The Modulus and Argument of a Complex Number | 复数的模与辐角

📚 The Modulus and Argument of a Complex Number | 复数的模与辐角

In A-Level Mathematics, every complex number can be represented in two fundamental ways: the Cartesian form z = a + bi, and the modulus-argument form z = r(cos θ + i sin θ). The modulus and argument are the polar coordinates of a complex number on the Argand diagram, and mastering them is essential for solving equations, multiplying and dividing complex numbers, and interpreting their geometric meaning.

在 A-Level 数学中，每个复数都可以用两种基本方式表示：笛卡尔形式 z = a + bi，以及模辐角形式 z = r(cos θ + i sin θ)。模与辐角是复数在阿甘图上的极坐标，掌握它们对于解方程、复数的乘除运算以及理解其几何意义至关重要。

1. What Is the Modulus? | 什么是模？

The modulus of a complex number z = a + bi, denoted |z|, is the distance from the origin to the point (a, b) on the Argand diagram. It is always a non-negative real number, calculated using Pythagoras’ theorem:

复数 z = a + bi 的模，记作 |z|，是阿甘图上原点 (0,0) 到点 (a, b) 的距离。它始终是一个非负实数，通过勾股定理计算：

|z| = |a + bi| = √(a² + b²)

For example, if z = 3 + 4i, then |z| = √(3² + 4²) = √25 = 5. The modulus is never negative, and |z| = 0 only when z = 0. Geometrically, this is exactly the same idea as the magnitude of a vector.

例如，若 z = 3 + 4i，则 |z| = √(3² + 4²) = √25 = 5。模永远不为负数，且 |z| = 0 当且仅当 z = 0。从几何角度看，这与向量的模长是同一个概念。

2. What Is the Argument? | 什么是辐角？

The argument of a complex number z = a + bi, denoted arg(z), is the angle θ that the line connecting the origin to the point (a, b) makes with the positive real axis. It is measured anticlockwise from the positive real axis, and is usually given in radians.

复数 z = a + bi 的辐角，记作 arg(z)，是原点与点 (a, b) 的连线与正实轴之间的夹角 θ。它从正实轴逆时针方向测量，通常以弧度为单位。

tan θ = b / a, therefore θ = tan⁻¹(b / a)

However, this formula alone is not enough: the value of tan⁻¹(b/a) only gives the reference angle in the range (-π/2, π/2). The actual argument depends on which quadrant the complex number lies in on the Argand diagram.

然而，仅靠这个公式是不够的：tan⁻¹(b/a) 只给出 (-π/2, π/2) 范围内的参考角。实际的辐角取决于复数在阿甘图上位于哪个象限。

3. The Principal Argument | 主辐角

Because angles can repeat every 2π radians, the argument of a complex number is not unique. To make it unique, we define the principal argument, written as Arg(z) with a capital A, which lies in the range:

由于角度每 2π 弧度循环一次，复数的辐角并不唯一。为了使它唯一，我们定义主辐角，用大写 A 写为 Arg(z)，其取值范围为：

-π < Arg(z) ≤ π

For a complex number in the first quadrant (a > 0, b > 0), Arg(z) = tan⁻¹(b/a). In the second quadrant (a < 0, b > 0), Arg(z) = tan⁻¹(b/a) + π. In the third quadrant (a < 0, b < 0), Arg(z) = tan⁻¹(b/a) - π. In the fourth quadrant (a > 0, b < 0), Arg(z) = tan⁻¹(b/a).

对于第一象限的复数（a > 0, b > 0），Arg(z) = tan⁻¹(b/a)。第二象限（a < 0, b > 0）时，Arg(z) = tan⁻¹(b/a) + π。第三象限（a < 0, b < 0）时，Arg(z) = tan⁻¹(b/a) - π。第四象限（a > 0, b < 0）时，Arg(z) = tan⁻¹(b/a)。

Quadrant	Condition	Principal Argument
First (第一象限)	a > 0, b > 0	θ = tan⁻¹(b/a)
Second (第二象限)	a < 0, b > 0	θ = tan⁻¹(b/a) + π
Third (第三象限)	a < 0, b < 0	θ = tan⁻¹(b/a) – π
Fourth (第四象限)	a > 0, b < 0	θ = tan⁻¹(b/a)

4. Special Cases on the Axes | 坐标轴上的特殊情形

When a complex number lies exactly on one of the axes, the argument is easy to identify directly. For any positive real number x, arg(x) = 0. For any negative real number -x, arg(-x) = π. For any positive imaginary number bi, arg(bi) = π/2. For any negative imaginary number -bi, arg(-bi) = -π/2.

当复数恰好位于某条坐标轴上时，辐角可以直接判断。对于任意正实数 x，arg(x) = 0。对于任意负实数 -x，arg(-x) = π。对于任意正虚数 bi，arg(bi) = π/2。对于任意负虚数 -bi，arg(-bi) = -π/2。

These special cases are worth memorising because they frequently appear in exam questions, especially when working with complex roots and loci. For instance, the complex number -4 has modulus 4 and principal argument π, so -4 = 4(cos π + i sin π).

这些特殊情形值得牢记，因为它们经常出现在考题中，尤其是在处理复根和轨迹问题时。例如，复数 -4 的模为 4，主辐角为 π，因此 -4 = 4(cos π + i sin π)。

5. The Modulus-Argument Form | 模辐角形式

Once the modulus and argument of a complex number are known, we can write it in modulus-argument form, also called polar form. If z = a + bi has modulus r and argument θ, then:

一旦知道复数的模和辐角，我们就可以将其写成模辐角形式，也称为极坐标形式。若 z = a + bi 的模为 r，辐角为 θ，则：

z = r(cos θ + i sin θ)

This form is particularly powerful because multiplication and division of complex numbers become much simpler. If z₁ = r₁(cos θ₁ + i sin θ₁) and z₂ = r₂(cos θ₂ + i sin θ₂), then z₁z₂ = r₁r₂(cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)). The moduli multiply, and the arguments add.

这种形式非常强大，因为复数的乘法和除法会变得简单得多。若 z₁ = r₁(cos θ₁ + i sin θ₁)，z₂ = r₂(cos θ₂ + i sin θ₂)，则 z₁z₂ = r₁r₂(cos(θ₁ + θ₂) + i sin(θ₁ + θ₂))。模相乘，辐角相加。

This result, known as de Moivre’s theorem in its generalised form, means that raising a complex number to a power n simply raises the modulus to the power n and multiplies the argument by n:

这个结果，即推广形式的棣莫弗定理，意味着将复数提升到 n 次幂时，只需将模提升到 n 次幂并将辐角乘以 n：

zⁿ = rⁿ(cos(nθ) + i sin(nθ))

6. Converting from Cartesian to Mod-Arg Form | 从笛卡尔形式转换为模辐角形式

To convert a complex number z = a + bi into modulus-argument form, follow these steps. First, compute the modulus: r = √(a² + b²). Second, determine the quadrant based on the signs of a and b. Third, calculate the reference angle α = tan⁻¹(|b|/|a|), and then adjust to find the principal argument according to the quadrant.

要将复数 z = a + bi 转换为模辐角形式，请遵循以下步骤。首先，计算模：r = √(a² + b²)。其次，根据 a 和 b 的符号确定象限。第三，计算参考角 α = tan⁻¹(|b|/|a|)，然后根据象限调整以求得主辐角。

Worked Example: Express z = -1 + √3 i in modulus-argument form.

例题：将 z = -1 + √3 i 表示为模辐角形式。

The modulus is r = √((-1)² + (√3)²) = √(1 + 3) = 2. The number lies in the second quadrant since a < 0 and b > 0. The reference angle is α = tan⁻¹(√3/1) = π/3. Therefore Arg(z) = π/3 + π = 2π/3. Hence z = 2(cos(2π/3) + i sin(2π/3)).

模为 r = √((-1)² + (√3)²) = √(1 + 3) = 2。该数位于第二象限，因为 a < 0 且 b > 0。参考角为 α = tan⁻¹(√3/1) = π/3。因此 Arg(z) = π/3 + π = 2π/3。所以 z = 2(cos(2π/3) + i sin(2π/3))。

7. Converting from Mod-Arg to Cartesian Form | 从模辐角形式转换为笛卡尔形式

Converting in the opposite direction is straightforward: given z = r(cos θ + i sin θ), the real part is a = r cos θ and the imaginary part is b = r sin θ. Simply evaluate the sine and cosine using known exact values or a calculator.

反向转换很简单：给定 z = r(cos θ + i sin θ)，实部为 a = r cos θ，虚部为 b = r sin θ。只需使用已知的精确值或计算器计算正弦和余弦即可。

Worked Example: Express z = 4(cos(5π/6) + i sin(5π/6)) in Cartesian form.

例题：将 z = 4(cos(5π/6) + i sin(5π/6)) 表示为笛卡尔形式。

We know cos(5π/6) = -√3/2 and sin(5π/6) = 1/2. Therefore a = 4 × (-√3/2) = -2√3, and b = 4 × (1/2) = 2. Hence z = -2√3 + 2i.

我们知道 cos(5π/6) = -√3/2，sin(5π/6) = 1/2。因此 a = 4 × (-√3/2) = -2√3，b = 4 × (1/2) = 2。所以 z = -2√3 + 2i。

This conversion is frequently tested in AQA exams together with exact trigonometric values. Make sure you are confident with the sine and cosine of π/6, π/4, π/3, and their multiples.

这种转换在 AQA 考试中经常与精确三角函数值一起考查。请确保你对 π/6、π/4、π/3 及其倍数的正弦和余弦值非常熟悉。

8. Geometric Interpretation on the Argand Diagram | 阿甘图上的几何解释

The modulus-argument form gives a beautiful geometric interpretation of complex numbers. The modulus r is the distance from the origin to the point representing z, and the argument θ is the angle between the vector Oz and the positive real axis. Thus, complex numbers are exactly like polar coordinates on a plane.

模辐角形式为复数提供了优美的几何解释。模 r 是原点与表示 z 的点之间的距离，辐角 θ 是向量 Oz 与正实轴之间的夹角。因此，复数本质上就是平面上的极坐标。

For example, the set of points satisfying |z| = 3 is a circle of radius 3 centred at the origin. The set of points satisfying arg(z) = π/4 is a half-line starting from the origin, making an angle of π/4 with the positive real axis, excluding the origin itself. These loci appear frequently in A-Level exam questions.

例如，满足 |z| = 3 的点集是一个以原点为圆心、半径为 3 的圆。满足 arg(z) = π/4 的点集是从原点出发、与正实轴成 π/4 角的半直线，且不包含原点本身。这些轨迹经常出现在 A-Level 考试题目中。

Moreover, the expression |z – z₁| represents the distance between the point z and the fixed point z₁. This is a powerful tool for describing circles and regions on the Argand diagram.

此外，表达式 |z – z₁| 表示点 z 与固定点 z₁ 之间的距离。这是描述阿甘图上圆和区域的强大工具。

9. Properties of the Modulus | 模的性质

The modulus satisfies several important algebraic properties that are essential for problem solving. For any complex numbers z₁ and z₂:

模满足几个重要的代数性质，对于解题至关重要。对于任意复数 z₁ 和 z₂：

The modulus of a product: |z₁z₂| = |z₁| × |z₂|. The modulus of a product equals the product of the moduli.
模的乘积性质： |z₁z₂| = |z₁| × |z₂|。乘积的模等于模的乘积。
The modulus of a quotient: |z₁ / z₂| = |z₁| / |z₂|, provided z₂ ≠ 0.
模的商性质： |z₁ / z₂| = |z₁| / |z₂|，前提是 z₂ ≠ 0。
The modulus of a conjugate: |z̄| = |z|, where z̄ is the complex conjugate of z.
共轭的模性质： |z̄| = |z|，其中 z̄ 是 z 的共轭复数。
Triangle inequality: |z₁ + z₂| ≤ |z₁| + |z₂|. Equality holds when z₁ and z₂ are in the same direction.
三角不等式： |z₁ + z₂| ≤ |z₁| + |z₂|。当 z₁ 和 z₂ 方向相同时等号成立。

These properties allow us to simplify complex expressions without fully expanding them. For example, if z = (3 + 4i) / (1 – i), we can find |z| = |3 + 4i| / |1 – i| = 5 / √2 without performing the full division.

这些性质使我们无需完全展开即可简化复杂的表达式。例如，若 z = (3 + 4i) / (1 – i)，我们可以直接求得 |z| = |3 + 4i| / |1 – i| = 5 / √2，而不必进行完整的除法。

10. Properties of Arguments | 辐角的性质

Similar properties hold for arguments, but special care must be taken because angles are periodic. For any complex numbers z₁ and z₂:

辐角也有类似的性质，但必须特别小心，因为角度是周期性的。对于任意复数 z₁ 和 z₂：

Argument of a product: arg(z₁z₂) = arg(z₁) + arg(z₂), possibly adjusted by adding or subtracting 2π to obtain the principal argument.
乘积的辐角： arg(z₁z₂) = arg(z₁) + arg(z₂)，可能需要加上或减去 2π 以得到主辐角。
Argument of a quotient: arg(z₁ / z₂) = arg(z₁) – arg(z₂), with the same periodic adjustment.
商的辐角： arg(z₁ / z₂) = arg(z₁) – arg(z₂)，同样需要周期性调整。
Argument of a conjugate: arg(z̄) = -arg(z), provided z is not on the negative real axis.
共轭的辐角： arg(z̄) = -arg(z)，前提是 z 不在负实轴上。
Argument of a negative: arg(-z) = arg(z) + π, adjusted to lie in (-π, π].
负数的辐角： arg(-z) = arg(z) + π，调整到 (-π, π] 范围内。

Because of the periodic nature, statements involving arguments should always be interpreted as equal modulo 2π. In exam solutions, always state the final argument in its principal value form.

由于角度的周期性，涉及辐角的结论都应理解为模 2π 意义下的相等。在考试解答中，始终以主值形式给出最终辐角。

11. Worked Exam-Style Examples | 考试风格例题

Example 1: Given that z = 2 – 2i, find |z| and Arg(z).

例题 1：已知 z = 2 – 2i，求 |z| 和 Arg(z)。

First, |z| = √(2² + (-2)²) = √8 = 2√2. The number lies in the fourth quadrant since a > 0 and b < 0. The reference angle is tan⁻¹(2/2) = tan⁻¹(1) = π/4. Therefore Arg(z) = -π/4. Hence z = 2√2(cos(-π/4) + i sin(-π/4)).

首先，|z| = √(2² + (-2)²) = √8 = 2√2。该数位于第四象限，因为 a > 0 且 b < 0。参考角为 tan⁻¹(2/2) = tan⁻¹(1) = π/4。因此 Arg(z) = -π/4。所以 z = 2√2(cos(-π/4) + i sin(-π/4))。

Example 2: The complex number w has modulus 3 and argument 2π/3. Write w in the form a + bi, where a and b are exact values.

例题 2：复数 w 的模为 3，辐角为 2π/3。将 w 写成 a + bi 的形式，其中 a 和 b 为精确值。

We have w = 3(cos(2π/3) + i sin(2π/3)) = 3(-1/2 + i × √3/2) = -3/2 + (3√3/2)i. Therefore a = -3/2 and b = 3√3/2.

我们有 w = 3(cos(2π/3) + i sin(2π/3)) = 3(-1/2 + i × √3/2) = -3/2 + (3√3/2)i。因此 a = -3/2，b = 3√3/2。

Example 3: If z₁ = 1 + i and z₂ = √3 – i, find the modulus and argument of z₁z₂.

例题 3：若 z₁ = 1 + i，z₂ = √3 – i，求 z₁z₂ 的模与辐角。

Using the properties: |z₁z₂| = |z₁| × |z₂| = √2 × 2 = 2√2. Now Arg(z₁) = π/4 since z₁ is in the first quadrant. For z₂, |z₂| = √(3 + 1) = 2, and z₂ is in the fourth quadrant, so Arg(z₂) = -π/6. Therefore Arg(z₁z₂) = π/4 + (-π/6) = π/12. This lies within the principal range, so the answer is π/12.

利用性质：|z₁z₂| = |z₁| × |z₂| = √2 × 2 = 2√2。因为 z₁ 在第一象限，Arg(z₁) = π/4。对于 z₂，|z₂| = √(3 + 1) = 2，且 z₂ 在第四象限，所以 Arg(z₂) = -π/6。因此 Arg(z₁z₂) = π/4 + (-π/6) = π/12。该值在主值范围内，所以答案为 π/12。

12. Common Mistakes and Exam Tips | 常见错误与考试技巧

Mistake 1: Incorrect quadrant adjustment. Many students simply compute tan⁻¹(b/a) and forget to adjust for the quadrant. Always sketch the point on the Argand diagram first, or at least check the signs of a and b carefully.

错误 1：象限调整错误。许多学生只计算 tan⁻¹(b/a) 而忘记根据象限调整。请务必先在阿甘图上画点，或者至少仔细检查 a 和 b 的符号。

Mistake 2: Mixing degrees and radians. A-Level exams require arguments in radians unless the question explicitly states otherwise. Always set your calculator to radian mode when working with arguments.

错误 2：混用角度制和弧度制。A-Level 考试要求辐角使用弧度制，除非题目明确说明。在使用辐角计算时，务必将计算器设为弧度模式。

Mistake 3: Forgetting the range of the principal argument. The principal argument lies in (-π, π], not [0, 2π). If your computed angle is outside this range, add or subtract 2π until it falls within the interval.

错误 3：忘记主辐角的范围。主辐角位于 (-π, π]，而不是 [0, 2π)。如果计算出的角度超出此范围，请加上或减去 2π 直到它落在该区间内。

Mistake 4: Sign errors in the imaginary part. When converting from polar form, remember that b = r sin θ can be negative. The sine of a negative argument is negative, which correctly places the point below the real axis.

错误 4：虚部符号错误。从极坐标形式转换时，记住 b = r sin θ 可以是负数。负辐角的正弦值为负，这正确地将点放置在实轴下方。

Exam Tip: Always write down the modulus and argument of each complex number separately before combining them. This structured approach reduces errors and earns method marks even if the final answer is incorrect.

考试技巧：在合并之前，务必分别写出每个复数的模和辐角。这种结构化的方法可以减少错误，即使最终答案不正确，也能获得方法分。

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The Modulus and Argument of a Complex Number | 复数的模与辐角