Complex Numbers in Cartesian Form | 复数的直角坐标形式

📚 Complex Numbers in Cartesian Form | 复数的直角坐标形式

Complex numbers extend the real number system by including an imaginary unit, enabling solutions to equations like x² = -1. In Cartesian (or algebraic) form, each complex number is written as a+bi, where a and b are real numbers. This article covers key definitions, operations, and applications required in IB Mathematics, focusing entirely on the Cartesian representation.

复数通过引入虚数单位扩展了实数系统，使 x² = -1 这类方程有了明确的解。在直角坐标形式（也称代数形式）下，每个复数都可以写成 a+bi，其中 a 和 b 为实数。本文涵盖 IB 数学中所要求的定义、运算及应用，全部围绕直角坐标表示展开。

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1. The Imaginary Unit and Definition | 虚数单位与定义

The imaginary unit, denoted by i, is defined by the property that its square equals -1:

虚数单位记作 i，其定义由平方等于 -1 给出：

i² = −1

From this definition, we can form complex numbers. A complex number in Cartesian form is any number of the type a+bi, where a and b are real numbers (a, b ∈ ℝ). The value a is called the real part and b is called the imaginary part.

由此定义即可构造复数。直角坐标形式下的复数是指任何形如 a+bi 的数，其中 a 和 b 为实数（a, b ∈ ℝ）。a 称为实部，b 称为虚部。

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2. Real and Imaginary Parts | 实部与虚部

For a complex number z = a+bi, we write Re(z) = a and Im(z) = b. Note that the imaginary part is b, not bi. When b = 0, the number is purely real; when a = 0 but b ≠ 0, the number is purely imaginary, such as 4i or -√2 i.

对于复数 z = a+bi，记 Re(z) = a，Im(z) = b。注意虚部是 b 而不是 bi。当 b=0 时，该数为纯实数；当 a=0 而 b≠0 时，该数为纯虚数，例如 4i 或 -√2 i。

This distinction is essential when comparing, adding, or multiplying complex numbers, as the real and imaginary parts are treated independently in the Cartesian system.

这一区分在比较、相加或相乘复数时至关重要，因为在直角坐标系中实部与虚部需独立处理。

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3. Equality and the Zero Complex Number | 复数相等与零复数

Two complex numbers z₁ = a+bi and z₂ = c+di are equal if and only if their real parts are equal and their imaginary parts are equal:

两个复数 z₁ = a+bi 与 z₂ = c+di 相等，当且仅当它们的实部相等且虚部相等：

a+bi = c+di ⇔ a = c and b = d

The zero complex number is 0+0i, often written simply as 0. Its real and imaginary parts are both zero. This definition is used to solve equations involving complex unknowns, equating real and imaginary coefficients.

零复数为 0+0i，常简写为 0。其实部与虚部均为零。这一相等条件常用于求解含有复数未知数的方程，即分别令实部与虚部系数相等。

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4. Addition and Subtraction | 加法与减法

Addition and subtraction of complex numbers in Cartesian form are performed component‑wise. Given z₁ = a+bi and z₂ = c+di:

直角坐标形式下复数的加法和减法按分量进行。设 z₁ = a+bi，z₂ = c+di：

z₁ + z₂ = (a+c) + (b+d)i

z₁ − z₂ = (a−c) + (b−d)i

These operations are commutative and associative, just like real addition. Subtracting a complex number means subtracting both its real and imaginary parts.

这些运算满足交换律和结合律，与实数加法一致。减去一个复数即分别减去其实部和虚部。

For example, (3+2i)+(1-5i) = 4-3i, and (6+4i)-(2-7i) = 4+11i.

例如，(3+2i)+(1-5i) = 4-3i，而 (6+4i)-(2-7i) = 4+11i。

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5. Multiplication and Powers of i | 乘法与 i 的幂

To multiply two complex numbers, expand brackets using the distributive law and replace i² with -1. If z₁ = a+bi and z₂ = c+di, then:

两个复数相乘时，利用分配律展开括号并将 i² 替换为 -1。若 z₁ = a+bi，z₂ = c+di，则：

z₁ z₂ = (a+bi)(c+di) = ac + adi + bci + bdi² = (ac−bd) + (ad+bc)i

A crucial part of multiplication is handling powers of i efficiently. The cycle of i is:

乘法的关键在于高效处理 i 的幂。i 的幂具有周期性：

i¹ = i, i² = −1, i³ = i²·i = −i, i⁴ = (i²)² = 1

After i⁴ = 1, the pattern repeats: i⁵ = i, i⁶ = -1, etc. To simplify a high power like i¹⁷, divide the exponent by 4 and use the remainder: 17 ÷ 4 leaves remainder 1, so i¹⁷ = i¹ = i.

自 i⁴ = 1 后，规律重复：i⁵ = i，i⁶ = -1 等等。简化高次幂如 i¹⁷ 时，将指数除以 4 取余数：17 ÷ 4 余 1，因此 i¹⁷ = i¹ = i。

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6. The Complex Conjugate | 共轭复数

The conjugate of a complex number z = a+bi is obtained by reversing the sign of the imaginary part. It is denoted by z* or sometimes z. In IB Mathematics, z* is commonly used:

复数 z = a+bi 的共轭通过改变虚部符号得到，记作 z*（IB 数学中常使用星号表示），有时也记作 z：

z = a+bi ⇒ z* = a−bi

The conjugate has several important properties that are heavily used in division and in quadratic equations:

共轭具有若干重要性质，在除法和二次方程中频繁运用：

• z + z* = 2a = 2 Re(z) (纯实数)
• z − z* = 2bi (纯虚数)
• z z* = a² + b² (实数，且非负)

对应的中文为：

• z + z* = 2a = 2 Re(z)（纯实数）
• z − z* = 2bi（纯虚数）
• z z* = a² + b²（非负实数）

The product z z* is especially useful because it eliminates the imaginary unit, giving a real, positive number (unless z=0). It helps to rationalise denominators when dividing complex numbers.

乘积 z z* 尤为重要，因为它消除了虚数单位，给出一个实数（且为正，除非 z=0）。这有助于在复数除法中对分母进行实数化。

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7. Division | 除法

To divide two complex numbers z₁ = a+bi by z₂ = c+di, we multiply the numerator and denominator by the conjugate of the denominator, z₂* = c−di. This converts the denominator to the real number c²+d²:

将两个复数 z₁ = a+bi 除以 z₂ = c+di 时，分子分母同乘分母的共轭 z₂* = c−di，从而将分母转化为实数 c²+d²：

z₁ / z₂ = (a+bi)(c−di) / ((c+di)(c−di)) = (ac+bd) + i(bc−ad) / (c²+d²)

The result is then expressed as a real part plus an imaginary part: (ac+bd)/(c²+d²) + i(bc−ad)/(c²+d²).

结果可写成实部加虚部的标准形式：(ac+bd)/(c²+d²) + i(bc−ad)/(c²+d²)。

For example, (3+2i)/(1+i) = (3+2i)(1−i)/(1²+1²) = (3−3i+2i−2i²)/2 = (3−i+2)/2 = (5−i)/2 = 2.5 − 0.5i.

例如，(3+2i)/(1+i) = (3+2i)(1−i)/(1²+1²) = (3−3i+2i−2i²)/2 = (3−i+2)/2 = (5−i)/2 = 2.5 − 0.5i。

Division in Cartesian form is thus straightforward as long as the conjugate is used correctly.

因此，只要正确使用共轭，直角坐标形式下的除法可直接操作。

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8. Solving Quadratic Equations | 解二次方程

The original motivation for introducing complex numbers was to solve quadratic equations with negative discriminants. For ax²+bx+c = 0, the solutions are given by the quadratic formula:

引入复数的最初动机正是为了解判别式为负的二次方程。对于 ax²+bx+c = 0，求根公式为：

x = (−b ± √(b²−4ac)) / (2a)

When the discriminant Δ = b²−4ac is negative, √Δ becomes an imaginary number. We rewrite √(−d) as i√d, where d > 0. The solutions are then a pair of complex conjugates.

当判别式 Δ = b²−4ac 为负时，√Δ 为虚数。我们将 √(−d) 写作 i√d（其中 d>0），解即为共轭复数对。

Example: Solve x²+4x+13 = 0. Here a=1, b=4, c=13. Δ = 16−52 = −36. So x = (−4 ± √(−36))/2 = (−4 ± 6i)/2 = −2 ± 3i. The roots are −2+3i and −2-3i, which are conjugates.

例题：解 x²+4x+13=0。此处 a=1, b=4, c=13。Δ=16−52=−36。故 x = (−4 ± √(−36))/2 = (−4 ± 6i)/2 = −2 ± 3i。两根为 −2+3i 与 −2-3i，互为共轭。

IB problems often ask students to find the real and imaginary parts of the roots, or to verify that they satisfy the original equation.

IB 考题常要求学生写出根的实部和虚部，或验证这些根满足原方程。

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9. Properties of Complex Numbers (Summary) | 复数的性质总结

The Cartesian form satisfies all the familiar algebraic laws: commutative, associative, and distributive properties for addition and multiplication. However, the ordering relation (

直角坐标形式满足所有熟悉的代数法则：加法与乘法的交换律、结合律和分配律。然而，实数中使用的序关系（

The following table summarises the key operations for z₁ = a+bi and z₂ = c+di:

下表总结了 z₁ = a+bi 与 z₂ = c+di 的重要运算：

Operation	Result in Cartesian Form	中文形式
Addition	(a+c) + (b+d)i	(a+c) + (b+d)i
Subtraction	(a−c) + (b−d)i	(a−c) + (b−d)i
Multiplication	(ac−bd) + (ad+bc)i	(ac−bd) + (ad+bc)i
Conjugate	z* = a−bi	z* = a−bi
Division	(ac+bd)/(c²+d²) + i(bc−ad)/(c²+d²)	(ac+bd)/(c²+d²) + i(bc−ad)/(c²+d²)

Mastering these forms ensures accuracy when simplifying expressions or solving equations that involve complex numbers in Cartesian form.

掌握这些形式能够确保在化简含有复数的表达式或解方程时的准确性。

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10. Worked Examples | 典型例题

Example 1: Given z = 2+5i and w = 3−4i, compute (a) z+w, (b) z−w, (c) z·w, (d) z / w.

例题1：已知 z = 2+5i, w = 3−4i，计算 (a) z+w, (b) z−w, (c) z·w, (d) z / w。

(a) z+w = (2+3)+(5−4)i = 5 + 1i = 5+i.
(b) z−w = (2−3)+(5+4)i = −1+9i.
(c) z·w = (2·3 − 5·(−4)) + (2·(−4)+5·3)i = (6+20) + (−8+15)i = 26+7i.
(d) z / w = (2+5i)/(3−4i) = (2+5i)(3+4i)/(3²+4²) = (6+8i+15i+20i²)/25 = (6+23i−20)/25 = (−14+23i)/25 = −0.56 + 0.92i.

(a) z+w = (2+3)+(5−4)i = 5+i；
(b) z−w = (2−3)+(5+4)i = −1+9i；
(c) z·w = (2·3 − 5·(−4)) + (2·(−4)+5·3)i = (6+20) + (−8+15)i = 26+7i；
(d) z / w = (2+5i)/(3−4i) = (2+5i)(3+4i)/(9+16) = (6+8i+15i−20)/25 = (−14+23i)/25 = −0.56 + 0.92i。

Example 2: Solve the equation 2x²−3x+5=0 over the complex numbers.

例题2：在复数范围内解方程 2x²−3x+5=0。

Δ = (−3)²−4·2·5 = 9−40 = −31. Then x = [3 ± √(−31)] / (4) = (3 ± i√31)/4 = 0.75 ± (√31/4)i. The solutions are a conjugate pair.

Δ = 9−40 = −31。于是 x = [3 ± √(−31)] / 4 = (3 ± i√31)/4 = 0.75 ± (√31/4)i。解为一对共轭复数。

Example 3: If (x+yi)(1+2i) = 3+4i, find the real numbers x and y.

例题3：若 (x+yi)(1+2i) = 3+4i，求实数 x 和 y。

Expand LHS: (x+yi)(1+2i) = x+2xi+yi+2yi² = x+2xi+yi−2y = (x−2y) + (2x+y)i.
Equating real and imaginary parts: x−2y = 3 and 2x+y = 4. Solving simultaneously gives x = 11/5, y = −2/5.

展开左边： (x+yi)(1+2i) = x+2xi+yi−2y = (x−2y) + (2x+y)i。
令实部和虚部分别相等：x−2y=3，2x+y=4。联立解得 x = 11/5，y = −2/5。

These examples illustrate the typical IB approach: arithmetic proficiency and equation solving using real-imaginary parts equality.

以上例题体现了 IB 的典型思路：复数运算的熟练度以及利用实部虚部相等来解方程。

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