The Polar Form of a Complex Number | 复数的极坐标形式

📚 The Polar Form of a Complex Number | 复数的极坐标形式

In AQA A-Level Mathematics, complex numbers are introduced first in Cartesian form \( x + yi \), but the polar form — also called modulus-argument form — reveals a deeper geometric structure. It simplifies multiplication, division, powers, and roots, and connects algebra to trigonometry in a powerful way.

在 AQA A-Level 数学中，复数首先以笛卡尔形式 \( x + yi \) 引入，但极坐标形式（也称模-辐角形式）揭示了更深层的几何结构。它简化了乘法、除法、幂和求根运算，并将代数与三角学紧密相连。

1. Representing a Complex Number Geometrically | 复数的几何表示

Every complex number \( z = x + yi \) corresponds to a point \( (x, y) \) in the Argand diagram. The horizontal axis is the real axis and the vertical axis is the imaginary axis. This representation is known as the complex plane.

每一个复数 \( z = x + yi \) 都对应复平面（阿甘图）中的一个点 \( (x, y) \)。水平轴为实轴，垂直轴为虚轴，这种表示方法称为复平面。

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

The modulus of \( z = x + yi \), denoted \( |z| \), is the distance from the origin to the point \( (x, y) \). It is given by \( |z| = \sqrt{x^2 + y^2} \). The modulus is always a non-negative real number.

复数 \( z = x + yi \) 的模，记作 \( |z| \)，是原点与点 \( (x, y) \) 之间的距离。其计算公式为 \( |z| = \sqrt{x^2 + y^2} \)。模永远是非负实数。

|z| = √(x² + y²)

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

The argument of \( z \) is the angle \( \theta \) that the line from the origin to the point makes with the positive real axis, measured in radians. It is usually taken in the principal range \( -\pi < \theta \leq \pi \). The principal argument is written as \( \arg(z) \).

复数 \( z \) 的辐角是原点指向该点的线段与正实轴之间的夹角 \( \theta \)，以弧度为单位。通常取主值范围 \( -\pi < \theta \leq \pi \)，主辐角记作 \( \arg(z) \)。

tan θ = y / x

Care must be taken when determining \( \theta \) from \( \tan \theta = y / x \), because the inverse tangent function only returns values in \( (-\pi/2, \pi/2) \). You must adjust the angle according to the quadrant of \( (x, y) \).

由 \( \tan \theta = y / x \) 求 \( \theta \) 时必须小心，因为反正切函数仅返回 \( (-\pi/2, \pi/2) \) 内的值。你需要根据 \( (x, y) \) 所在的象限对角度进行调整。

4. The Polar Form Definition | 极坐标形式的定义

For a complex number with modulus \( r \) and argument \( \theta \), the polar form is written as \( z = r(\cos \theta + i \sin \theta) \). This is often abbreviated as \( z = r \operatorname{cis} \theta \).

对于模为 \( r \)、辐角为 \( \theta \) 的复数，其极坐标形式写作 \( z = r(\cos \theta + i \sin \theta) \)，常简写为 \( z = r \operatorname{cis} \theta \)。

z = r(cos θ + i sin θ) = r cis θ

This form directly connects the Cartesian coordinates with the polar coordinates \( r \) and \( \theta \): \( x = r \cos \theta \), \( y = r \sin \theta \).

这种形式直接将笛卡尔坐标与极坐标 \( r \) 和 \( \theta \) 联系了起来：\( x = r \cos \theta \)，\( y = r \sin \theta \)。

5. Converting from Cartesian to Polar | 从笛卡尔形式转换为极坐标形式

To convert \( z = x + yi \) into polar form, first compute \( r = |z| = \sqrt{x^2 + y^2} \). Then determine \( \theta \) by combining \( \tan \theta = y / x \) with the quadrant information.

将 \( z = x + yi \) 转换为极坐标形式，首先计算 \( r = |z| = \sqrt{x^2 + y^2} \)，然后结合 \( \tan \theta = y / x \) 及象限信息确定 \( \theta \)。

If \( x > 0, y > 0 \): \( \theta = \arctan(y/x) \), first quadrant.
If \( x < 0, y > 0 \): \( \theta = \pi – \arctan(|y/x|) \), second quadrant.
If \( x < 0, y < 0 \): \( \theta = -\pi + \arctan(|y/x|) \), third quadrant.
If \( x > 0, y < 0 \): \( \theta = -\arctan(|y/x|) \), fourth quadrant.

例如：\( z = 1 – i \sqrt{3} \)，则 \( r = \sqrt{1 + 3} = 2 \)，\( \tan \theta = -\sqrt{3} \)，且位于第四象限，所以 \( \theta = -\pi/3 \)。故极坐标形式为 \( z = 2 \operatorname{cis}(-\pi/3) \)。

For example: for \( z = 1 – i \sqrt{3} \), \( r = \sqrt{1 + 3} = 2 \), \( \tan \theta = -\sqrt{3} \), and the point lies in the fourth quadrant, so \( \theta = -\pi/3 \). Hence the polar form is \( z = 2 \operatorname{cis}(-\pi/3) \).

6. Converting from Polar to Cartesian | 从极坐标形式转换为笛卡尔形式

Given \( z = r \operatorname{cis} \theta \), the Cartesian form is \( z = r \cos \theta + i r \sin \theta \). Use known values of sine and cosine to evaluate the expression exactly or approximately.

已知 \( z = r \operatorname{cis} \theta \)，其笛卡尔形式为 \( z = r \cos \theta + i r \sin \theta \)。利用已知的三角函数值可以精确或近似地计算该表达式。

x = r cos θ, y = r sin θ

例如：\( z = 4 \operatorname{cis}(\pi/2) \)。则 \( x = 4 \cos(\pi/2) = 0 \)，\( y = 4 \sin(\pi/2) = 4 \)，所以 \( z = 4i \)。

For instance: \( z = 4 \operatorname{cis}(\pi/2) \). Then \( x = 4 \cos(\pi/2) = 0 \), \( y = 4 \sin(\pi/2) = 4 \), so \( z = 4i \).

7. Multiplication and Division in Polar Form | 极坐标形式的乘法与除法

The polar form makes multiplication and division extremely elegant. If \( z_1 = r_1 \operatorname{cis} \theta_1 \) and \( z_2 = r_2 \operatorname{cis} \theta_2 \), then:

极坐标形式使乘法和除法变得极为简洁。若 \( z_1 = r_1 \operatorname{cis} \theta_1 \)，\( z_2 = r_2 \operatorname{cis} \theta_2 \)，则：

z₁ z₂ = r₁ r₂ cis(θ₁ + θ₂)

z₁ / z₂ = (r₁ / r₂) cis(θ₁ − θ₂)

So the moduli are multiplied (or divided) and the arguments are added (or subtracted). This is a direct consequence of the compound angle identities for sine and cosine.

也就是说，模相乘（或相除），辐角相加（或相减）。这是正弦、余弦的合角公式的直接结果。

8. De Moivre’s Theorem | 棣莫弗定理

De Moivre’s theorem states that for any integer \( n \), \( (\operatorname{cis} \theta)^n = \operatorname{cis}(n\theta) \). More generally, for \( z = r \operatorname{cis} \theta \), \( z^n = r^n \operatorname{cis}(n\theta) \).

棣莫弗定理指出：对于任意整数 \( n \)，有 \( (\operatorname{cis} \theta)^n = \operatorname{cis}(n\theta) \)。更一般地，若 \( z = r \operatorname{cis} \theta \)，则 \( z^n = r^n \operatorname{cis}(n\theta) \)。

(r cis θ)ⁿ = rⁿ cis(nθ)

This theorem is extremely useful for expanding \( \cos(n\theta) \) or \( \sin(n\theta) \) in terms of powers of \( \cos \theta \) and \( \sin \theta \), and for computing powers of complex numbers without first converting to Cartesian form.

该定理在将 \( \cos(n\theta) \) 或 \( \sin(n\theta) \) 展开为 \( \cos \theta \) 和 \( \sin \theta \) 的幂，以及直接计算复数幂而无需先转换为笛卡尔形式时非常有用。

9. Finding Roots of Complex Numbers | 求复数的方根

To find all \( n \)-th roots of a complex number \( z = r \operatorname{cis} \theta \), we use the general formula:

要求复数 \( z = r \operatorname{cis} \theta \) 的所有 \( n \) 次方根，我们使用一般公式：

wₖ = r^(1/n) cis((θ + 2kπ) / n), k = 0, 1, 2, …, n−1

The \( n \) distinct roots are equally spaced around a circle of radius \( r^{1/n} \) in the Argand diagram. Each successive root is rotated by \( 2\pi / n \) radians.

这 \( n \) 个不同的根在复平面上均匀分布在半径为 \( r^{1/n} \) 的圆上，相邻根之间的辐角相差 \( 2\pi / n \) 弧度。

例如，求 \( z^3 = 8i \) 的根。首先 \( 8i = 8 \operatorname{cis}(\pi/2) \)。则 \( w_k = 2 \operatorname{cis}((\pi/2 + 2k\pi)/3) \)，\( k = 0, 1, 2 \)。根分别为 \( 2\operatorname{cis}(\pi/6) \)，\( 2\operatorname{cis}(5\pi/6) \)，\( 2\operatorname{cis}(3\pi/2) \)。

For example, find the cube roots of \( 8i \). First write \( 8i = 8 \operatorname{cis}(\pi/2) \). Then \( w_k = 2 \operatorname{cis}((\pi/2 + 2k\pi)/3) \), \( k = 0, 1, 2 \). The roots are \( 2\operatorname{cis}(\pi/6) \), \( 2\operatorname{cis}(5\pi/6) \), and \( 2\operatorname{cis}(3\pi/2) \).

10. Euler’s Formula and Exponential Form | 欧拉公式与指数形式

Euler’s formula states \( e^{i\theta} = \cos \theta + i \sin \theta \). This gives another compact way to write a complex number: \( z = r e^{i\theta} \).

欧拉公式表明 \( e^{i\theta} = \cos \theta + i \sin \theta \)。这给出了复数的另一种紧凑写法：\( z = r e^{i\theta} \)。

z = r e^{iθ}

This exponential form obeys the usual index laws, so \( z_1 z_2 = r_1 r_2 e^{i(\theta_1+\theta_2)} \) and \( z^n = r^n e^{i n \theta} \). It also provides a direct link to \( e^{i\pi} + 1 = 0 \), the famous identity.

这种指数形式遵循通常的指数运算法则，因此 \( z_1 z_2 = r_1 r_2 e^{i(\theta_1+\theta_2)} \)，\( z^n = r^n e^{i n \theta} \)。它还直接联系了著名的恒等式 \( e^{i\pi} + 1 = 0 \)。

11. Geometric Interpretation and Loci | 几何解释与轨迹

The polar form naturally describes circular motion. The equation \( |z – a| = r \) represents a circle with centre \( a \) and radius \( r \). The equation \( \arg(z – a) = \theta \) represents a half-line starting at \( a \), excluding the point \( a \) itself.

极坐标形式自然地描述了圆周运动。方程 \( |z – a| = r \) 表示以 \( a \) 为圆心、\( r \) 为半径的圆。方程 \( \arg(z – a) = \theta \) 表示从点 \( a \) 出发（不含 \( a \)）的半射线。

These loci are often tested in AQA exams alongside polar form. Understanding how multiplication by a complex number rotates and scales another complex number is essential for such problems.

在 AQA 考试中，这些轨迹常常与极坐标形式一起考查。理解复数乘法是如何旋转和缩放另一个复数，对解决这类问题至关重要。

12. Worked Exam-Style Example | 考试风格例题

Given \( z = 1 + i \), find the modulus and principal argument, then write \( z \) in polar form and compute \( z^{10} \).

已知 \( z = 1 + i \)，求模与主辐角，写出极坐标形式，并计算 \( z^{10} \)。

Solution: \( r = \sqrt{1^2 + 1^2} = \sqrt{2} \). Since the point is in the first quadrant, \( \theta = \tan^{-1}(1) = \pi/4 \). So \( z = \sqrt{2} \operatorname{cis}(\pi/4) \).

解：\( r = \sqrt{1^2 + 1^2} = \sqrt{2} \)。该点在第一象限，故 \( \theta = \tan^{-1}(1) = \pi/4 \)。因此 \( z = \sqrt{2} \operatorname{cis}(\pi/4) \)。

Then \( z^{10} = (\sqrt{2})^{10} \operatorname{cis}(10 \cdot \pi/4) = 2^{5} \operatorname{cis}(5\pi/2) = 32 \operatorname{cis}(\pi/2) = 32i \).

于是 \( z^{10} = (\sqrt{2})^{10} \operatorname{cis}(10 \cdot \pi/4) = 2^{5} \operatorname{cis}(5\pi/2) = 32 \operatorname{cis}(\pi/2) = 32i \)。

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The Polar Form of a Complex Number | 复数的极坐标形式