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 AQA A-level Mathematics, a complex number is often first written in Cartesian form z = x + iy. However, the same number can also be described by how far it is from the origin and by the angle it makes with the positive real axis. These two quantities are called the modulus and the argument, and they lead to a very useful way of writing complex numbers: the modulus-argument form.

在 AQA A-level 数学中，复数通常先写成笛卡儿形式 z = x + iy。但同一个复数也可以用它与原点的距离以及与正实轴所成的角度来描述。这两个量分别称为模和辐角，它们引出了复数的模-辐角形式，这是一种非常有用的表示方法。

This article covers the definitions, conversion methods, arithmetic rules and common mistakes involving the modulus and argument of a complex number.

本文涵盖复数的模和辐角的定义、转换方法、运算规则以及常见错误。

1. Cartesian Form and the Argand Diagram | 笛卡儿形式与阿甘特图

A complex number can be written as z = x + iy, where x is the real part Re(z) and y is the imaginary part Im(z). We represent z as the point (x, y) on an Argand diagram, where the horizontal axis is the real axis and the vertical axis is the imaginary axis.

复数可以写成 z = x + iy，其中 x 是实部 Re(z)，y 是虚部 Im(z)。我们在阿甘特图上把 z 表示为点 (x, y)，其中横轴是实轴，纵轴是虚轴。

This geometric view is essential because the modulus measures the distance from the origin to the point (x, y), and the argument measures the angle from the positive real axis to the line joining the origin to (x, y).

这种几何观点非常重要，因为模度量的是原点到点 (x, y) 的距离，而辐角度量的是从正实轴到连接原点与 (x, y) 的直线的角度。

2. The Modulus of a Complex Number | 复数的模

For a complex number z = x + iy, the modulus of z is denoted by |z| and is defined as:

对于复数 z = x + iy，z 的模记作 |z|，定义为：

|z| = √(x² + y²)

The modulus is always a non-negative real number. Geometrically, |z| is the distance from the origin to the point (x, y) on the Argand diagram, so it behaves like the length of the vector representing z.

模始终是一个非负实数。从几何上看，|z| 是原点到阿甘特图上点 (x, y) 的距离，因此它就像表示 z 的向量的长度。

For example, if z = 3 – 4i, then |z| = √(3² + (-4)²) = √(9 + 16) = √25 = 5. This tells us that the point (3, -4) is 5 units from the origin.

例如，如果 z = 3 – 4i，那么 |z| = √(3² + (-4)²) = √(9 + 16) = √25 = 5。这说明点 (3, -4) 到原点的距离是 5 个单位。

Two useful properties are that |z| = 0 only when z = 0, and the modulus of a complex number is equal to the modulus of its conjugate: |z̄| = |z|.

两个常用性质是：只有当 z = 0 时才有 |z| = 0；复数与其共轭复数具有相同的模，即 |z̄| = |z|。

3. The Argument of a Complex Number | 复数的辐角

The argument of a non-zero complex number z = x + iy is the angle θ measured from the positive real axis to the vector representing z, usually measured in radians. It is often written as arg z.

非零复数 z = x + iy 的辐角是从正实轴到表示 z 的向量的角度 θ，通常以弧度为单位。它常写作 arg z。

The argument is not unique: adding or subtracting any multiple of 2π gives the same direction. For example, π/4, 9π/4 and -7π/4 all describe the same direction.

辐角不是唯一的：加上或减去 2π 的任意整数倍都会得到相同的方向。例如，π/4、9π/4 和 -7π/4 都表示同一个方向。

To remove this ambiguity, AQA uses the principal argument, written Arg z with a capital A. The principal argument satisfies:

为了消除这种不唯一性，AQA 使用主辐角，用大写字母 A 写作 Arg z。主辐角满足：

-π < Arg z ≤ π

For z = 1 + i, the point (1, 1) lies on the line y = x in the first quadrant, so Arg z = π/4.

对于 z = 1 + i，点 (1, 1) 位于第一象限的直线 y = x 上，因此 Arg z = π/4。

4. Finding the Argument: Quadrant Rules | 求辐角：象限规则

If z = x + iy with x ≠ 0 and y ≠ 0, a useful first step is to calculate the acute reference angle:

如果 z = x + iy 且 x ≠ 0、y ≠ 0，一个有用的第一步是计算锐角参考角：

α = arctan(|y| / |x|)

The quadrant in which the point (x, y) lies then determines the principal argument. The table below summarises the four cases.

点 (x, y) 所在的象限决定了主辐角。下表总结了四种情况。

Quadrant \| 象限	Sign of x and y \| x、y 的符号	Principal argument \| 主辐角
I	x > 0, y > 0	Arg z = α
II	x < 0, y > 0	Arg z = π – α
III	x < 0, y < 0	Arg z = α – π
IV	x > 0, y < 0	Arg z = -α

For example, if z = -1 – i, then α = arctan(|-1| / |-1|) = π/4. The point (-1, -1) is in the third quadrant, so Arg z = π/4 – π = -3π/4.

例如，如果 z = -1 – i，则 α = arctan(|-1| / |-1|) = π/4。点 (-1, -1) 在第三象限，因此 Arg z = π/4 – π = -3π/4。

Special cases on the axes are:

坐标轴上的特殊情况如下：

If x > 0 and y = 0, then Arg z = 0.

如果 x > 0 且 y = 0，那么 Arg z = 0。
If x = 0 and y > 0, then Arg z = π/2.

如果 x = 0 且 y > 0，那么 Arg z = π/2。
If x < 0 and y = 0, then Arg z = π.

如果 x < 0 且 y = 0，那么 Arg z = π。
If x = 0 and y < 0, then Arg z = -π/2.

如果 x = 0 且 y < 0，那么 Arg z = -π/2。

The argument of the zero complex number 0 is undefined, because the zero vector has no clear direction.

零复数 0 的辐角是未定义的，因为零向量没有明确的方向。

5. Modulus-Argument Form | 模-辐角形式

Once the modulus r = |z| and an argument θ = arg z are known, the complex number can be written in modulus-argument form:

一旦知道了模 r = |z| 和辐角 θ = arg z，复数就可以写成模-辐角形式：

z = r(cos θ + i sin θ)

This is sometimes called polar form. The real part is x = r cos θ and the imaginary part is y = r sin θ.

这有时也称为极坐标形式。实部为 x = r cos θ，虚部为 y = r sin θ。

For example, suppose z = 2(cos π/3 + i sin π/3). Since cos π/3 = 1/2 and sin π/3 = √3/2, we get z = 2(1/2 + i√3/2) = 1 + i√3.

例如，设 z = 2(cos π/3 + i sin π/3)。因为 cos π/3 = 1/2，sin π/3 = √3/2，所以得到 z = 2(1/2 + i√3/2) = 1 + i√3。

You can use either the principal argument or any equivalent argument in this form; the cosine and sine will produce the same Cartesian form.

在这种形式中，你可以使用主辐角或任何等价辐角；余弦和正弦会产生相同的笛卡儿形式。

6. Converting Between Cartesian and Modulus-Argument Form | 笛卡儿形式与模-辐角形式之间的转换

To convert from Cartesian form z = x + iy to modulus-argument form:

要从笛卡儿形式 z = x + iy 转换为模-辐角形式：

Calculate r = √(x² + y²).

计算 r = √(x² + y²)。
Find the principal argument θ = Arg z using the quadrant rules.

使用象限规则求出主辐角 θ = Arg z。
Write z = r(cos θ + i sin θ).

写出 z = r(cos θ + i sin θ)。

For example, convert z = -1 + i√3. We have r = √((-1)² + (√3)²) = √(1 + 3) = 2. The reference angle is α = arctan(|√3 / -1|) = π/3. Since x < 0 and y > 0, the point is in the second quadrant, so Arg z = π – π/3 = 2π/3. Therefore z = 2(cos 2π/3 + i sin 2π/3).

例如，将 z = -1 + i√3 进行转换。我们有 r = √((-1)² + (√3)²) = √(1 + 3) = 2。参考角为 α = arctan(|√3 / -1|) = π/3。因为 x < 0 且 y > 0，该点位于第二象限，所以 Arg z = π – π/3 = 2π/3。因此 z = 2(cos 2π/3 + i sin 2π/3)。

To convert from modulus-argument form to Cartesian form, simply expand r cos θ and r sin θ to obtain the real and imaginary parts.

要从模-辐角形式转换为笛卡儿形式，只需展开 r cos θ 和 r sin θ 得到实部和虚部即可。

7. Multiplication and Division Using Modulus and Argument | 用模和辐角进行乘法与除法

Let z₁ = r₁(cos θ₁ + i sin θ₁) and z₂ = r₂(cos θ₂ + i sin θ₂). Then their product is:

设 z₁ = r₁(cos θ₁ + i sin θ₁)，z₂ = r₂(cos θ₂ + i sin θ₂)。那么它们的积为：

z₁z₂ = r₁r₂ [cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)]

This means the moduli multiply and the arguments add:

这意味着模相乘，辐角相加：

|z₁z₂| = |z₁| |z₂| and arg

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