The General Complex Number | 一般复数

📚 The General Complex Number | 一般复数

The general complex number is one of the most elegant extensions of the real number system. It allows us to solve equations that have no real solutions, such as x² = −1, and provides a unifying framework for many areas of mathematics, physics and engineering. In this article we explore the definition, representation and arithmetic of complex numbers at A-Level standard.

一般复数是实数系统最优雅的扩展之一。它使我们能够求解没有实数解的方程，例如 x² = −1，并为数学、物理学和工程学的许多领域提供了统一的框架。本文在 A-Level 标准下探讨复数的定义、表示与运算。

1. What is a Complex Number? | 什么是复数？

A complex number is a number of the form z = a + bi, where a and b are real numbers, and i is the imaginary unit defined by i² = −1. The letter z is conventionally used to denote a complex number, while a and b are often replaced by x and y in coordinate contexts. Every real number is also a complex number with b = 0.

复数是指形如 z = a + bi 的数，其中 a 和 b 是实数，而 i 是由 i² = −1 定义的虚数单位。习惯上用字母 z 表示复数，在坐标语境中 a 和 b 常被替换为 x 和 y。每个实数也都是复数，此时 b = 0。

2. Real and Imaginary Parts | 实部与虚部

For a complex number z = a + bi, we call a the real part and b the imaginary part. These are written as Re(z) = a and Im(z) = b. It is important to note that the imaginary part is the real number b, not bi. For example, if z = 3 − 4i, then Re(z) = 3 and Im(z) = −4.

对于复数 z = a + bi，我们称 a 为实部，b 为虚部，记作 Re(z) = a 和 Im(z) = b。需要特别注意，虚部是实数 b，而不是 bi。例如，若 z = 3 − 4i，则 Re(z) = 3，Im(z) = −4。

3. The Argand Diagram | 阿根图

A complex number z = a + bi can be represented as a point on a plane called the Argand diagram, with the horizontal axis representing the real part and the vertical axis representing the imaginary part. This is analogous to Cartesian coordinates (a, b), and it enables us to visualise complex numbers geometrically.

复数 z = a + bi 可以在称为阿根图的平面上表示为一个点，其中横轴表示实部，纵轴表示虚部。这类似于直角坐标 (a, b)，使我们能够在几何上直观地理解复数。

For example, the point (2, 3) corresponds to the complex number 2 + 3i, and the point (−1, 2) corresponds to −1 + 2i. The distance of the point from the origin is called the modulus, and the angle it makes with the positive real axis is called the argument.

例如，点 (2, 3) 对应复数 2 + 3i，点 (−1, 2) 对应复数 −1 + 2i。该点到原点的距离称为模，它与正实轴所成的角称为辐角。

4. Modulus and Argument | 模与辐角

The modulus of z = a + bi is defined as |z| = √(a² + b²). It measures the distance from the origin to the point (a, b) in the Argand diagram. The argument, denoted arg(z) or θ, is the angle between the positive real axis and the line joining the point to the origin, measured in radians in the interval (−π, π] for the principal value.

复数 z = a + bi 的模定义为 |z| = √(a² + b²)，它衡量阿根图中点 (a, b) 到原点的距离。辐角记作 arg(z) 或 θ，是正实轴与连接原点和该点的直线之间的夹角，主值以弧度表示，取值范围为 (−π, π]。

Using trigonometry, we can write a = |z| cos θ and b = |z| sin θ, giving the polar form z = r(cos θ + i sin θ), where r = |z|. For example, the complex number 1 + i has modulus √2 and argument π/4, so its polar form is √2(cos π/4 + i sin π/4).

利用三角函数，可以写成 a = |z| cos θ，b = |z| sin θ，从而得到极坐标形式 z = r(cos θ + i sin θ)，其中 r = |z|。例如，复数 1 + i 的模为 √2，辐角为 π/4，因此其极坐标形式为 √2(cos π/4 + i sin π/4)。

5. Complex Conjugation | 共轭复数

The complex conjugate of z = a + bi is denoted z̄ = a − bi. It is obtained by changing the sign of the imaginary part. Geometrically, z̄ is the reflection of z in the real axis of the Argand diagram. For example, the conjugate of 5 + 2i is 5 − 2i, and the conjugate of −3i is 3i.

复数 z = a + bi 的共轭复数记作 z̄ = a − bi，通过改变虚部的符号得到。在几何上，z̄ 是 z 关于阿根图实轴的镜像。例如，5 + 2i 的共轭为 5 − 2i，−3i 的共轭为 3i。

Conjugation is extremely useful because z + z̄ = 2a and z − z̄ = 2bi, which allow us to isolate the real and imaginary parts. Also, zz̄ = a² + b² = |z|², a real positive number, which is the key to dividing complex numbers.

共轭的用途非常广泛，因为 z + z̄ = 2a，z − z̄ = 2bi，可用于分离实部和虚部。同时，zz̄ = a² + b² = |z|²，是一个非负实数，这是进行复数除法的关键。

6. Addition and Subtraction | 加法与减法

To add or subtract complex numbers, we combine their real parts and imaginary parts separately. If z₁ = a + bi and z₂ = c + di, then z₁ + z₂ = (a + c) + (b + d)i and z₁ − z₂ = (a − c) + (b − d)i.

复数相加或相减时，分别合并实部和虚部。若 z₁ = a + bi，z₂ = c + di，则 z₁ + z₂ = (a + c) + (b + d)i，z₁ − z₂ = (a − c) + (b − d)i。

For instance, (3 + 4i) + (2 − i) = 5 + 3i, and (3 + 4i) − (2 − i) = 1 + 5i. Vector addition of the corresponding points in the Argand diagram exactly matches this operation.

例如，(3 + 4i) + (2 − i) = 5 + 3i，而 (3 + 4i) − (2 − i) = 1 + 5i。阿根图中相应点的向量加法与这一运算完全一致。

7. Multiplication | 乘法

Multiplication of complex numbers follows the ordinary rules of algebra, with the key replacement i² = −1. If z₁ = a + bi and z₂ = c + di, then z₁z₂ = (ac − bd) + (ad + bc)i. This result is obtained by expanding (a + bi)(c + di) and simplifying.

复数乘法遵循通常的代数法则，关键是代换 i² = −1。若 z₁ = a + bi，z₂ = c + di，则 z₁z₂ = (ac − bd) + (ad + bc)i。该结果由展开 (a + bi)(c + di) 并化简得到。

For example, (2 + 3i)(1 − i) = 2 − 2i + 3i − 3i² = 2 + i + 3 = 5 + i. Note that the product of a complex number and its conjugate is always real: (a + bi)(a − bi) = a² + b².

例如，(2 + 3i)(1 − i) = 2 − 2i + 3i − 3i² = 2 + i + 3 = 5 + i。注意，复数与其共轭的乘积总是实数：(a + bi)(a − bi) = a² + b²。

8. Division | 除法

To divide complex numbers, we multiply both the numerator and the denominator by the conjugate of the denominator. This rationalises the denominator, turning it into a real number. If z₁ = a + bi and z₂ = c + di (with z₂ ≠ 0), then z₁/z₂ = (a + bi)/(c + di) × (c − di)/(c − di) = ((ac + bd) + (bc − ad)i)/(c² + d²).

复数除法需要将分子和分母同时乘以分母的共轭复数，从而将分母化为实数。若 z₁ = a + bi，z₂ = c + di（且 z₂ ≠ 0），则 z₁/z₂ = (a + bi)/(c + di) × (c − di)/(c − di) = ((ac + bd) + (bc − ad)i)/(c² + d²)。

As an example, (3 + i)/(1 − i) = ((3 + i)(1 + i))/((1 − i)(1 + i)) = (3 + 3i + i + i²)/(1 + 1) = (2 + 4i)/2 = 1 + 2i.

例如，(3 + i)/(1 − i) = ((3 + i)(1 + i))/((1 − i)(1 + i)) = (3 + 3i + i + i²)/(1 + 1) = (2 + 4i)/2 = 1 + 2i。

9. Equality of Complex Numbers | 复数相等

Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. That is, a + bi = c + di ⇔ a = c and b = d. This principle is often used to solve equations involving complex unknowns.

两个复数相等当且仅当它们的实部相等且虚部相等。即 a + bi = c + di ⇔ a = c 且 b = d。这一原理常用于求解含复未知数的方程。

For example, if (x + yi)(2 + i) = 5 − i, we can expand the left-hand side to obtain (2x − y) + (x + 2y)i = 5 − i. Then equating real and imaginary parts gives the simultaneous equations 2x − y = 5 and x + 2y = −1, which yield x = 9/5 and y = −7/5.

例如，若 (x + yi)(2 + i) = 5 − i，展开左边得到 (2x − y) + (x + 2y)i = 5 − i。然后令实部、虚部分别相等，得到联立方程 2x − y = 5 和 x + 2y = −1，解得 x = 9/5，y = −7/5。

10. Properties of Modulus and Conjugate | 模与共轭的性质

There are several key properties that greatly simplify calculations with complex numbers. For any complex numbers z and w, the conjugate of a sum is the sum of the conjugates: z̄ + w̄ = z̄ + w̄. Similarly, z̄w̄ = z̄ × w̄, and (z/w)̄ = z̄/w̄ for w ≠ 0.

有一些重要性质能极大简化复数运算。对任意复数 z 和 w，和的共轭等于共轭的和：z̄ + w̄ = z̄ + w̄。类似地，z̄w̄ = z̄ × w̄，且当 w ≠ 0 时，(z/w)̄ = z̄/w̄。

For the modulus, |zw| = |z||w| and |z/w| = |z|/|w|. Also, |z̄| = |z|, and |z|² = zz̄. These identities are frequently used in A-Level problems to avoid expanding complex expressions fully.

关于模，|zw| = |z||w|，|z/w| = |z|/|w|。同时，|z̄| = |z|，且 |z|² = zz̄。这些恒等式在 A-Level 题目中经常用来避免完全展开复杂表达式。

11. Square Roots of a Complex Number | 复数的平方根

Finding the square root of a complex number z = a + bi involves solving (x + yi)² = a + bi. Expanding gives (x² − y²) + 2xyi = a + bi, so x² − y² = a and 2xy = b. These two equations can be solved simultaneously, together with the relation x² + y² = √(a² + b²).

求复数 z = a + bi 的平方根需要解 (x + yi)² = a + bi。展开得 (x² − y²) + 2xyi = a + bi，因此 x² − y² = a，2xy = b。这两个方程可联立求解，并利用关系 x² + y² = √(a² + b²)。

For instance, to find √(3 + 4i), set x² − y² = 3 and 2xy = 4. Using x² + y² = 5, we find x² = 4 and y² = 1. Since xy = 2 > 0, x and y have the same sign, so the roots are 2 + i and −2 − i.

例如，求 √(3 + 4i) 时，令 x² − y² = 3，2xy = 4。利用 x² + y² = 5，可得 x² = 4，y² = 1。由于 xy = 2 > 0，x 与 y 同号，因此平方根为 2 + i 和 −2 − i。

12. Solving Quadratic Equations | 解二次方程

Complex numbers allow every quadratic equation to have solutions. For ax² + bx + c = 0, the formula x = (−b ± √(b² − 4ac))/(2a) remains valid. When the discriminant b² − 4ac is negative, the solutions are complex conjugates of each other.

复数使每个二次方程都有解。对于 ax² + bx + c = 0，求根公式 x = (−b ± √(b² − 4ac))/(2a) 仍然成立。当判别式 b² − 4ac 为负时，解为一对共轭复数。

For example, x² + 4x + 5 = 0 has discriminant 16 − 20 = −4. Thus x = (−4 ± √(−4))/2 = (−4 ± 2i)/2 = −2 ± i. The two solutions are −2 + i and −2 − i, confirming the conjugate pair property when coefficients are real.

例如，x² + 4x + 5 = 0 的判别式为 16 − 20 = −4。因此 x = (−4 ± √(−4))/2 = (−4 ± 2i)/2 = −2 ± i。两个解为 −2 + i 和 −2 − i，验证了实系数方程的解成共轭对出现的性质。

In summary, the general complex number z = a + bi provides a complete and consistent system for solving equations, performing algebraic operations and modelling rotations and oscillations. Mastering its definition, geometric representation, conjugation and arithmetic is essential for success in A-Level Mathematics and beyond.

总之，一般复数 z = a + bi 提供了一个完备且自洽的系统，用于求解方程、进行代数运算以及建模旋转和振荡。掌握其定义、几何表示、共轭及运算是 A-Level 数学及更深入学习中取得成功的必要条件。

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The General Complex Number | 一般复数