Complex Numbers | 复数

📚 Complex Numbers | 复数

In A-Level Mathematics, complex numbers extend the real number system to solve equations that have no real solutions, such as x² + 1 = 0. They are fundamental in many areas of mathematics and physics, from fluid dynamics to electrical engineering. This guide covers the key concepts you need for the AQA syllabus, including arithmetic, conjugates, modulus and argument, polar and exponential forms, De Moivre’s theorem, and finding roots of complex numbers.

在A-Level数学中，复数将实数系扩展，以解决没有实数解的方程，例如x² + 1 = 0。复数在数学和物理的许多领域都至关重要，从流体力学到电气工程。本文涵盖AQA考纲所需的核心概念，包括复数的运算、共轭、模与辐角、极坐标与指数形式、棣莫弗定理以及复数求根。

1. The Imaginary Unit and Why We Need Complex Numbers | 虚数单位与为何需要复数

The imaginary unit is denoted by i and is defined by the property i² = -1. Equivalently, i = √(-1). This definition allows us to write solutions to quadratic equations that have a negative discriminant.

虚数单位用i表示，定义性质为i² = -1，亦即 i = √(-1)。这一定义使我们能够写出判别式为负的二次方程的解。

For example, the equation x² + 1 = 0 has no real roots, because no real number squares to -1. But using complex numbers, we can write x = ±i as solutions. A complex number is an expression of the form z = a + bi, where a and b are real numbers. Here a is called the real part, Re(z), and b is called the imaginary part, Im(z).

例如，方程x² + 1 = 0没有实数根，因为任何实数的平方都不是-1。但使用复数，我们可以写出解x = ±i。复数的一般形式为z = a + bi，其中a和b为实数。a称为实部，记作Re(z)；b称为虚部，记作Im(z)。

A complex number like a + bi is in Cartesian form. Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. This property is often used to solve equations involving complex unknowns.

形如a + bi的复数称为笛卡尔形式。两个复数相等当且仅当它们的实部相等且虚部相等。这一性质常用于求解含有复数未知量的方程。

2. Arithmetic of Complex Numbers | 复数的运算

Addition and subtraction of complex numbers are performed by combining the real parts and the imaginary parts separately. For example, (a + bi) + (c + di) = (a + c) + (b + d)i, and similarly (a + bi) – (c + di) = (a – c) + (b – d)i.

复数的加法和减法分别合并实部和虚部。例如，(a + bi) + (c + di) = (a + c) + (b + d)i；类似地，(a + bi) – (c + di) = (a – c) + (b – d)i。

Multiplication follows the distributive property, remembering that i² = -1. Thus (a + bi)(c + di) = ac + adi + bci + bdi² = (ac – bd) + (ad + bc)i.

乘法遵循分配律，并记住i² = -1。因此(a + bi)(c + di) = ac + adi + bci + bdi² = (ac – bd) + (ad + bc)i。

Division is a bit more subtle. To divide by a complex number, we multiply the numerator and denominator by the complex conjugate of the denominator, so that the denominator becomes a real number. This will be discussed in more detail in Section 3.

除法略为复杂。为了除以一个复数，我们将分子和分母同时乘以分母的共轭复数，使分母变为实数。这将在第3节中详细讨论。

Matrix representation of complex numbers is sometimes used, but the additive and multiplicative structures remain the same. For AQA, arithmetic fluency is expected, especially simplifying expressions that involve powers of i.

有时会使用复数的矩阵表示，但加法和乘法的结构保持不变。对于AQA，要求熟练掌握运算，尤其是化简含有i的幂的表达式。

The powers of i follow a cycle: i¹ = i, i² = -1, i³ = -i, i⁴ = 1, and then the pattern repeats every four powers. This is useful for simplifying large exponents.

i的幂遵循周期：i¹ = i, i² = -1, i³ = -i, i⁴ = 1，之后每四个幂重复一次。这在化简大指数时很有用。

3. Complex Conjugate and Division | 复共轭与除法

The complex conjugate of z = a + bi is denoted by z̄ = a – bi. Geometrically, it is a reflection of z across the real axis. The conjugate is fundamental for division, modulus, and solving equations with real coefficients.

复数z = a + bi的共轭记作z̄ = a – bi。在几何上，它是z关于实轴的反射。共轭在除法、模的计算以及求解实系数方程中至关重要。

Key properties of conjugates include: (z₁ + z₂)̄ = z̄₁ + z̄₂, (z₁z₂)̄ = z̄₁z̄₂, and (z̄) = z. Also, z + z̄ = 2a (real) and z – z̄ = 2bi (pure imaginary).

共轭的关键性质包括：(z₁ + z₂)̄ = z̄₁ + z̄₂，(z₁z₂)̄ = z̄₁z̄₂，以及(z̄)的共轭为z。此外，z + z̄ = 2a（实数），z – z̄ = 2bi（纯虚数）。

To compute a quotient, multiply numerator and denominator by the conjugate of the denominator:

为计算商，将分子和分母同时乘以分母的共轭：

(a + bi) / (c + di) = ((a + bi)(c – di)) / ((c + di)(c – di)) = ((ac + bd) + (bc – ad)i) / (c² + d²)

The denominator becomes t = c² + d², which is always a non-negative real number. This makes division possible and produces a result in Cartesian form.

分母变为c² + d²，这是一个非负实数。这使得除法可以完成并得到笛卡尔形式的结果。

For example, (2 + 3i)/(1 – i) = (2 + 3i)(1 + i)/(1 – i)(1 + i) = (2 + 2i + 3i + 3i²)/(1 + 1) = (-1 + 5i)/2 = -0.5 + 2.5i.

例如，(2 + 3i)/(1 – i) = (2 + 3i)(1 + i)/(1 – i)(1 + i) = (2 + 2i + 3i + 3i²)/(1 + 1) = (-1 + 5i)/2 = -0.5 + 2.5i。

4. Modulus and Argument | 模与辐角

For a complex number z = a + bi, the modulus, written |z|, is defined as the distance from the origin to the point (a, b) in the complex plane. Thus |z| = √(a² + b²). The modulus is always a non-negative real number.

对于复数z = a + bi，模记作|z|，定义为从原点到复平面点(a, b)的距离。因此|z| = √(a² + b²)。模总是非负实数。

The argument of z, written arg(z), is the angle θ between the positive real axis and the line segment joining the origin to the point (a, b). Usually we take the principal argument, which lies in the interval (-π, π] (or sometimes [0, 2π) depending on convention).

z的辐角记作arg(z)，是从正实轴到连接原点与点(a, b)的线段之间的夹角。通常我们取主辐角，位于区间(-π, π]（或有些约定使用[0, 2π)）。

From trigonometry, we have a = |z| cos θ and b = |z| sin θ, so tan θ = b/a, with care to choose the correct quadrant. For example, if z = 1 + i, then |z| = √2 and arg(z) = π/4.

由三角学可知，a = |z| cos θ，b = |z| sin θ，因此tan θ = b/a，注意要选择正确的象限。例如，若z = 1 + i，则|z| = √2，arg(z) = π/4。

Modulus and argument are essential for the polar form of a complex number, and they also help in interpreting multiplication and division geometrically. Specifically, |z₁z₂| = |z₁||z₂| and arg(z₁z₂) = arg(z₁) + arg(z₂) (mod 2π).

模和辐角对于复数的极坐标形式至关重要，也帮助我们几何地理解乘法和除法。具体来说，|z₁z₂| = |z₁||z₂|，且arg(z₁z₂) = arg(z₁) + arg(z₂)（模2π）。

5. Argand Diagrams | 阿尔甘图

An Argand diagram is a graphical representation of complex numbers. The horizontal axis is the real axis and the vertical axis is the imaginary axis. Each complex number corresponds to a point (a, b) in the plane.

阿尔甘图是复数的图形表示。横轴为实轴，纵轴为虚轴。每个复数对应平面上的一个点(a, b)。

Addition of complex numbers corresponds to vector addition in the Argand diagram. Multiplication by a complex number with modulus r and argument θ is equivalent to rotating the point by θ and scaling its distance from the origin by r.

复数的加法对应阿尔甘图中的向量加法。乘以一个模为r、辐角为θ的复数，等价于将该点旋转θ并以r缩放其到原点的距离。

Shading regions on an Argand diagram can represent sets of complex numbers, such as |z – z₀| = r, which is a circle of radius r centred at z₀. Inequalities like |z – z₀| < r represent the interior of the circle.

在阿尔甘图上涂色可以表示复数集合，例如|z – z₀| = r表示以z₀为圆心、半径为r的圆。不等式如|z – z₀| < r表示圆的内部。

Another common locus is arg(z – z₀) = θ, which is a half-line starting from z₀ making an angle θ with the positive real axis. Being able to sketch these loci is a standard exam skill.

另一个常见轨迹是arg(z – z₀) = θ，这是从z₀出发、与正实轴成θ角的射线。能够画出这些轨迹是标准考试技能。

6. Polar Form and Exponential Form | 极坐标形式与指数形式

The polar form of a complex number z is written as z = r(cos θ + i sin θ), where r = |z| and θ = arg(z). This form is especially useful for multiplication, division, and powers.

复数的极坐标形式写作z = r(cos θ + i sin θ)，其中r = |z|，θ = arg(z)。这种形式在乘法、除法和幂运算中特别有用。

Using Euler’s formula, e^(iθ) = cos θ + i sin θ, we obtain the exponential form: z = r e^(iθ). This compact notation simplifies many analytic calculations and is widely used in advanced mathematics.

利用欧拉公式e^(iθ) = cos θ + i sin θ，我们得到指数形式：z = r e^(iθ)。这种简洁的记法简化了许多解析计算，在高等数学中广泛使用。

Converting from Cartesian to polar form involves finding r and θ, while converting back uses a = r cos θ and b = r sin θ. The principal argument must satisfy -π < θ ≤ π unless otherwise stated.

从笛卡尔形式转换到极坐标形式需要求出r和θ，而转换回来则使用a = r cos θ和b = r sin θ。除非另有说明，主辐角应满足-π < θ ≤ π。

For multiplication in polar form: 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)]. Division gives z₁/z₂ = (r₁/r₂)[cos(θ₁ – θ₂) + i sin(θ₁ – θ₂)].

极坐标形式下的乘法：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)]。除法给出z₁/z₂ = (r₁/r₂)[cos(θ₁ – θ₂) + i sin(θ₁ – θ₂)]。

7. De Moivre’s Theorem | 棣莫弗定理

De Moivre’s theorem states that for any complex number z = r(cos θ + i sin θ) and any integer n, we have zⁿ = rⁿ[cos(nθ) + i sin(nθ)]. This is an extremely powerful tool for computing powers and roots.

棣莫弗定理指出，对于任意复数z = r(cos θ + i sin θ)和任意整数n，有zⁿ = rⁿ[cos(nθ) + i sin(nθ)]。这是计算幂和根的有力工具。

The theorem can be derived from Euler’s formula, since (r e^(iθ))ⁿ = rⁿ e^(inθ). It holds for positive integers, negative integers, and even rational values, though we must be careful with principal arguments in the latter cases.

该定理可由欧拉公式推导，因为(r e^(iθ))ⁿ = rⁿ e^(inθ)。它对正整数、负整数乃至有理数成立，但在有理数情形需要小心处理主辐角。

A classic use of De Moivre’s theorem is to derive trigonometric identities. For example, cos(3θ) can be written in terms of cos θ by expanding (cos θ + i sin θ)³ and equating real parts.

棣莫弗定理的一个经典用途是推导三角恒等式。例如，通过展开(cos θ + i sin θ)³并取实部，可以将cos(3θ)表示为cos θ的代数式。

Another important application is calculating powers of complex numbers. For instance, to compute (1 + i)⁶, first write 1 + i = √2(cos π/4 + i sin π/4). Then (1 + i)⁶ = (√2)⁶(cos 6π/4 + i sin 6π/4) = 8(cos 3π/2 + i sin 3π/2) = -8i.

另一个重要应用是计算复数的幂。例如，计算(1 + i)⁶，先将1 + i写作√2(cos π/4 + i sin π/4)。然后(1 + i)⁶ = (√2)⁶(cos 6π/4 + i sin 6π/4) = 8(cos 3π/2 + i sin 3π/2) = -8i。

8. Finding Roots of Complex Numbers | 复数的求根

The nth roots of a complex number z = r(cos θ + i sin θ) are given by the formula:

复数z = r(cos θ + i sin θ)的n次方根由以下公式给出：

z^(1/n) = r^(1/n) [cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)], k = 0, 1, 2, …, n-1

Here r^(1/n) is the positive real nth root of r. The term 2kπ accounts for the periodicity of the trigonometric functions, and gives n distinct roots equally spaced around the Argand diagram.

其中r^(1/n)是r的正实n次方根。2kπ项考虑了三角函数的周期性，并给出n个在阿尔甘图上等距分布的不同的根。

For example, the square roots of i are found by writing i = cos(π/2) + i sin(π/2). Then for k = 0 and k = 1, we obtain roots at angles π/4 and 5π/4, both with modulus 1. Hence the roots are (1 + i)/√2 and -(1 + i)/√2.

例如，求i的平方根时，先写i = cos(π/2) + i sin(π/2)。然后取k = 0和k = 1，得到辐角为π/4和5π/4、模均为1的根。因此根为(1 + i)/√2和-(1 + i)/√2。

On the Argand diagram, the n roots of a number all lie on a circle of radius |z|^(1/n) centred at the origin, and the angles between successive roots are 2π/n. This symmetry is a favourite exam question.

在阿尔甘图上，一个数的n个根都位于以原点为圆心、半径为|z|^(1/n)的圆上，并且相邻根之间的角度差为2π/n。这种对称性是考试的热门问题。

To solve equations like z⁴ = -16, express -16 in polar form as 16(cos π + i sin π), then apply the formula with n = 4. The four roots will be 2(cos(π/4 + kπ/2) + i sin(π/4 + kπ/2)) for k = 0, 1, 2, 3.

要解如z⁴ = -16的方程，先将-16表示为16(cos π + i sin π)，然后应用n = 4的公式。四个根为2(cos(π/4 + kπ/2) + i sin(π/4 + kπ/2))，其中k = 0, 1, 2, 3。

9. Solving Polynomial Equations | 解多项式方程

Complex numbers allow us to find all roots of polynomial equations, not just the real ones. A polynomial of degree n has exactly n roots in the complex numbers, counting multiplicity. This is the Fundamental Theorem of Algebra.

复数使我们能够找到多项式方程的所有根，而不仅仅是实数根。在复数范围内，n次多项式恰好有n个根（重根按重数计算）。这就是代数基本定理。

If a polynomial has real coefficients, then any non-real roots must occur in conjugate pairs. For example, if 2 + i is a root, then 2 – i is also a root. This is because conjugating the polynomial equation preserves equality.

如果多项式具有实系数，那么任何非实根必定成对共轭出现。例如，若2 + i是根，则2 – i也是根。这是因为对多项式方程取共轭不会改变等式的成立。

To find roots of a cubic like z³ + z² + 2z + 4 = 0, we can first test real factors. If none exist, we may need to find one complex root and then use the fact that the sum of the roots equals -b/a (coefficient relation) to determine the others.

求解三次方程如z³ + z² + 2z + 4 = 0的根时，可以先测试实数因子。如果没有，我们可能需要找到一个复数根，然后利用根与系数关系（根之和等于-b/a）来确定其余的根。

Given one root α, the product of roots and sum of roots give linear equations for the unknown roots. Alternatively, divide the polynomial by the corresponding quadratic factor (z – α)(z – ᾱ) to obtain a linear factor.

给定一个根α，利用根的乘积之和可以得到关于未知根的线性方程。或者，用二次因子(z – α)(z – ᾱ)去除多项式，得到线性因子。

Be careful when a polynomial is of degree four or higher: you may need to combine the conjugate root theorem with synthetic division or factorisation to reduce the problem step by step.

当多项式次数为四或更高时，要谨慎：你可能需要将共轭根定理与综合除法或因式分解结合，逐步降低问题难度。

10. Applications and Exam Tips | 应用与考试提示

Complex numbers appear not only in pure mathematics but also in alternating current circuits, where impedance is expressed as a complex quantity, and in quantum mechanics, where wave functions use complex exponentials. Understanding the algebra and geometry of complex numbers is essential for these applications.

复数不仅出现在纯数学中，还出现在交流电路中（阻抗以复数表示）以及量子力学中（波函数使用复指数）。理解复数的代数与几何对上述应用至关重要。

For AQA A-Level exams, you should be able to:

对于AQA A-Level考试，你应该能够：

Perform arithmetic with complex numbers, including division using conjugates.
Represent complex numbers on an Argand diagram and sketch loci such as circles and half-lines.
Convert between Cartesian, polar and exponential forms.
Apply De Moivre’s theorem to compute powers and roots.
Solve polynomial equations with real coefficients, using conjugate root pairs.

进行复数的运算，包括使用共轭进行除法。
在阿尔甘图上表示复数，并绘制诸如圆和半线的轨迹。
在笛卡尔形式、极坐标形式和指数形式之间转换。
应用棣莫弗定理计算幂和根。
利用共轭根对求解实系数多项式方程。

A common mistake is choosing the wrong quadrant for the argument when converting from Cartesian to polar form. Always sketch the point first or use the signs of a and b to determine the correct range.

一个常见错误是在从笛卡尔形式转换成极坐标形式时选错辐角的象限。应先在草稿上画出点，或利用a和b的符号确定正确范围。

Another common error is forgetting to add 2kπ when finding roots, which leads to missing solutions. Remember that n distinct roots require k = 0, 1, …, n-1.

另一个常见错误是求根时忘记加2kπ，导致遗漏解。记住要得到n个不同根，需要k = 0, 1, …, n-1。

When using De Moivre’s theorem for negative powers, treat it as the reciprocal of the corresponding positive power. For example, z⁻² = (1/r²)[cos(2θ) – i sin(2θ)] by taking the conjugate of the positive power.

当使用棣莫弗定理计算负幂时，将其视为相应正幂的倒数。例如，z⁻² = (1/r²)[cos(2θ) – i sin(2θ)]，通过对正幂取共轭得到。

Finally, always check your final answer by substituting back into the original equation, or by verifying that the modulus and argument are consistent with the operations performed. A good sketch can often reveal algebraic errors.

最后，始终通过代入原方程或检验模与辐角是否与所执行运算一致来检查答案。好的草图往往能揭示代数错误。

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Complex Numbers | 复数