📚 Exercise I.4: Complex Numbers – Operations and Geometry | 练习 I.4：复数的运算与几何
This article provides a comprehensive review of the algebraic and geometric properties of complex numbers, designed to support your work on Exercise I.4 in the IB Mathematics course. You will find clear explanations of operations, polar form, De Moivre's theorem, and practical problem-solving methods.
本文全面复习了复数的代数与几何性质，旨在帮助你在 IB 数学课程中完成练习 I.4。文中清晰解释了运算、极坐标形式、棣莫弗定理以及实用的解题方法。
1. Introducing Complex Numbers | 复数简介
A complex number z is written as z = a + bi, where a and b are real numbers and i is the imaginary unit defined by i² = –1. The real part is Re(z) = a, the imaginary part is Im(z) = b. Complex numbers extend the real line to the complex plane, allowing every algebraic equation to have a solution.
复数 z 可写作 z = a + bi，其中 a 与 b 为实数，i 是满足 i² = –1 的虚数单位。实部为 Re(z) = a，虚部为 Im(z) = b。复数将实数线扩展为复平面，使得每个代数方程都有解。
Two complex numbers are equal only when their real parts are equal and their imaginary parts are equal. The complex plane represents z = a + bi as the point (a, b) or as a position vector from the origin to that point.
两个复数相等当且仅当实部相等且虚部相等。复平面将 z = a + bi 表示为点 (a, b) 或从原点指向该点的位置向量。
2. Addition and Subtraction | 加法与减法
Addition and subtraction of complex numbers follow component‑wise rules: if z₁ = a + bi and z₂ = c + di, then z₁ ± z₂ = (a ± c) + (b ± d)i. This mirrors vector addition in the plane.
复数的加法与减法遵循分量逐项运算：若 z₁ = a + bi，z₂ = c + di，则 z₁ ± z₂ = (a ± c) + (b ± d)i。这与平面内向量的加减一致。
Geometrically, the sum z₁ + z₂ is obtained by placing the tail of the second vector at the head of the first, completing a parallelogram. Subtraction z₁ – z₂ can be viewed as adding the opposite vector.
从几何上看，和 z₁ + z₂ 可通过将第二个向量的尾部置于第一个向量的头部、完成平行四边形得到。减法 z₁ – z₂ 可看作加上反向向量。
3. Multiplication of Complex Numbers | 复数的乘法
Using the distributive property and i² = –1, multiplication acts as (a + bi)(c + di) = ac + adi + bci + bdi² = (ac – bd) + (ad + bc)i. The result is another complex number.
利用分配律和 i² = –1，乘法运算为 (a + bi)(c + di) = ac + adi + bci + bdi² = (ac – bd) + (ad + bc)i。结果仍为复数。
It is often helpful to remember that multiplication by i rotates a complex number by 90° counter‑clockwise about the origin. This geometric insight will be fully explained in the polar form section.
记住乘以 i 相当于将复数绕原点逆时针旋转 90° 常常很有用。这一几何直观将在极坐标一节详细说明。
4. The Complex Conjugate | 共轭复数
The complex conjugate of z = a + bi is denoted by z̄ (or z*) and is defined as z̄ = a – bi. It reflects the number across the real axis in the complex plane.
z = a + bi 的共轭复数记作 z̄（或 z*），定义为 z̄ = a – bi。它在复平面上是原数关于实轴的镜像。
Key properties: z + z̄ = 2 Re(z) is real; z – z̄ = 2i Im(z) is purely imaginary; z · z̄ = a² + b² is a non‑negative real number. The conjugate of a sum, product, or quotient is the sum, product, or quotient of the conjugates.
关键性质：z + z̄ = 2 Re(z) 为实数；z – z̄ = 2i Im(z) 为纯虚数；z · z̄ = a² + b² 为非负实数。和、积、商的共轭等于共轭的和、积、商。
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:
(a + bi) / (c + di) = [(a + bi)(c – di)] / (c² + d²)
进行复数除法时，将分子分母同乘分母的共轭：
(a + bi) / (c + di) = [(a + bi)(c – di)] / (c² + d²)
This process eliminates the imaginary part from the denominator, leaving a standard complex number with a real denominator.
这一过程消去了分母中的虚部，得到一个分母为实数的标准复数。
6. Modulus of a Complex Number | 复数的模
The modulus (or absolute value) of z = a + bi is the distance from the origin to the point (a, b):
|z| = √a² + b²
z = a + bi 的模（绝对值）是原点到点 (a, b) 的距离：
|z| = √a² + b²
Important modulus properties include: |z₁z₂| = |z₁|·|z₂|; |z₁ / z₂| = |z₁| / |z₂| for z₂ ≠ 0; the triangle inequality |z₁ + z₂| ≤ |z₁| + |z₂|.
模的重要性质包括：|z₁z₂| = |z₁|·|z₂|；当 z₂ ≠ 0 时，|z₁ / z₂| = |z₁| / |z₂|；以及三角不等式 |z₁ + z₂| ≤ |z₁| + |z₂|。
7. Argument and Polar Form | 辐角与极坐标形式
The argument of z, arg(z), is the angle θ measured from the positive real axis to the line joining the origin to (a, b). The principal value is usually taken in (–π, π] or [0, 2π). Using trigonometry, a = |z| cos θ and b = |z| sin θ.
z 的辐角 arg(z) 是从正实轴到连接原点与点 (a, b) 的线段所成的角度 θ。主值通常取在 (–π, π] 或 [0, 2π)。由三角函数得 a = |z| cos θ，b = |z| sin θ。
The polar (or modulus‑argument) form is:
z = r (cos θ + i sin θ) = r cis θ
where r = |z| and θ = arg(z). The abbreviation cis θ stands for cos θ + i sin θ.
极坐标（模‑辐角）形式为：
z = r (cos θ + i sin θ) = r cis θ
其中 r = |z|，θ = arg(z)。简写 cis θ 表示 cos θ + i sin θ。
8. Multiplication and Division in Polar Form | 极坐标形式的乘除
If z₁ = r₁ cis θ₁ and z₂ = r₂ cis θ₂, then their product is z₁z₂ = r₁r₂ cis (θ₁ + θ₂). Moduli multiply while arguments add.
若 z₁ = r₁ cis θ₁，z₂ = r₂ cis θ₂，则乘积为 z₁z₂ = r₁r₂ cis (θ₁ + θ₂)。模相乘，辐角相加。
For division, z₁ / z₂ = (r₁ / r₂) cis (θ₁ – θ₂), provided r₂ ≠ 0. Geometrically, division scales the modulus and subtracts the argument.
除法为 z₁ / z₂ = (r₁ / r₂) cis (θ₁ – θ₂)，前提是 r₂ ≠ 0。几何上，除法缩放模并减去辐角。
9. De Moivre's Theorem | 棣莫弗定理
For any integer n, De Moivre's theorem states:
(r cis θ)n = rn cis (nθ)
This also holds for rational powers when used carefully for finding roots. For n = 2, (r cis θ)² = r² cis (2θ), which gives double‑angle identities.
对任意整数 n，棣莫弗定理指出：
(r cis θ)n = rn cis (nθ)
当谨慎处理求根问题时，它同样适用于有理指数。如当 n = 2 时，(r cis θ)² = r² cis (2θ)，可据此导出倍角公式。
De Moivre's theorem is a powerful tool for simplifying powers of complex numbers and for solving equations of the form zn = k.
棣莫弗定理是化简复数幂以及求解形如 zn = k 方程的有力工具。
10. Roots of Complex Numbers | 复数的根
To find the nth roots of a complex number w = R cis φ, we set zn = w and write z = r cis θ. Then rn cis (nθ) = R cis φ. Hence r = R1/n and nθ = φ + 2πk, k = 0, 1, …, n–1.
要求复数 w = R cis φ 的 n 次方根，设 zn = w 并令 z = r cis θ。则 rn cis (nθ) = R cis φ，于是得 r = R1/n，且 nθ = φ + 2πk，k = 0, 1, …, n–1。
The n distinct roots are given by zk = R1/n cis ((φ + 2πk) / n). They lie on a circle of radius R1/n and are equally spaced by an angle of 2π/n.
n 个不同的根为 zk = R1/n cis ((φ + 2πk) / n)。它们分布在半径为 R1/n 的圆上，且等角度间隔 2π/n。
11. Geometric Representation in the Complex Plane | 复平面上的几何表示
Every complex number corresponds to a unique point in the plane, making geometric interpretations natural. The distance between z₁ and z₂ is |z₁ – z₂|. A circle with centre z₀ and radius r is described by |z – z₀| = r.
每个复数对应平面上唯一的点，几何解释非常自然。z₁ 与 z₂ 之间的距离为 |z₁ – z₂|。以 z₀ 为圆心、r 为半径的圆可用 |z – z₀| = r 描述。
A perpendicular bisector can be expressed as |z – z₁| = |z – z₂|, and a ray from the origin at angle θ is described by arg(z oci problems are often tackled by translating geometric conditions into complex equations.
垂直平分线可表示为 |z – z₁| = |z – z₂|，从原点出发角度为 θ 的射线则为 arg(z) = θ。轨迹问题常通过将几何条件转为复数方程来求解。
12. Problem‑Solving Tips and Common Pitfalls | 解题技巧与常见误区
Always write the complex number exactly as needed for the operation: use a + bi for addition, polar form for multiplication and powers. When finding arguments, pay attention to the quadrant; using arctan alone may give the wrong angle.
务必根据运算选择合适形式：加减用 a + bi，乘除和乘方用极坐标形式。求辐角时要注意象限，单纯依靠 arctan 可能会得到错误的角度。
Remember that i² = –1 is the only reduction rule; never write i = √–1 loosely without considering branch cuts. For roots, list all n values and avoid stopping after the first one. Check your answers by substitution or by verifying that the modulus of the product equals the product of moduli.
记住 i² = –1 是唯一的化简法则；不要随意写 i = √–1 而不考虑分支切割。求根时须列出全部 n 个值，切勿得到一个就停止。通过代回或验证乘积的模等于模的乘积来检查答案。
Finally, using a diagram often reveals symmetries and saves algebraic work, especially in locus and root problems from Exercise I.4.
最后，借助图形往往能揭示对称性，减少代数运算，特别是在练习 I.4 的轨迹和求根问题中。
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