The Modulus of Complex Numbers and Its Properties | 复数的模及其性质

📚 The Modulus of Complex Numbers and Its Properties | 复数的模及其性质

The modulus of a complex number is one of the most fundamental concepts in complex analysis. It measures the “size” of a complex number, just as the absolute value measures the size of a real number, and it plays a central role in everything from geometry to polynomial equations.

复数的模是复分析中最基本的概念之一。它度量一个复数的“大小”，正如绝对值度量实数的大小一样，并且从几何到多项式方程等各个方面都发挥着核心作用。

1. Definition of the Modulus | 模的定义

For a complex number z = a + b i, where a and b are real numbers, the modulus of z is denoted by |z| and is defined as the non-negative real number:

对于复数 z = a + b i，其中 a 和 b 为实数，z 的模记为 |z|，定义为非负实数：

|z| = √(a² + b²)

In words, the modulus is the square root of the sum of the squares of the real part and the imaginary part.

也就是说，模等于实部平方与虚部平方之和的算术平方根。

Example: |3 + 4 i| = √(3² + 4²) = √25 = 5
例：|3 + 4 i| = √(3² + 4²) = √25 = 5
Example: |-2 – 5 i| = √(4 + 25) = √29
例：|-2 – 5 i| = √(4 + 25) = √29

2. Geometric Interpretation | 模的几何意义

In the complex plane, a complex number z = a + b i is represented by the point (a, b). The modulus |z| is exactly the distance from this point to the origin (0, 0).

在复平面中，复数 z = a + b i 对应点 (a, b)。模 |z| 恰好就是该点到原点 (0, 0) 的距离。

Consequently, the distance between two complex numbers z₁ and z₂ is given by |z₁ – z₂|.

因此，两个复数 z₁ 与 z₂ 之间的距离为 |z₁ – z₂|。

If z = 0, then |z| = 0 and the point is the origin.
若 z = 0，则 |z| = 0，对应原点。
If |z| = r, then z lies on a circle of radius r centered at the origin.
若 |z| = r，则 z 位于以原点为圆心、半径为 r 的圆上。

3. Basic Properties of the Modulus | 模的基本性质

The following basic properties follow directly from the definition and from the geometry of the complex plane.

以下基本性质可直接由定义和复平面的几何性质得到。

|z| ≥ 0 for every complex number z.
对任意复数 z，有 |z| ≥ 0。
|z| = 0 if and only if z = 0.
|z| = 0 当且仅当 z = 0。
|z| = |z̄|, where z̄ is the conjugate of z.
|z| = |z̄|，其中 z̄ 是 z 的共轭复数。
|Re(z)| ≤ |z| and |Im(z)| ≤ |z|.
|Re(z)| ≤ |z| 且 |Im(z)| ≤ |z|。
|-z| = |z|.
|-z| = |z|。

These properties are often used without comment in more advanced work, so they should become automatic.

这些性质在较高级的内容中常常不加说明地使用，因此应当熟练掌握。

4. The Modulus and the Conjugate | 模与共轭复数

The conjugate of z = a + b i is z̄ = a – b i. A crucial identity connects the modulus and the conjugate:

z = a + b i 的共轭复数是 z̄ = a – b i。一个重要恒等式将模与共轭联系起来：

|z|² = z z̄

This identity is extremely useful because it allows a modulus to be replaced by a product of complex numbers.

这个恒等式非常有用，因为它能把模转化为复数的乘积。

Since z z̄ = |z|², we can write 1/z = z̄ / |z|² for z ≠ 0.
因为 z z̄ = |z|²，当 z ≠ 0 时可写成 1/z = z̄ / |z|²。
Example: For z = 3 + 4 i, |z|² = 25 and z z̄ = (3 + 4 i)(3 – 4 i) = 9 + 16 = 25.
例：对 z = 3 + 4 i，|z|² = 25，且 z z̄ = (3 + 4 i)(3 – 4 i) = 9 + 16 = 25。

5. Modulus of a Product | 积的模

If z₁ and z₂ are any complex numbers, then the modulus of their product equals the product of their moduli:

若 z₁ 和 z₂ 是任意复数，则它们乘积的模等于各自模的乘积：

|z₁ z₂| = |z₁||z₂|

This can be proved by writing |z₁ z₂|² = (z₁ z₂)(z₁ z₂)⁻ = |z₁|²|z₂|² and taking square roots.

证明可写为 |z₁ z₂|² = (z₁ z₂)(z₁ z₂)⁻ = |z₁|²|z₂|²，再开平方即可。

The rule extends to any finite product: |z₁ z₂ … zₙ| = |z₁||z₂| … |zₙ|.
该规则可推广到任意有限乘积：|z₁ z₂ … zₙ| = |z₁||z₂| … |zₙ|。

6. Modulus of a Quotient and Powers | 商的模与幂的模

For division, provided z₂ ≠ 0, we have:

对于除法，当 z₂ ≠ 0 时，有：

|z₁ / z₂| = |z₁| / |z₂|

For positive integer powers, the product rule gives:

对于正整数幂，由乘积规则可得：

|zⁿ| = |z|ⁿ

Example: |(1 + i)⁶| = |1 + i|⁶ = (√2)⁶ = 2³ = 8.
例：|(1 + i)⁶| = |1 + i|⁶ = (√2)⁶ = 2³ = 8。
This property is especially useful in combination with De Moivre’s Theorem for rapidly calculating powers.
这一性质结合棣莫弗定理，可以快速计算复数的幂。

7. The Triangle Inequality | 三角不等式

One of the most important inequalities involving the modulus is the triangle inequality:

关于模的最重要不等式之一是三角不等式：

|z₁ + z₂| ≤ |z₁| + |z₂|

Geometrically, this says that the length of one side of a triangle cannot exceed the sum of the lengths of the other two sides.

从几何上看，这表示三角形一边的长度不能超过另外两边长度之和。

A related form is the reverse triangle inequality:

8. Solving Equations Involving the Modulus | 含模方程的求解

Many equations can be solved by replacing |z|² with z z̄, or by using geometric interpretations.

许多方程可以通过用 z z̄ 替换 |z|²，或利用几何意义来求解。

Example: Solve z z̄ + |z|² = 8. Since z z̄ = |z|², the equation becomes 2|z|² = 8, so |z| = 2. The solution set is the whole circle of radius 2 centered at the origin.

例如：解 z z̄ + |z|² = 8。因为 z z̄ = |z|²，方程化为 2|z|² = 8，所以 |z| = 2。解集是以原点为圆心、半径为 2 的整个圆。

When solving, remember that |z| is a real number, so only non-negative solutions are acceptable after taking square roots.

求解时要注意 |z| 是实数，开平方后只取非负解。

9. Loci in the Complex Plane | 复平面上的轨迹

Using the modulus, we can describe important geometric sets, called loci, in the complex plane.

利用模可以在复平面上描述重要的几何集合，称为轨迹。

|z – z₀| = r is a circle with centre z₀ and radius r.
|z – z₀| = r 表示以 z₀ 为圆心、半径为 r 的圆。
|z – z₀| < r is the interior of that circle.
|z – z₀| < r 表示该圆的内部。
|z – z₁| = |z – z₂| is the perpendicular bisector of the segment joining z₁ and z₂.
|z – z₁| = |z – z₂| 表示连接 z₁ 和 z₂ 的线段的垂直平分线。
|z – z₁| + |z – z₂| = 2a, with 2a greater than the distance between z₁ and z₂, is an ellipse with foci z₁ and z₂.
|z – z₁| + |z – z₂| = 2a，当 2a 大于 z₁ 与 z₂ 之间的距离时，表示以 z₁ 和 z₂ 为焦点的椭圆。

10. Worked Examples | 典型例题

Example 1: Find the modulus of (2 + i) / (1 – 2 i).

例 1：求 (2 + i) / (1 – 2 i) 的模。

Using the quotient property,

利用商的模的性质，

|(2 + i) / (1 – 2 i)| = |2 + i| / |1 – 2 i| = √5 / √5 = 1

Example 2: Describe the set of points satisfying |z – 1| = |z + i|.

例 2：描述满足 |z – 1| = |z + i| 的点集。

This is the perpendicular bisector of the segment between 1 and -i. Hence the locus is the straight line y = -x.

这是连接 1 和 -i 的线段的垂直平分线。因此轨迹是直线 y = -x。

Example 3: If |z| = 1, show that |z² – 1| = |z – z̄|.

例 3：若 |z| = 1，证明 |z² – 1| = |z – z̄|。

Since z̄ = 1/z, we have z – z̄ = z – 1/z = (z² – 1)/z, so |z – z̄| = |z² – 1| / |z| = |z² – 1|.

因为 z̄ = 1/z，所以 z – z̄ = z – 1/z = (z² – 1)/z，因此 |z – z̄| = |z² – 1| / |z| = |z² – 1|。

11. Common Mistakes and Pitfalls | 常见错误与易错点

Even strong students often make the following mistakes when working with moduli.

即使是能力较强的学生，在处理模时也常犯以下错误。

Writing |z₁ + z₂| = |z₁| + |z₂| without checking the equality condition. This is false in general.
不经检查等号条件就写 |z₁ + z₂| = |z₁| + |z₂|。这在一般情况下是错误的。
Confusing z² with |z|². For z = i, z² = -1 but |z|² = 1.
混淆 z² 与 |z|²。例如 z = i 时，z² = -1，但 |z|² = 1。
Forgetting that |z| is a real number when solving equations.
求解方程时忘记 |z| 是实数。
Using |zⁿ| = zⁿ when n is not an integer. The power property is valid for integer exponents but must be handled carefully otherwise.
当 n 不是整数时使用 |zⁿ| = zⁿ。幂的性质对整数指数有效，其他情形需谨慎处理。

12. Summary | 小结

The modulus of a complex number is its distance from the origin, and it satisfies a rich set of algebraic and geometric properties. The key identities are |z| = √(a² + b²), |z|² = z z̄, |z₁ z₂| = |z₁||z₂|, and the triangle inequality |z₁ + z₂| ≤ |z₁| + |z₂|.

复数的模是它到原点的距离，并且满足一系列丰富的代数与几何性质。关键恒等式包括 |z| = √(a² + b²)、|z|² = z z̄、|z₁ z₂| = |z₁||z₂| 以及三角不等式 |z₁ + z₂| ≤ |z₁| + |z₂|。

Mastering the modulus and its properties is essential for solving equations, describing loci, and working with complex powers and roots.

掌握模及其性质，对于求解方程、描述轨迹以及处理复数的幂与方根都至关重要。

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The Modulus of Complex Numbers and Its Properties | 复数的模及其性质