The Conjugate of a Complex Number and Division of Complex Numbers of the Form x + iy | 复数的共轭与 x + iy 形式复数的除法

📚 The Conjugate of a Complex Number and Division of Complex Numbers of the Form x + iy | 复数的共轭与 x + iy 形式复数的除法

In AQA A-Level Mathematics, working confidently with complex numbers means more than just adding and subtracting. You must be able to find the conjugate of a complex number and use it to divide one complex number by another. This topic underpins many later ideas, including modulus-argument form, loci and polynomial equations.

在 AQA A-Level 数学中，熟练处理复数不仅仅是加减运算。你必须能够求出复数的共轭，并利用它进行复数之间的除法。这一主题是后续许多内容的基础，包括模-辐角形式、轨迹以及多项式方程。

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1. Review: Complex Numbers in the Form x + iy | 复习：x + iy 形式的复数

A complex number is written in Cartesian form as z = x + iy, where x and y are real numbers and i is the imaginary unit with i² = -1. The real part is Re(z) = x and the imaginary part is Im(z) = y.

复数以笛卡尔形式写作 z = x + iy，其中 x 和 y 是实数，i 是虚数单位且 i² = -1。实部为 Re(z) = x，虚部为 Im(z) = y。

Two complex numbers are equal only when their real parts are equal and their imaginary parts are equal. This idea is used frequently in AQA exam questions.

两个复数相等，当且仅当它们的实部相等且虚部相等。这一思想在 AQA 考试题中经常用到。

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2. Defining the Complex Conjugate | 定义复共轭

For z = x + iy, the complex conjugate is z* = x – iy. The conjugate is formed by changing the sign of the imaginary part while keeping the real part unchanged.

对于 z = x + iy，其复共轭为 z* = x – iy。共轭是通过改变虚部的符号而保持实部不变得到的。

On an Argand diagram, z* is the reflection of z in the real axis. If z = 3 + 4i, then z* = 3 – 4i.

在阿尔冈图上，z* 是 z 关于实轴的反射。如果 z = 3 + 4i，那么 z* = 3 – 4i。

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3. Key Properties of Conjugates | 共轭的关键性质

Conjugation distributes over addition, subtraction, multiplication and division. For complex numbers z and w, the following identities hold:

共轭运算对加法、减法、乘法和除法都具有分配性。对于复数 z 和 w，以下恒等式成立：

(z + w)* = z* + w*, (z – w)* = z* – w*, (zw)* = z* w*, (z / w)* = z* / w* (w ≠ 0)

Also, (z*)* = z, so applying conjugation twice returns the original complex number.

此外，(z*)* = z，因此连续两次共轭运算将返回原来的复数。

The real and imaginary parts can be recovered using conjugates: Re(z) = (z + z*)/2 and Im(z) = (z – z*)/(2i). This is useful in algebraic manipulations.

实部和虚部可以利用共轭恢复：Re(z) = (z + z*)/2，Im(z) = (z – z*)/(2i)。这在代数运算中很有用。

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4. Multiplying a Number by Its Conjugate | 复数与其共轭相乘

For z = x + iy, the product z z* is always a non-negative real number. Multiplying gives (x + iy)(x – iy) = x² – i² y² = x² + y².

对于 z = x + iy，乘积 z z* 总是一个非负实数。相乘得到 (x + iy)(x – iy) = x² – i² y² = x² + y²。

z z* = x² + y² = |z|²

This real number is the square of the modulus of z. In particular, if z ≠ 0, then z z* > 0.

这个实数就是 z 的模的平方。特别地，如果 z ≠ 0，那么 z z* > 0。

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5. Why the Conjugate Enables Division | 为什么共轭可以实现除法

To divide by a complex number w = c + id, we need to make the denominator real. Multiplying the numerator and denominator by w* achieves this because w w* is real.

要除以一个复数 w = c + id，我们需要使分母变为实数。将分子和分母同时乘以 w* 就能实现，因为 w w* 是实数。

This technique is analogous to rationalising a surd denominator, such as multiplying 1/(√2 – 1) by (√2 + 1)/(√2 + 1).

这种方法类似于将带根式的分母有理化，例如将 1/(√2 – 1) 乘以 (√2 + 1)/(√2 + 1)。

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6. The Division Formula for x + iy | x + iy 形式的除法公式

For z = a + ib and w = c + id with w ≠ 0, write the division as z / w = (a + ib)/(c + id). Multiply numerator and denominator by c – id.

对于 z = a + ib 和 w = c + id 且 w ≠ 0，将除法写作 z / w = (a + ib)/(c + id)。将分子和分母同时乘以 c – id。

(a + ib)/(c + id) = [(a + ib)(c – id)] / (c² + d²)

Expanding the numerator gives the standard result:

展开分子得到标准结果：

(a + ib)/(c + id) = (ac + bd)/(c² + d²) + i(bc – ad)/(c² + d²)

This expression has a real denominator and separates the real and imaginary parts clearly.

该表达式的分母为实数，并清楚地分开了实部和虚部。

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7. Worked Example: Dividing Two Complex Numbers | 例题：两个复数相除

Divide z = 3 + 2i by w = 1 – 4i. Multiply numerator and denominator by the conjugate of w, which is 1 + 4i.

用 w = 1 – 4i 除 z = 3 + 2i。将分子和分母同时乘以 w 的共轭，即 1 + 4i。

(3 + 2i)/(1 – 4i) = [(3 + 2i)(1 + 4i)] / [(1 – 4i)(1 + 4i)]

Expand the numerator: (3 + 2i)(1 + 4i) = 3 + 12i + 2i + 8i² = 3 – 8 + 14i = -5 + 14i. The denominator is 1² + 4² = 17.

展开分子：(3 + 2i)(1 + 4i) = 3 + 12i + 2i + 8i² = 3 – 8 + 14i = -5 + 14i。分母为 1² + 4² = 17。

(3 + 2i)/(1 – 4i) = -5/17 + (14/17)i

Always check that the final answer is in the form x + iy with x and y expressed as fractions if necessary.

始终检查最终答案是否为 x + iy 形式，必要时 x 和 y 用分数表示。

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8. Worked

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