Working with Complex Numbers in AQA A-Level Mathematics | AQA A-Level 数学中的复数运算

📚 Working with Complex Numbers in AQA A-Level Mathematics | AQA A-Level 数学中的复数运算

-1. This article covers the core AQA techniques for adding, subtracting, multiplying and dividing complex numbers, solving polynomial equations with complex roots, and representing complex numbers on an Argand diagram.

复数通过引入虚数单位 i（满足 i² = -1）扩展了实数系统。本文涵盖 AQA 核心技巧：复数的加减乘除、求解具有复数根的多项式方程，以及在阿干特图上表示复数。

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1. The Imaginary Unit and Standard Form | 虚数单位与标准形式

a and b are real numbers. The real part is Re(z) = a and the imaginary part is Im(z) = b.

虚数单位 i 由性质 i² = -1 定义。复数写作标准形式 z = a + bi，其中 a 与 b 为实数。实部为 Re(z) = a，虚部为 Im(z) = b。

is 3 and the imaginary part is -4. The imaginary part is the real coefficient of i, not including i itself.

例如，在 z = 3 - 4i 中，实部为 3，虚部为 -4。虚部是 i 的实数系数，不包括 i 本身。

i² = -1, z = a + bi, a, b ∈ ℝ

Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. This is useful when comparing coefficients in equations.

两个复数相等，当且仅当它们的实部相等且虚部相等。这在方程中比较系数时非常有用。

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2. Adding and Subtracting Complex Numbers | 复数的加法与减法

ombine the real parts and the imaginary parts separately. For z₁ = a + bi and z₂ = c + di:

复数相加或相减时，分别合并实部与虚部。对于 z₁ = a + bi 和 z₂ = c + di：

z₁ + z₂ = (a + c) + (b + d)i

z₁ − z₂ = (a − c) + (b − d)i

This works because addition is commutative and associative over the real and imaginary components. You cannot simplify a real term and an imaginary term into one single term, so 2 + 3i stays as two separate parts.

这是因为实部与虚部分别满足加法交换律与结合律。你不能将一个实数项与一个虚数项合并为一项，因此 2 + 3i 仍然保留为两个独立部分。

Example: (5 + 2i) + (3 − 7i) = (5 + 3) + (2 − 7)i = 8 − 5i.

示例：(5 + 2i) + (3 − 7i) = (5 + 3) + (2 − 7)i = 8 − 5i。

egative sign appears in front of brackets: (2 + 3i) − (1 − 4i) = 2 + 3i − 1 + 4i = 1 + 7i.

减法在括号前出现负号时必须小心处理：(2 + 3i) − (1 − 4i) = 2 + 3i − 1 + 4i = 1 + 7i。

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3. Multiplying Complex Numbers and Powers of i | 复数的乘法与 i 的幂

ry algebra and then replace every occurrence of i² by −1. For z₁ = a + bi and z₂ = c + di:

复数相乘时，如同普通代数一样展开括号，然后将每一个 i² 替换为 −1。对于 z₁ = a + bi 和 z₂ = c + di：

z₁z₂ = (ac − bd) + (ad + bc)i

Example: (2 + 3i)(1 − 4i) = 2 − 8i + 3i − 12i² = 2 − 5i + 12 = 14 − 5i.

示例：(2 + 3i)(1 − 4i) = 2 − 8i + 3i − 12i² = 2 − 5i + 12 = 14 − 5i。

, i³ = −i, i⁴ = 1, i⁵ = i, and so on. This is tested in many AQA simplification questions.

i 的幂每四次循环一次。你可以利用以下规律快速计算：i² = −1，i³ = −i，i⁴ = 1，i⁵ = i，依此类推。许多 AQA 化简题都会考查这一点。

Power | 幂	Value | 值
i¹	i
i²	−1
i³	−i
i⁴	1

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4. The Complex Conjugate | 共轭复数

a − bi. Geometrically, it reflects the point (a, b) in the real axis on an Argand diagram.

复数 z = a + bi 的共轭写作 z* = a − bi。在几何上，它在阿干特图中将点 (a, b) 关于实轴进行镜像反射。

A very important property is that z z* is always real and non-negative. For z = a + bi:

一个非常重要的性质是 z z* 总是实数且非负。对于 z = a + bi：

z z* = (a + bi)(a − bi) = a² + b²

Example: if z = 3 + 4i, then z* = 3 − 4i and z z* = 3² + 4² = 25.

示例：如果 z = 3 + 4i，则 z* = 3 − 4i，且 z z* = 3² + 4² = 25。

lows the same operation: (z₁ + z₂)* = z₁* + z₂*, (z₁z₂)* = z₁*z₂*, and (z₁/z₂)* = z₁*/z₂*.

共轭运算对加法、乘法和除法都保持相同结构：(z₁ + z₂)* = z₁* + z₂*，(z₁z₂)* = z₁*z₂*，(z₁/z₂)* = z₁*/z₂*。

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5. Division of Complex Numbers | 复数的除法

To divide by a complex number, multiply the numerator and denominator by the conjugate of the denominator. This turns the denominator into a real number.

除以复数时，将分子与分母同时乘以分母的共轭。这样可将分母变为实数。

(a + bi)/(c + di) = [(a + bi)(c − di)] / [(c + di)(c − di)]

= [(ac + bd) + (bc − ad)i] / (c² + d²)

Example: divide 3 + 2i by 1 − i.

示例：计算 (3 + 2i) ÷ (1 − i)。

(3 + 2i)/(1 − i) = (3 + 2i)(1 + i) / (1 − i)(1 + i)

= (3 + 3i + 2i + 2i²) / (1 + 1) = (1 + 5i)/2 = 1/2 + (5/2)i

Always simplify the final answer into the standard form a + bi. If the numerator is purely real or purely imaginary, the same method still works.

最后答案务必化简为标准形式 a + bi。如果分子为纯实数或纯虚数，同样的方法仍然适用。

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6. Solving Quadratic Equations with Complex Roots | 求解具有复数根的二次方程

= b² − 4ac decides the nature of the roots. If Δ < 0, the roots are complex conjugates.

对于实系数二次方程 ax² + bx + c = 0，判别式 Δ = b² − 4ac 决定根的类型。如果 Δ < 0，则根为一对共轭复数。

x = [−b ± √b² − 4ac] / (2a)

When the discriminant is negative, write √b² − 4ac = i√4ac − b². Do not leave a negative number under the square root in a final answer.

当判别式为负时，将 √b² − 4ac 写为 i√4ac − b²。最终答案中不要把负数留在平方根号下。

Example: solve x² − 4x + 13 = 0.

示例：求解 x² − 4x + 13 = 0。

Δ = (−4)² − 4(1)(13) = 16 − 52 = −36

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