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

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

A complex number can be written in Cartesian form as z = a + bi, where a and b are real numbers and i satisfies i² = −1. While this algebraic notation is very useful, it does not immediately reveal the geometric meaning of the number. The modulus and argument provide a way to describe a complex number using a distance from the origin and an angle from the positive real axis.

复数可以写成直角坐标形式 z = a + bi，其中 a 和 b 为实数，i 满足 i² = −1。这种代数表示非常有用，但不能直接体现复数的几何意义。模与辐角通过从原点到点的距离以及与正实轴的夹角来描述一个复数。

1. The Argand Diagram | 阿干特图（复数平面）

An Argand diagram is a two-dimensional coordinate system in which the horizontal axis represents the real part a, and the vertical axis represents the imaginary part b. The complex number z = a + bi is plotted as the point (a, b). This visual representation makes the modulus and argument intuitive.

阿干特图是平面直角坐标系，其中水平轴表示实部 a，垂直轴表示虚部 b。复数 z = a + bi 被绘制为点 (a, b)。这种图形表示使得模与辐角的意义变得直观。

2. The Modulus | 模

The modulus of a complex number z = a + bi is the distance from the origin to the point (a, b) on the Argand diagram. It is denoted by |z| and is defined as:

|z| = |a + bi| = √(a² + b²)

Since it is a distance, the modulus is always a non-negative real number. For example, if z = 3 + 4i, then |z| = √(3² + 4²) = √25 = 5.

复数 z = a + bi 的模是从原点到阿干特图上点 (a, b) 的距离，记作 |z|，定义为：

|z| = |a + bi| = √(a² + b²)

由于模表示距离，它总是非负实数。例如，若 z = 3 + 4i，则 |z| = √(3² + 4²) = √25 = 5。

3. The Argument | 辐角

The argument of a complex number is the angle θ measured from the positive real axis to the line segment joining the origin and the point (a, b). It is denoted by arg z. For a non-zero complex number, the argument is not unique, because adding any multiple of 2π gives the same direction.

复数辐角是从正实轴到连接原点与点 (a, b) 的线段所成的角 θ，记作 arg z。对非零复数而言，辐角不唯一，因为加上 2π 的任何整数倍都表示同一方向。

tan θ = b / a

However, the value of θ must be chosen in the correct quadrant, which we discuss in the next sections.

tan θ = b / a

然而，必须根据象限选择正确的 θ 值，这一点将在后续小节讨论。

4. Principal Argument | 主辐角

To make the argument unique, we define the principal argument, usually written as Arg z (with a capital A). In AQA A-Level Mathematics, the principal argument is chosen in the interval:

−π < Arg z ≤ π

For example, the number 1 has argument 0, the number i has principal argument π/2, and the number −1 has principal argument π. The number 0 is the only complex number for which the argument is not defined.

为了使辐角唯一，我们定义主辐角，通常记作 Arg z（首字母大写）。在 AQA A-Level 数学中，主辐角选取在区间：

−π < Arg z ≤ π

例如，数 1 的辐角为 0，数 i 的主辐角为 π/2，数 −1 的主辐角为 π。只有复数 0 的辐角无定义。

5. Calculating the Argument from Cartesian Coordinates | 由直角坐标计算辐角

To calculate the principal argument of z = a + bi, we first compute the acute angle φ using the absolute values of a and b:

φ = arctan( |b| / |a| ) , a ≠ 0

Then the principal argument depends on which quadrant the point lies in. The table below summarises the rules.

要计算 z = a + bi 的主辐角，我们先用 a 和 b 的绝对值计算锐角 φ：

φ = arctan( |b| / |a| ) , a ≠ 0

然后根据点所在的象限确定主辐角，下表总结了规则。

Quadrant \| 象限	Signs of a, b \| a, b 的符号	Principal Argument \| 主辐角
I	a > 0, b > 0	θ = arctan(b/a)
II	a < 0, b > 0	θ = arctan(b/a) + π
III	a < 0, b < 0	θ = arctan(b/a) − π
IV	a > 0, b < 0	θ = arctan(b/a)

For points on the axes: if b = 0 and a > 0, then θ = 0; if b = 0 and a < 0, then θ = π; if a = 0 and b > 0, then θ = π/2; if a = 0 and b < 0, then θ = −π/2.

对于坐标轴上的点：若 b = 0 且 a > 0，则 θ = 0；若 b = 0 且 a < 0，则 θ = π；若 a = 0 且 b > 0，则 θ = π/2；若 a = 0 且 b < 0，则 θ = −π/2。

6. Properties of the Modulus | 模的性质

The modulus satisfies a number of important algebraic properties. Let z₁ and z₂ be arbitrary complex numbers. Then:

模满足若干重要的代数性质。设 z₁ 和 z₂ 为任意复数，则：

|z| ≥ 0, and |z| = 0 if and only if z = 0.

|z| ≥ 0，且 |z| = 0 当且仅当 z = 0。
|z₁ z₂| = |z₁| |z₂|

|z₁ z₂| = |z₁| |z₂|
|z₁ / z₂| = |z₁| / |z₂|, provided z₂ ≠ 0.

|z₁ / z₂| = |z₁| / |z₂|，其中 z₂ ≠ 0。
|z̄| = |z|, where z̄ is the complex conjugate of z.

|z̄| = |z|，其中 z̄ 是 z 的共轭复数。
|zⁿ| = |z|ⁿ for any positive integer n.

|zⁿ| = |z|ⁿ，对任意正整数 n 成立。
|z₁ + z₂| ≤ |z₁| + |z₂| (triangle inequality).

|z₁ + z₂| ≤ |z₁| + |z₂|（三角不等式）。

7. Properties of the Argument | 辐角的性质

Just like the modulus, the argument behaves nicely under multiplication and division. Because the argument is an angle, these results are only true up to adding multiples of 2π.

与模类似，辐角在乘法和除法下也有很好的性质。因为辐角是角度，这些结果在相差 2π 整数倍的意义下成立。

arg(z₁ z₂) = arg z₁ + arg z₂ (mod 2π)

arg(z₁ z₂) = arg z₁ + arg z₂（模 2π 意义下）
arg(z₁ / z₂) = arg z₁ − arg z₂ (mod 2π)

arg(z₁ / z₂) = arg z₁ − arg z₂（模 2π 意义下）
arg(z̄) = − arg z (mod 2π)

arg(z̄) = − arg z（模 2π 意义下）
arg(zⁿ) = n arg z (mod 2π)

arg(zⁿ) = n arg z（模 2π 意义下）

When using these rules, it is often necessary to add or subtract 2π so that the final answer is in the principal interval (−π, π].

使用这些法则时，通常需要加减 2π，使最终答案落在主值区间 (−π, π] 内。

8. Modulus-Argument Form | 模辐角形式（极形式）

Using the modulus r and the argument θ, any non-zero complex number can be written in modulus-argument form:

z = r (cos θ + i sin θ)

where r = |z| and θ = arg z. This form is also called the polar form of a complex number.

利用模 r 和辐角 θ，任何非零复数都可以写成模辐角形式：

z = r (cos θ + i sin θ)

其中 r = |z|，θ = arg z。这种形式也称为复数的极形式。

To convert from Cartesian to modulus-argument form, use:

r = √(a² + b²), θ = arg z

To convert from modulus-argument form back to Cartesian form, use:

a = r cos θ, b = r sin θ

从直角坐标形式转换为模辐角形式，使用：

r = √(a² + b²), θ = arg z

从模辐角形式转换回直角坐标形式，使用：

a = r cos θ, b = r sin θ

9. Worked Examples | 实例解析

Example 1: Find the modulus and principal argument of z = 1 + i.

例 1：求 z = 1 + i 的模与主辐角。

Here a = 1 and b = 1, so r = √(1² + 1²) = √2. Since the point lies in the first quadrant, θ = arctan(1/1) = π/4. Therefore |z| = √2 and Arg z = π/4.

这里 a = 1，b = 1，所以 r = √(1² + 1²) = √2。由于点在第一象限，θ = arctan(1/1) = π/4。因此 |z| = √2，Arg z = π/4。

Example 2: Find the modulus and principal argument of z = −√3 − i.

例 2：求 z = −√3 − i 的模与主辐角。

We have a = −√3 and b = −1, so r = √((−√3)² + (−1)²) = √(3 + 1) = 2. The point is in the third quadrant. Compute arctan(b/a) = arctan((−1)/(−√3)) = arctan(1/√3) = π/6. Then subtract π to obtain θ = π/6 − π = −5π/6. Hence Arg z = −5π/6.

我们有 a = −√3，b = −1，所以 r = √((−√3)² + (−1)²) = √(3 + 1) = 2。点在第三象限。计算 arctan(b/a) = arctan((−1)/(−√3)) = arctan(1/√3) = π/6，然后减去 π 得 θ = π/6 − π = −5π/6。因此 Arg z = −5π/6。

10. Exam Tips and Common Mistakes | 考试技巧与常见错误

When working with the modulus and argument, careful attention to detail will help you avoid common pitfalls.

在处理模与辐角时，细心关注细节有助于避免常见错误。

Always use the principal argument interval (−π, π] unless the question says otherwise.

除非题目另有说明，始终使用主辐角区间 (−π, π]。
Do not simply compute arctan(b/a) without considering the quadrant. Calculators usually return a value in (−π/2, π/2), which is only correct for quadrants I and IV.

不要不看象限就直接计算 arctan(b/a)。计算器通常返回 (−π/2, π/2) 内的值，这只在第一和第四象限正确。
Remember that 0 has no argument. It is the only complex number without a defined argument.

记住 0 没有辐角。它是唯一没有定义辐角的复数。
When using properties of arguments, convert the final answer back to the principal value.

使用辐角性质后，要将最终答案转换回主值。
Draw a quick sketch on an Argand diagram. This helps you verify the quadrant and the approximate size of the angle.

在阿干特图上快速画草图。这有助于你验证象限和角的大致大小。

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