) / (r₂cisθ₂) = (r₁/r₂) cis(θ₁ – θ₂)
These rules make products and quotients extremely efficient, especially when dealing with multiple rotations or power calculations.
这些规则使乘除运算极为高效，尤其是在处理多次旋转或乘方计算时。
10. De Moivre's Theorem | 德莫弗定理
De Moivre's theorem extends the polar multiplication rule to powers: for any integer n, (r cisθ)ⁿ = rⁿ cis(nθ). This provides a quick way to raise a complex number to any integer power and to find roots of complex numbers.
德莫弗定理将极坐标乘法规则推广到乘方：对任意整数 n，(r cisθ)ⁿ = rⁿ cis(nθ)。这为求复数的整数次幂和开方提供了快捷方法。
For example, (1 + i)⁴: first write 1 + i = √2 cis(π/4), then (√2 cis π/4)⁴ = (√2)⁴ cis(π) = 4 cis π = –4.
例如 (1 + i)⁴：先写 1 + i = √2 cis(π/4)，则 (√2 cis π/4)⁴ = (√2)⁴ cis(π) = 4 cis π = –4。
11. Solving Quadratic Equations with Complex Roots | 求解具有复根的二次方程
When a quadratic equation ax² + bx + c = 0 has a negative discriminant (Δ = b² – 4ac < 0), the roots are complex conjugates. Using the quadratic formula gives:
当二次方程 ax² + bx + c = 0 的判别式 Δ = b² – 4ac < 0 时，根为一对共轭复数。利用求根公式可得：
x = [ –b ± i√4ac – b² ] / 2a
For instance, x² – 2x + 5 = 0 gives x = [2 ± i√20 – 4]/2 = 1 ± 2i. The roots are always a ± bi form with the real part being –b/(2a).
例如 x² – 2x + 5 = 0 解得 x = [2 ± i√20 – 4]/2 = 1 ± 2i。根总是 a ± bi 的形式，实部为 –b/(2a)。
12. Geometric Applications and Loci | 几何应用与轨迹
On an Argand diagram, the set of points satisfying |z – z₀| = r is a circle with centre z₀ and radius r. The argument condition arg(z – z₀) = θ is a half‑ starting from z₀ (excluding z₀ itself) making an angle θ with the positive real direction.
在阿尔冈图上，满足 |z – z₀| = r 的点集表示以 z₀ 为圆心、r 为半径的圆。辐角条件 arg(z – z₀) = θ 表示从 z₀ 出发（不含 z₀）与正实轴成角度 θ 的射线。
The perpendicular bisector of the segment joining two points z₁ and z₂ is given by |z – z₁| = |z – z₂| tional Mathematics to check understanding of modulus and argument as distances and angles.
两点 z₁ 与 z₂ 连线的垂直平分线由 |z – z₁| = |z – z₂| 给出。这类轨迹常出现在 IGCSE 附加数学考试中，用以考查模表示距离、辐角表示角度的理解。
Published by TutorHao | Additional Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图