The Conjugate of a Complex Number and the Division of Complex Numbers

📚 The Conjugate of a Complex Number and the Division of Complex Numbers | 复数的共轭与除法

A complex number is written in the form z = a + bi, where a and b are real numbers and i satisfies i² = −1. The conjugate of z, denoted z̄ (or z*), is defined as z̄ = a − bi. The conjugate is a powerful tool because it lets us simplify expressions involving division by a complex number, turning the denominator into a real number.

复数写作 z = a + bi，其中 a、b 为实数，并且 i 满足 i² = −1。复数 z 的共轭记作 z̄（或 z*），定义为 z̄ = a − bi。共轭是一个强大工具，因为它能让我们化简涉及复数除法的表达式，将分母变为实数。

1. Definition and Basic Properties | 定义与基本性质

The conjugate of z = a + bi is z̄ = a − bi. Geometrically, the conjugate represents a reflection of z across the real axis in the Argand diagram.

z = a + bi 的共轭为 z̄ = a − bi。从几何上看，共轭表示复平面上点 z 关于实轴的对称。

Several properties follow directly from the definition. Let z and w be complex numbers:

以下性质可直接从定义推出。设 z、w 为复数：

(z̄)̄ = z.

(z̄)̄ = z。
z + z̄ = 2Re(z) and z − z̄ = 2i Im(z).

z + z̄ = 2Re(z)，且 z − z̄ = 2i Im(z)。
(z ± w)̄ = z̄ ± w̄.

(z ± w)̄ = z̄ ± w̄。
(zw)̄ = z̄ w̄.

(zw)̄ = z̄ w̄。
If w ≠ 0, then (z/w)̄ = z̄ / w̄.

若 w ≠ 0，则 (z/w)̄ = z̄ / w̄。
z is real if and only if z = z̄.

z 为实数当且仅当 z = z̄。

2. The Product of a Complex Number and Its Conjugate | 复数与其共轭的乘积

For any complex number z = a + bi, the product z z̄ is a real number:

对任意复数 z = a + bi，乘积 z z̄ 是一个实数：

z z̄ = (a + bi)(a − bi) = a² − (bi)² = a² + b²

Thus z z̄ = |z|², where |z| = √(a² + b²) is the modulus of z. This property is the key to dividing complex numbers.

因此 z z̄ = |z|²，其中 |z| = √(a² + b²) 是 z 的模。此性质是复数除法的关键。

3. Division of Complex Numbers: The Idea | 复数除法：基本思想

Suppose we want to compute z/w, where z = a + bi and w = c + di, with w ≠ 0. The denominator is complex, so we cannot simplify directly. Instead, we multiply the numerator and denominator by the conjugate of w:

假设需要计算 z/w，其中 z = a + bi，w = c + di，且 w ≠ 0。由于分母为复数，无法直接化简。我们改为将分子与分母同时乘以 w 的共轭：

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

The denominator becomes c² + d², a real number, so the fraction can be split into real and imaginary parts.

分母变为 c² + d²（实数），因此分数可分解为实部和虚部。

4. The Division Formula | 除法公式

Expanding the numerator (a + bi)(c − di) gives:

展开分子 (a + bi)(c − di) 得到：

(a + bi)(c − di) = ac − adi + bci − bdi² = (ac + bd) + (bc − ad)i

Therefore,

因此，

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

So the real part is (ac + bd)/(c² + d²) and the imaginary part is (bc − ad)/(c² + d²). This formula can be used directly, but it is often safer to perform the multiplication step by step.

因此实部为 (ac + bd)/(c² + d²)，虚部为 (bc − ad)/(c² + d²)。可以直接使用该公式，但通常逐步相乘更加稳妥。

5. Worked Example 1 | 例题详解 1

Compute (3 + 2i)/(1 − i).

计算 (3 + 2i)/(1 − i)。

Multiply numerator and denominator by the conjugate of the denominator, 1 + i:

分子分母同乘分母的共轭 1 + i：

(3 + 2i)(1 + i) / [(1 − i)(1 + i)]

Numerator: (3 + 2i)(1 + i) = 3 + 3i + 2i + 2i² = 3 + 5i − 2 = 1 + 5i.
Denominator: (1 − i)(1 + i) = 1² + 1² = 2.

分子：(3 + 2i)(1 + i) = 3 + 3i + 2i + 2i² = 3 + 5i − 2 = 1 + 5i。
分母：(1 − i)(1 + i) = 1² + 1² = 2。

Hence (3 + 2i)/(1 − i) = (1 + 5i)/2 = 0.5 + 2.5i.

因此 (3 + 2i)/(1 − i) = (1 + 5i)/2 = 0.5 + 2.5i。

6. Worked Example 2 | 例题详解 2

Compute (4 − i)/(2 + 3i).

计算 (4 − i)/(2 + 3i)。

Use the conjugate of the denominator, 2 − 3i:

使用分母的共轭 2 − 3i：

(4 − i)(2 − 3i) / [(2 + 3i)(2 − 3i)]

Numerator: (4 − i)(2 − 3i) = 8 − 12i − 2i + 3i² = 8 − 14i − 3 = 5 − 14i.
Denominator: (2 + 3i)(2 − 3i) = 4 + 9 = 13.

分子：(4 − i)(2 − 3i) = 8 − 12i − 2i + 3i² = 8 − 14i − 3 = 5 − 14i。
分母：(2 + 3i)(2 − 3i) = 4 + 9 = 13。

Therefore (4 − i)/(2 + 3i) = 5/13 − (14/13)i.

因此 (4 − i)/(2 + 3i) = 5/13 − (14/13)i。

7. Finding Real and Imaginary Parts | 求实部与虚部

When z/w is written in the form A + Bi, we must identify A and B correctly. Using the general formula:

当 z/w 写成 A + Bi 形式时，必须正确识别 A 和 B。使用一般公式：

Re(z/w) = (ac + bd)/(c² + d²), Im(z/w) = (bc − ad)/(c² + d²)

For example, from Example 2, z = 4 − i and w = 2 + 3i, so a = 4, b = −1, c = 2, d = 3. Then Re = (4×2 + (−1)×3)/(4 + 9) = 5/13, Im = ((−1)×2 − 4×3)/13 = −14/13.

例如，在例题 2 中，z = 4 − i，w = 2 + 3i，所以 a = 4，b = −1，c = 2，d = 3。于是 Re = (4×2 + (−1)×3)/(4 + 9) = 5/13，Im = ((−1)×2 − 4×3)/13 = −14/13。

8. Conjugate Pairs in Polynomial Equations | 多项式方程中的共轭根对

If a polynomial has real coefficients, then any non-real roots must occur in conjugate pairs. For instance, the quadratic equation x² − 2x + 5 = 0 has roots 1 + 2i and 1 − 2i.

若多项式系数全为实数，则非实根必然成共轭对出现。例如，二次方程 x² − 2x + 5 = 0 的根为 1 + 2i 和 1 − 2i。

This fact is useful when solving equations: once one complex root is known, its conjugate is also a root, and we can use this to factorise the polynomial.

这个事实在解方程时十分有用：已知一个复数根，其共轭也是根，于是可以据此对多项式进行因式分解。

9. Common Mistakes and Tips | 常见错误与技巧

Forgetting that i² = −1 when expanding products. Always replace i² with −1.

在展开乘积时忘记 i² = −1。务必用 −1 替换 i²。
Sign errors in the numerator: (a + bi)(c − di) = ac + bd + (bc − ad)i, not ac − bd.

分子符号错误：(a + bi)(c − di) = ac + bd + (bc − ad)i，而非 ac − bd。
Using the conjugate of the numerator instead of the denominator. The denominator must be multiplied by its own conjugate.

误用了分子的共轭而非分母的共轭。分母必须乘以其自身的共轭。
Simplifying z/(wi) incorrectly: remember that 1/i = −i, so division by i often leads to a sign change.

化简 z/(wi) 时出错：记住 1/i = −i，因此除以 i 通常会导致符号变化。

10. Practice Problems | 练习

Try the following exercises before checking the answers.

先尝试以下练习，再核对答案。

Compute (1 + 2i)/(3 + 4i).

计算 (1 + 2i)/(3 + 4i)。
Compute (2 − 3i)/(2 + i).

计算 (2 − 3i)/(2 + i)。
Find z if z/(1 − i) = 2 + i.

求 z，已知 z/(1 − i) = 2 + i。
Express (1 + i)/(1 − i) in the form A + Bi.

将 (1 + i)/(1 − i) 写成 A + Bi 的形式。

Answers: 1) 11/25 + (2/25)i 2) 1/5 − (8/5)i 3) z = 3 − i 4) i.

答案：1) 11/25 + (2/25)i 2) 1/5 − (8/5)i 3) z = 3 − i 4) i。

11. Summary | 总结

The conjugate of z = a + bi is z̄ = a − bi, and the key identity is z z̄ = a² + b² = |z|². To divide complex numbers, multiply both numerator and denominator by the conjugate of the denominator. This converts the denominator into a real number, allowing us to write the result as A + Bi. Conjugate pairs also play a central role in solving polynomial equations with real coefficients. Master this technique, and complex division becomes a straightforward algebraic operation.

z = a + bi 的共轭为 z̄ = a − bi，关键恒等式是 z z̄ = a² + b² = |z|²。复数除法时，将分子分母同时乘以分母的共轭，使分母化为实数，从而将结果写成 A + Bi。共轭对在解实系数多项式方程中也非常重要。掌握这一技巧后，复数除法就变成一种直接的代数运算。

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The Conjugate of a Complex Number and the Division of Complex Numbers | 复数的共轭与除法